NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry

 

NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry: In class 11th you have learnt about 2-D geometry. The 3-D geometry is an extension of 2-D geometry with taking consideration of 3 orthogonal axes. It requires more of your imagination power to visualize 3-D geometry concepts. In NCERT solutions for class 12 maths chapter 11 three dimensional geometry article, you will learn how to use vector algebra to study three dimensional geometry. It will be easy for you to solve three dimensional problems with help of vector algebra. The purpose of this approach in the CBSE NCERT solutions for class 12 maths chapter 11 three dimensional geometry is to make the study simple and effective. In this chapter there are many formulas to solve the problems. You are advised to write formulas when you are solving the problems. So, you can remember formulas very easily. Solutions of NCERT for class 12 maths chapter 11 three dimensional geometry will build your base for many other higher-level concepts like tensors and manifolds. Check all NCERT solutions, which will help you get a better understanding of concepts in a much easy way.

The important topics like direction cosines and direction ratios of a line joining two points, equations of lines in space, the angle between two lines, the angle between two planes, angle between a line and a plane, the shortest distance between two skew lines, equation of a plane in the normal form, etc. are covered in this chapter. Questions from all the topics are covered in the CBSE NCERT solutions for class 12 maths chapter 11 three dimensional geometry.  A total of 36 questions in 3 exercises are given in this chapter. All these NCERT questions are solved and explained in the CBSE NCERT solutions for class 12 maths chapter 11 three dimensional geometry article to clear your doubts.

 
 

In this chapter we deal with formulas like-

  • If l, m, n are the direction cosines of a line, then l^2+m^2+n^2=1.
  • Q(x_{2}, y_{2}, z_{2}) and P(x_{1}, y_{1}, z_{1})Direction cosines of a line joining two points PQ=\sqrt{(X_2-X_1)^2+(Y_2-Y_1)^2+(Z_2-Z_1)^2}are \frac{X_2-X_1}{PQ},\:\frac{Y_2-Y_1}{PQ},\:\frac{Z_2-Z_1}{PQ}, where
  • If l, m, n are the direction cosines and a, b, c are the direction of a line then-

             \dpi{80} l=\frac{a}{\sqrt{a^2+b^2+c^2}} ; m=\frac{b}{\sqrt{a^2+b^2+c^2}}; n=\frac{c}{\sqrt{a^2+b^2+c^2}}

Topics and sub-topics of NCERT class 12 maths chapter 11 Three Dimensional Geometry

11.1 Introduction

11.2 Direction Cosines and Direction Ratios of a Line

11.2.1 Relation between the direction cosines of a line

11.2.2 Direction cosines of a line passing through two points

11.3 Equation of a Line in Space

11.3.1Equation of a line through a given point and parallel to a given vector b

11.3.2 Equation of a line passing through two given points

11.4 Angle between Two Lines

11.5 Shortest Distance between Two Lines

11.5.1 Distance between two skew lines

11.5.2 Distance between parallel lines

11.6 Plane

11.6.1 Equation of a plane in normal form

11.6.2 Equation of a plane perpendicular to a given vector and passing through a given point

11.6.3 Equation of a plane passing through three noncollinear points

11.6.4 Intercept form of the equation of a plane

11.6.5 Plane passing through the intersection of two given planes

11.7 Coplanarity of Two Lines

11.8 Angle between Two Planes

11.9 Distance of a Point from a Plane

11.10 Angle between a Line and a Plane

NCERT solutions for class 12 maths chapter 11 Three Dimensional Geometry- Exercise Questions

Solutions of NCERT for class 12 maths chapter 11 three dimensional geometry-Exercise: 11.1 

Question:1 If a line makes angles 90^{\circ}, 135^{\circ},45^{\circ} with the x, y and z-axes respectively, find its direction cosines.

Answer:

Let the direction cosines of the line be l,m, and n.

So, we have

l = \cos90^{\circ}=0       

m = \cos135^{\circ}=-\frac{1}{\sqrt2}

n= \cos45^{\circ}=\frac{1}{\sqrt2}

Therefore the direction cosines of the lines are 0,\ -\frac{1}{\sqrt2},and\ \ \frac{1}{\sqrt2}.

Question:2 Find the direction cosines of a line which makes equal angles with the coordinate axes.

Answer:

If the line is making equal angle with the coordinate axes. Then,

Let the common angle made is \alpha with each coordinate axes.

Therefore, we can write;

l = \cos \alpha,\ m= \cos \alpha,and\ n= \cos \alpha

And as we know the relation; l^2+m^2+n^2 = 1

\Rightarrow \cos^2 \alpha +\cos^2 \alpha+\cos^2 \alpha = 1

\Rightarrow \cos^2 \alpha = \frac{1}{3}

or \cos \alpha =\pm \frac{1}{\sqrt3}

Thus the direction cosines of the line are \pm \frac{1}{\sqrt3},\ \pm \frac{1}{\sqrt3},and\ \pm \frac{1}{\sqrt3}  

Question:3 If a line has the direction ratios –18, 12, – 4, then what are its direction cosines ?

Answer:

GIven a line has direction ratios of -18, 12, – 4 then its direction cosines are;

Line having direction ratio -18 has direction cosine: 

\frac{-18}{\sqrt{(-18)^2+(12)^2+(-4)^2}} = \frac{-18}{22} = \frac{-9}{11}

Line having direction ratio 12 has direction cosine:

\frac{12}{\sqrt{(-18)^2+(12)^2+(-4)^2}} = \frac{12}{22} =\frac{6}{11}

Line having direction ratio -4 has direction cosine:

\frac{12}{\sqrt{(-4)^2+(12)^2+(-4)^2}} = \frac{-4}{22} = \frac{-2}{11}

Thus, the direction cosines are \frac{-9}{11},\ \frac{6}{11},\ \frac{-2}{11}.

Question:4 Show that the points (2, 3, 4), (– 1, – 2, 1), (5, 8, 7) are collinear.

Answer:

We have the points,  A (2, 3, 4),B (– 1, – 2, 1),C (5, 8, 7);

And as we can find the direction ratios of the line joining the points (x_{1},y_{1},z_{1}) \ and\ (x_{2},y_{2},z_{2}) is given by x_{2}-x_{1}, y_{2}-y_{1}, \ and\ z_{2}-z_{1}. 

The direction ratios of AB are (-1-2), (-2-3),\ and\ (1-4) i.e., -3,\ -5,\ and\ -3

The direction ratios of BC are (5-(-1)), (8-(-2)),\ and\ (7-1) i.e., 6,\ 10,\ and\ 6.

We can see that the direction ratios of AB and BC are proportional to each other and is -2 times.

\therefore AB is parallel to BC. and as point B is common to both AB and BC,

Hence the points A, B and C are collinear.

Question:5 Find the direction cosines of the sides of the triangle whose vertices are (3, 5, – 4), (– 1, 1, 2) and (– 5, – 5, – 2).

Answer:

Given vertices of the triangle \triangle ABC  (3, 5, – 4), (– 1, 1, 2) and (– 5, – 5, – 2).

Finding each side direction ratios;

\RightarrowDirection ratios of side AB are (-1-3), (1-5),\ and\ (2-(-4)) i.e.,

-4,-4,\ and\ 6.

Therefore its direction cosines values are;

\frac{-4}{\sqrt{(-4)^2+(-4)^2+(6)^2}},\ \frac{-4}{\sqrt{(-4)^2+(-4)^2+(6)^2}},\ \frac{6}{\sqrt{(-4)^2+(-4)^2+(6)^2}} or\ \frac{-4}{2\sqrt{17}},\frac{-4}{2\sqrt{17}},\frac{6}{2\sqrt{17}}\ or\ \frac{-2}{\sqrt{17}},\frac{-2}{\sqrt{17}},\frac{3}{\sqrt{17}}

SImilarly for side BC;

    \RightarrowDirection ratios of side BC are (-5-(-1)), (-5-1),\ and\ (-2-2) i.e.,

-4,-6,\ and\ -4.

Therefore its direction cosines values are;

\frac{-4}{\sqrt{(-4)^2+(-6)^2+(-4)^2}},\ \frac{-6}{\sqrt{(-4)^2+(-6)^2+(-4)^2}},\ \frac{-4}{\sqrt{(-4)^2+(-6)^2+(-4)^2}} or\ \frac{-4}{2\sqrt{17}},\frac{-6}{2\sqrt{17}},\frac{-4}{2\sqrt{17}}\ or\ \frac{-2}{\sqrt{17}},\frac{-3}{\sqrt{17}},\frac{-2}{\sqrt{17}}

  \RightarrowDirection ratios of side CA are (-5-3), (-5-5),\ and\ (-2-(-4)) i.e.,

-8,-10,\ and\ 2.

Therefore its direction cosines values are;

\frac{-8}{\sqrt{(-8)^2+(10)^2+(2)^2}},\ \frac{-5}{\sqrt{(-8)^2+(10)^2+(2)^2}},\ \frac{2}{\sqrt{(-8)^2+(10)^2+(2)^2}} or\ \frac{-8}{2\sqrt{42}},\frac{-10}{2\sqrt{42}},\frac{2}{2\sqrt{42}}\ or\ \frac{-4}{\sqrt{42}},\frac{-5}{\sqrt{42}},\frac{1}{\sqrt{42}}

CBSE NCERT solutions for class 12 maths chapter 11 three dimensional geometry-Exercise: 11.2 

Question:1 Show that the three lines with direction cosines

 \frac{12}{13}, \frac{-3}{13},\frac{-4}{13};\frac{4}{13},\frac{12}{13},\frac{3}{13};\frac{3}{13},\frac{-4}{13},\frac{12}{13}   are mutually perpendicular.

Answer:

GIven direction cosines of the three lines;

L_{1}\ \left ( \frac{12}{13}, \frac{-3}{13},\frac{-4}{13} \right )       L_{2}\ \left ( \frac{4}{13}, \frac{12}{13},\frac{3}{13} \right )      L_{3}\ \left ( \frac{3}{13}, \frac{-4}{13},\frac{12}{13} \right ) 

And we know that two lines with direction cosines l_{1},m_{1},n_{1}   and  l_{2},m_{2},n_{2} are perpendicular to each other, if l_{1}l_{2}+m_{1}m_{2}+n_{1}n_{2}=0

Hence we will check each pair of lines:

Lines L_{1}\ and\ L_{2};

l_{1}l_{2}+m_{1}m_{2}+n_{1}n_{2}= \left [ \frac{12}{13}\times\frac{4}{13} \right ]+\left [ \frac{-3}{13}\times\frac{12}{13} \right ]+\left [ \frac{-4}{13}\times \frac{3}{13} \right ]

= \left [ \frac{48}{169} \right ]-\left [ \frac{36}{169} \right ]-\left [ \frac{12}{169} \right ]= 0

\therefore the lines L_{1}\ and\ L_{2} are perpendicular.

Lines L_{2}\ and\ L_{3};

l_{1}l_{2}+m_{1}m_{2}+n_{1}n_{2}= \left [ \frac{4}{13}\times\frac{3}{13} \right ]+\left [ \frac{12}{13}\times\frac{-4}{13} \right ]+\left [ \frac{3}{13}\times \frac{12}{13} \right ]

= \left [ \frac{12}{169} \right ]-\left [ \frac{48}{169} \right ]+\left [ \frac{36}{169} \right ]= 0

\therefore the lines L_{2}\ and\ L_{3} are perpendicular.

Lines L_{3}\ and\ L_{1};

l_{1}l_{2}+m_{1}m_{2}+n_{1}n_{2}= \left [ \frac{3}{13}\times\frac{12}{13} \right ]+\left [ \frac{-4}{13}\times\frac{-3}{13} \right ]+\left [ \frac{12}{13}\times \frac{-4}{13} \right ]

= \left [ \frac{36}{169} \right ]+\left [ \frac{12}{169} \right ]-\left [ \frac{48}{169} \right ]= 0

\therefore the lines L_{3}\ and\ L_{1} are perpendicular.

Thus, we have all lines are mutually perpendicular to each other.

Question:2 Show that the line through the points (1, – 1, 2), (3, 4, – 2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6).

Answer:

We have given points where the line is passing through it;

Consider the line joining the points (1, – 1, 2) and (3, 4, – 2) is AB and line joining the points  (0, 3, 2) and (3, 5, 6).is CD.

So, we will find the direction ratios of the lines AB and CD;

Direction ratios of AB are a_{1},b_{1}, c_{1}

(3-1),\ (4-(-1)),\ and\ (-2-2)   or  2,\ 5,\ and\ -4

Direction ratios of CD are a_{2},b_{2}, c_{2}

(3-0),\ (5-3)),\ and\ (6-2)   or  3,\ 2,\ and\ 4.

Now, lines AB and CD will be perpendicular to each other if a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2} =0

a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2} =\left ( 2\times3 \right ) +\left ( 5\times2 \right )+ \left ( -4\times 4 \right )

= 6+10-16 = 0

Therefore, AB and CD are perpendicular to each other.

Question:3 Show that the line through the points (4, 7, 8), (2, 3, 4) is parallel to the line through the points (– 1, – 2, 1), (1, 2, 5).

Answer:

We have given points where the line is passing through it;

Consider the line joining the points (4, 7, 8) and (2, 3, 4) is AB and line joining the points  (– 1, – 2, 1) and (1, 2, 5)..is CD.

So, we will find the direction ratios of the lines AB and CD;

Direction ratios of AB are a_{1},b_{1}, c_{1}

(2-4),\ (3-7),\ and\ (4-8)   or  -2,\ -4,\ and\ -4

Direction ratios of CD are a_{2},b_{2}, c_{2}

(1-(-1)),\ (2-(-2)),\ and\ (5-1)   or  2,\ 4,\ and\ 4.

Now, lines AB and CD will be parallel to each other if \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}

Therefore we have now;

\frac{a_{1}}{a_{2}} = \frac{-2}{2}=-1        \frac{b_{1}}{b_{2}} = \frac{-4}{4}=-1       \frac{c_{1}}{c_{2}} = \frac{-4}{4}=-1

\therefore \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}

Hence we can say that AB is parallel to CD.

Question:4 Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector 3\widehat{i}+2\widehat{j}-2\widehat{k}.

Answer:

It is given that the line is passing through A (1, 2, 3) and is parallel to the vector \vec{b}=3\widehat{i}+2\widehat{j}-2\widehat{k}

We can easily find the equation of the line which passes through the point A and is parallel to the vector \vec{b} by the known relation;

\vec{r} = \vec{a} +\lambda\vec{b}, where \lambda is a constant.

So, we have now,

\\\mathrm{\Rightarrow \vec{r} = \widehat{i}+2\widehat{j}+3\widehat{k} + \lambda(3\widehat{i}+2\widehat{j}-2\widehat{k})} 

Thus the required equation of the line.

Answer:

Given that the line is passing through the point with position vector 2\widehat{i}-\widehat{j}+4\widehat{k} and is in the direction of the line \widehat{i}+2\widehat{j}-\widehat{k}.

And we know the equation of the line which passes through the point with the position vector \vec{a}and parallel to the vector \vec{b} is given by the equation,

\vec{r} = \vec{a} +\lambda\vec{b}

\Rightarrow \vec{r} =2\widehat{i}-\widehat{j}+4\widehat{k} + \lambda(\widehat{i}+2\widehat{j}-\widehat{k})

So, this is the required equation of the line in the vector form.

\vec{r} =x\widehat{i}+y\widehat{j}+z\widehat{k} = (\lambda+2)\widehat{i}+(2\lambda-1)\widehat{j}+(-\lambda+4)\widehat{k}

Eliminating \lambda, from the above equation we obtain the equation in the Cartesian form :

\frac{x-2}{1}= \frac{y+1}{2} =\frac{z-4}{-1}

Hence this is the required equation of the line in Cartesian form.

Question:6 Find the cartesian equation of the line which passes through the point (– 2, 4, – 5) and parallel to the line given by \frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}.

Answer:

Given a line which passes through the point (– 2, 4, – 5) and is parallel to the line given by the \frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6};

The direction ratios of the line, \frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6} are 3,5 and 6.

So, the required line is parallel to the above line.

Therefore we can take direction ratios of the required line as 3k, 5k, and 6k, where k is a non-zero constant.

And we know that the equation of line passing through the point (x_{1},y_{1},z_{1}) and with direction ratios a, b, c is written by: \frac{x-x_{1}}{a} = \frac{y-y_{1}}{b} = \frac{z-z_{1}}{c}.

Therefore we have the equation of the required line:

\frac{x+2}{3k} = \frac{y-4}{5k} = \frac{z+5}{6k}   

or   \frac{x+2}{3} = \frac{y-4}{5} = \frac{z+5}{6} = k

The required line equation. 

Question:7 The cartesian equation of a line is \frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{7}.  Write its vector form.

Answer:

Given the Cartesian equation of the line;

\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{7}

Here the given line is passing through the point (5,-4,6)

So, we can write the position vector of this point as;

\vec{a} = 5\widehat{i}-4\widehat{j}+6\widehat{k}

And the direction ratios of the line are 3, 7, and 2.

This implies that the given line is in the direction of the vector, \vec{b} = 3\widehat{i}+7\widehat{j}+2\widehat{k}.

Now, we can easily find the required equation of line:

As we know that the line passing through the position vector \vec{a} and in the direction of the vector \vec{b} is given by the relation,

\vec{r} = \vec{a} + \lambda \vec{b},\ \lambda \epsilon R

So, we get the equation.

\vec{r} = 5\widehat{i}-4\widehat{j}+6\widehat{k} + \lambda(3\widehat{i}+7\widehat{j}+2\widehat{k}),\ \lambda \epsilon R

This is the required equation of the line in the vector form.

Question:8 Find the vector and the cartesian equations of the lines that passes through the origin and (5, – 2, 3).

Answer:

GIven that the line is passing through the (0,0,0) and (5,-2,3)

Thus the required line passes through the origin. 

\therefore its position vector is given by,

\vec{a} = \vec{0}

So, the direction ratios of the line through  (0,0,0) and (5,-2,3) are,

(5-0) = 5, (-2-0) = -2, (3-0) = 3

The line is parallel to the vector given by the equation, \vec{b} = 5\widehat{i}-2\widehat{j}+3\widehat{k}

Therefore the equation of the line passing through the point with position vector \vec{a} and parallel to \vec{b} is given by;

\vec{r} = \vec{a}+\lambda\vec{b},\ where\ \lambda \epsilon R

\Rightarrow\vec{r} = 0+\lambda (5\widehat{i}-2\widehat{j}+3\widehat{k})

\Rightarrow\vec{r} = \lambda (5\widehat{i}-2\widehat{j}+3\widehat{k})

Now, the equation of the line through the point (x_{1},y_{1},z_{1}) and the direction ratios a, b, c is given by;

\frac{x-x_{1}}{a} = \frac{y-y_{1}}{b} =\frac{z-z_{1}}{c}

Therefore the equation of the required line in the Cartesian form will be;

\frac{x-0}{5} = \frac{y-0}{-2} =\frac{z-0}{3}

OR \frac{x}{5} = \frac{y}{-2} =\frac{z}{3}

Question:9 Find the vector and the cartesian equations of the line that passes through the points (3, – 2, – 5), (3, – 2, 6).

Answer:

Let the line passing through the points A(3,-2,-5) and B(3,-2,6) is AB;

Then as AB passes through through A so, we can write its position vector as;

\vec{a} =3\widehat{i}-2\widehat{j}-5\widehat{k}

Then direction ratios of PQ are given by,

(3-3)= 0,\ (-2+2) = 0,\ (6+5)=11

Therefore the equation of the vector in the direction of AB is given by,

\vec{b} =0\widehat{i}-0\widehat{j}+11\widehat{k} = 11\widehat{k}

We have then the equation of line AB in  vector form is given by,

\vec{r} =\vec{a}+\lambda\vec{b},\ where\ \lambda \epsilon R

\Rightarrow \vec{r} = (3\widehat{i}-2\widehat{j}-5\widehat{k}) + 11\lambda\widehat{k}

So, the equation of AB in Cartesian form is;

\frac{x-x_{1}}{a} =\frac{y-y_{1}}{b} =\frac{z-z_{1}}{c}

or \frac{x-3}{0} =\frac{y+2}{0} =\frac{z+5}{11}

Question:10 Find the angle between the following pairs of lines:

      (i)     \overrightarrow{r}=2\widehat{i}-5\widehat{j}+\widehat{k}+\lambda (3\widehat{i}+2\widehat{j}+6\widehat{k}) and \overrightarrow{r}=7\widehat{i}-6\widehat{k}+\mu (\widehat{i}+2\widehat{j}+2\widehat{k})

Answer:

To find the angle A between the pair of lines \vec{b_{1}}\ and\ \vec{b_{2}} we have the formula;

\cos A = \left | \frac{\vec{b_{1}}.\vec{b_{2}}}{|\vec{b_{1}}||\vec{b_{2}}|} \right |

We have two lines :

\overrightarrow{r}=2\widehat{i}-5\widehat{j}+\widehat{k}+\lambda (3\widehat{i}+2\widehat{j}+6\widehat{k}) and

\overrightarrow{r}=7\widehat{i}-6\widehat{k}+\mu (\widehat{i}+2\widehat{j}+2\widehat{k})

The given lines are parallel to the vectors \vec{b_{1}}\ and\ \vec{b_{2}};

where \vec{b_{1}}= 3\widehat{i}+2\widehat{j}+6\widehat{k}   and   \vec{b_{2}}= \widehat{i}+2\widehat{j}+2\widehat{k} respectively,

Then we have

\vec{b_{1}}.\vec{b_{2}} =(3\widehat{i}+2\widehat{j}+6\widehat{k}).(\widehat{i}+2\widehat{j}+2\widehat{k})

=3+4+12 = 19

and |\vec{b_{1}}| = \sqrt{3^2+2^2+6^2} = 7

|\vec{b_{2}}| = \sqrt{1^2+2^2+2^2} = 3

Therefore we have;

\cos A = \left | \frac{19}{7\times3} \right | = \frac{19}{21}

or A = \cos^{-1} \left ( \frac{19}{21} \right )

Question:10 Find the angle between the following pairs of lines:

     (ii)      \overrightarrow{r}= 3\widehat{i}+\widehat{j}-2\widehat{k}+\lambda (\widehat{i}-\widehat{j}-2\widehat{k}) and \overrightarrow{r}= 2\widehat{i}-\widehat{j}-56\widehat{k}+\mu (3\widehat{i}-5\widehat{j}-4\widehat{k})

Answer:

To find the angle A between the pair of lines \vec{b_{1}}\ and\ \vec{b_{2}} we have the formula;

\cos A = \left | \frac{\vec{b_{1}}.\vec{b_{2}}}{|\vec{b_{1}}||\vec{b_{2}}|} \right |

We have two lines :

\overrightarrow{r}= 3\widehat{i}+\widehat{j}-2\widehat{k}+\lambda (\widehat{i}-\widehat{j}-2\widehat{k}) and

\overrightarrow{r}= 2\widehat{i}-\widehat{j}-56\widehat{k}+\mu (3\widehat{i}-5\widehat{j}-4\widehat{k})

The given lines are parallel to the vectors \vec{b_{1}}\ and\ \vec{b_{2}};

where \vec{b_{1}}= \widehat{i}-\widehat{j}-2\widehat{k}   and   \vec{b_{2}}= 3\widehat{i}-5\widehat{j}-4\widehat{k} respectively,

Then we have

\vec{b_{1}}.\vec{b_{2}} =(\widehat{i}-\widehat{j}-2\widehat{k}).(3\widehat{i}-5\widehat{j}-4\widehat{k})

=3+5+8 = 16

and |\vec{b_{1}}| = \sqrt{1^2+(-1)^2+(-2)^2} = \sqrt{6}

|\vec{b_{2}}| = \sqrt{3^2+(-5)^2+(-4)^2} = \sqrt{50} = 5\sqrt2

Therefore we have;

\cos A = \left | \frac{16}{\sqrt6 \times5\sqrt2} \right | = \frac{16}{10\sqrt3}

or A = \cos^{-1} \left ( \frac{8}{5\sqrt3} \right )

Question:11 Find the angle between the following pair of lines:

     (i)     \frac{x-2}{2}= \frac{y-1}{5}= \frac{z+3}{-3} and \frac{x+2}{-1}= \frac{y-4}{8}= \frac{z-5}{4}

Answer:

Given lines are;

\frac{x-2}{2}= \frac{y-1}{5}= \frac{z+3}{-3}  and  \frac{x+2}{-1}= \frac{y-4}{8}= \frac{z-5}{4}

So, we two vectors \vec{b_{1}}\ and\ \vec{b_{2}} which are parallel to the pair of above lines respectively.

\vec{b_{1}}\ =2\widehat{i}+5\widehat{j}-3\widehat{k}   and  \vec{b_{2}}\ =-\widehat{i}+8\widehat{j}+4\widehat{k}

To find the angle A between the pair of lines \vec{b_{1}}\ and\ \vec{b_{2}} we have the formula;

\cos A = \left | \frac{\vec{b_{1}}.\vec{b_{2}}}{|\vec{b_{1}}||\vec{b_{2}}|} \right |

Then we have

\vec{b_{1}}.\vec{b_{2}} =(2\widehat{i}+5\widehat{j}-3\widehat{k}).(-\widehat{i}+8\widehat{j}+4\widehat{k})

=-2+40-12 = 26

and |\vec{b_{1}}| = \sqrt{2^2+5^2+(-3)^2} = \sqrt{38}

|\vec{b_{2}}| = \sqrt{(-1)^2+(8)^2+(4)^2} = \sqrt{81} = 9

Therefore we have;

\cos A = \left | \frac{26}{\sqrt{38} \times9} \right | = \frac{26}{9\sqrt{38}}

or A = \cos^{-1} \left ( \frac{26}{9\sqrt{38}} \right )

Question:11 Find the angle between the following pair of lines:

       (ii)     \frac{x}{2}= \frac{y}{2}=\frac{z}{1} and \frac{x -5}{4}= \frac{y-2}{1}=\frac{z-3}{8}

Answer:

Given lines are;

   \frac{x}{2}= \frac{y}{2}=\frac{z}{1} and \frac{x -5}{4}= \frac{y-2}{1}=\frac{z-3}{8}

So, we two vectors \vec{b_{1}}\ and\ \vec{b_{2}} which are parallel to the pair of above lines respectively.

\vec{b_{1}}\ =2\widehat{i}+2\widehat{j}+\widehat{k}   and  \vec{b_{2}}\ =4\widehat{i}+\widehat{j}+8\widehat{k}

To find the angle A between the pair of lines \vec{b_{1}}\ and\ \vec{b_{2}} we have the formula;

\cos A = \left | \frac{\vec{b_{1}}.\vec{b_{2}}}{|\vec{b_{1}}||\vec{b_{2}}|} \right |

Then we have

\vec{b_{1}}.\vec{b_{2}} =(2\widehat{i}+2\widehat{j}+\widehat{k}).(4\widehat{i}+\widehat{j}+8\widehat{k})

=8+2+8 = 18

and |\vec{b_{1}}| = \sqrt{2^2+2^2+1^2} = \sqrt{9} = 3

|\vec{b_{2}}| = \sqrt{(4)^2+(1)^2+(8)^2} = \sqrt{81} = 9

Therefore we have;

\cos A = \left | \frac{18}{ 3\times9} \right | = \frac{2}{3}

or A = \cos^{-1} \left ( \frac{2}{3} \right )

Question:12 Find the values of p so that the lines \frac{1-x}{3}=\frac{7y-14}{2p}= \frac{z-3}{2} and \frac{7-7x}{3p}=\frac{y-5}{1}= \frac{6-z}{5} are at right angles.

Answer:

First we have to write the given equation of lines in the standard form;

\frac{x-1}{-3}=\frac{y-2}{\frac{2p}{7}}= \frac{z-3}{2}   and  \frac{x-1}{\frac{-3p}{7}}=\frac{y-5}{1}= \frac{z-6}{-5}

Then we have the direction ratios of the above lines as;

-3,\ \frac{2p}{7},\ 2   and   \frac{-3p}{7},\ 1,\ -5   respectively..

Two lines with direction ratios a_{1},b_{1},c_{1} and a_{2},b_{2},c_{2}  are perpendicular to each other if,  a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}= 0

\therefore (-3).\left ( \frac{-3p}{7} \right )+(\frac{2p}{7}).(1) + 2.(-5) = 0

\Rightarrow \frac{9p}{7}+ \frac{2p}{7} =10

\Rightarrow 11p =70

\Rightarrow p =\frac{70}{11}

Thus, the value of p is \frac{70}{11}.

Question:13 Show that the lines \frac{x-5}{7}=\frac{y+2}{-3}=\frac{z}{1} and \frac{x}{1}=\frac{y}{2}=\frac{z}{3} are perpendicular to each other.

Answer:

First, we have to write the given equation of lines in the standard form;

 \frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1} and \frac{x}{1}=\frac{y}{2}=\frac{z}{3}

Then we have the direction ratios of the above lines as;

7,\ -5,\ 1   and   1,\ 2,\ 3   respectively..

Two lines with direction ratios a_{1},b_{1},c_{1} and a_{2},b_{2},c_{2}  are perpendicular to each other if,  a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}= 0

\therefore 7(1) + (-5)(2)+1(3) = 7-10+3 = 0

Therefore the two lines are perpendicular to each other.

Question:14 Find the shortest distance between the lines

              \overrightarrow{r}=(\widehat{i}+2\widehat{j}+\widehat{k})+\lambda (\widehat{i}-\widehat{j}+\widehat{k}) and  \overrightarrow{r}=(2\widehat{i}-\widehat{j}-\widehat{k})+\mu (2\widehat{i}+\widehat{j}+2\widehat{k})                 

Answer:

So given equation of lines;

 \overrightarrow{r}=(\widehat{i}+2\widehat{j}+\widehat{k})+\lambda (\widehat{i}-\widehat{j}+\widehat{k}) and  \overrightarrow{r}=(2\widehat{i}-\widehat{j}-\widehat{k})+\mu (2\widehat{i}+\widehat{j}+2\widehat{k}) in the vector form.

Now, we can find the shortest distance between the lines \vec{r} = \vec{a_{1}}+\lambda\vec{b_{1}} and \vec{r} = \vec{a_{2}}+\mu \vec{b_{2}}, is given by the formula,

d= \left | \frac{\left ( \vec{b_{1}}\times\vec{b_{2}} \right ).\left ( \vec{a_{2}}-\vec{a_{1}} \right )}{\left | \vec{b_{1}}\times \vec{b_{2}} \right |} \right |

Now comparing the values from the equation, we obtain

\vec{a_{1}} = \widehat{i}+2\widehat{j}+\widehat{k}                 \vec{b_{1}} = \widehat{i}-\widehat{j}+\widehat{k}

\vec{a_{2}} = 2\widehat{i}-\widehat{j}-\widehat{k}               \vec{b_{2}} = 2\widehat{i}+\widehat{j}+2\widehat{k}

\vec{a_{2}} -\vec{a_{1}} =\left ( 2\widehat{i}-\widehat{j}-\widehat{k} \right ) - \left ( \widehat{i}+2\widehat{j}+\widehat{k} \right ) = \widehat{i}-3\widehat{j}-2\widehat{k}

Then calculating 

\vec{b_{1}}\times \vec{b_{2}} = \begin{vmatrix} \widehat{i} &\widehat{j} &\widehat{k} \\ 1 & -1 &1 \\ 2& 1 &2 \end{vmatrix}

\vec{b_{1}}\times \vec{b_{2}} = (-2-1)\widehat{i} - (2-2) \widehat{j} +(1+2) \widehat{k} = -3\widehat{i}+3\widehat{k}

\Rightarrow \left | \vec{b_{1}}\times \vec{b_{2}} \right | = \sqrt{(-3)^2+(3)^2} = \sqrt{9+9} =\sqrt{18} =3\sqrt2

So, substituting the values now in the formula above we get;

d =\left | \frac{\left ( -3\widehat{i}+3\widehat{k} \right ).(\widehat{i}-3\widehat{j}-2\widehat{k})}{3\sqrt2} \right |

\Rightarrow d = \left | \frac{-3.1+3(-2)}{3\sqrt2} \right |

d = \left | \frac{-9}{3\sqrt2} \right | = \frac{3}{\sqrt2} = \frac{3\sqrt2}{2}

Therefore, the shortest distance between the two lines is \frac{3\sqrt2}{2} units.

Question:15 Find the shortest distance between the lines 

\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1} and \frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}

Answer:

We have given two lines: 

\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1} and \frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}

Calculating the shortest distance between the two lines,

\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}} and \frac{x-x_{2}}{a_{2}}=\frac{y-y_{2}}{b_{2}}=\frac{z-z_{2}}{c_{2}}

by the formula 

d = \frac{\begin{vmatrix} x_{2}-x_{1} &y_{2}-y_{1} &z_{2}-z_{1} \\ a_{1} &b_{1} &c_{1} \\ a_{2}&b_{2} &c_{2} \end{vmatrix}}{\sqrt{(b_{1}c_{2}-b_{2}c_{1})^2+(c_{1}a_{2}-c_{2}a_{1})^2+(a_{1}b_{2}-a_{2}b_{1})^2}}

Now, comparing the given equations, we obtain

x_{1} = -1,\ y_{1} =-1,\ z_{1} =-1

a_{1} = 7,\ b_{1} =-6,\ c_{1} =1

x_{2} = 3,\ y_{2} =5,\ z_{2} =7

a_{2} = 1,\ b_{2} =-2,\ c_{2} =1

Then calculating determinant 

\begin{vmatrix} x_{2}-x_{1} &y_{2}-y_{1} &z_{2}-z_{1} \\ a_{1} &b_{1} &c_{1} \\ a_{2}&b_{2} &c_{2} \end{vmatrix} = \begin{vmatrix} 4 &6 &8 \\ 7& -6& 1\\ 1& -2& 1 \end{vmatrix} 

= 4(-6+2)-6(7-1)+8(-14+6)

= -16-36-64

=-116

Now calculating the denominator,

\sqrt{(b_{1}c_{2}-b_{2}c_{1})^2+(c_{1}a_{2}-c_{2}a_{1})^2+(a_{1}b_{2}-a_{2}b_{1})^2} = \sqrt{(-6+2)^2+(1+7)^2+(-14+6)^2}= \sqrt{16+36+64}

= \sqrt{116} = 2\sqrt{29}

So, we will substitute all the values in the formula above to obtain,

d = \frac{-116}{2\sqrt{29}} = \frac{-58}{\sqrt{29}} = \frac{-2\times29}{\sqrt{29}} = -2\sqrt{29}

Since distance is always non-negative, the distance between the given lines is 

2\sqrt{29} units.

Question:16 Find the shortest distance between the lines whose vector equations are \overrightarrow{r}=(\widehat{i}+2\widehat{j}+3\widehat{k})+ \lambda (\widehat{i}-3\widehat{j}+2\widehat{k}) and 

             \overrightarrow{r}=(4\widehat{i}+5\widehat{j}+6\widehat{k})+ \mu (2\widehat{i}+3\widehat{j}+\widehat{k})

Answer:

Given two equations of line 

\overrightarrow{r}=(\widehat{i}+2\widehat{j}+3\widehat{k})+ \lambda (\widehat{i}-3\widehat{j}+2\widehat{k}) \overrightarrow{r}=(4\widehat{i}+5\widehat{j}+6\widehat{k})+ \mu (2\widehat{i}+3\widehat{j}+\widehat{k}) in the vector form.

So, we will apply the distance formula  for knowing the distance between two lines \vec{r} =\vec{a_{1}}+\lambda{b_{1}}   and   \vec{r} =\vec{a_{2}}+\lambda{b_{2}}

d= \left | \frac{\left ( \vec{b_{1}}\times\vec{b_{2}} \right ).\left ( \vec{a_{2}}-\vec{a_{1}} \right )}{\left | \vec{b_{1}}\times \vec{b_{2}} \right |} \right |

After comparing the given equations, we obtain

\vec{a_{1}} = \widehat{i}+2\widehat{j}+3\widehat{k}                \vec{b_{1}} = \widehat{i}-3\widehat{j}+2\widehat{k}

\vec{a_{2}} = 4\widehat{i}+5\widehat{j}+6\widehat{k}             \vec{b_{2}} = 2\widehat{i}+3\widehat{j}+\widehat{k}

\vec{a_{2}}-\vec{a_{1}} = (4\widehat{i}+5\widehat{j}+6\widehat{k}) - (\widehat{i}+2\widehat{j}+3\widehat{k})

                 = 3\widehat{i}+3\widehat{j}+3\widehat{k} 

Then calculating the determinant value numerator.

\vec{b_{1}}\times\vec{b_{2}} = \begin{vmatrix} \widehat{i} &\widehat{j} &\widehat{k} \\ 1& -3 &2 \\ 2& 3& 1 \end{vmatrix}

= (-3-6)\widehat{i}-(1-4)\widehat{j}+(3+6)\widehat{k} = -9\widehat{i}+3\widehat{j}+9\widehat{k}

That implies, \left | \vec{b_{1}}\times\vec{b_{2}} \right | = \sqrt{(-9)^2+(3)^2+(9)^2}

= \sqrt{81+9+81} = \sqrt{171} =3\sqrt{19}

\left ( \vec{b_{1}}\times\vec{b_{2}} \right ).\left ( \vec{a_{2}}-\vec{a_{1}} \right )=(-9\widehat{i}+3\widehat{j}+9\widehat{k})(3\widehat{i}+3\widehat{j}+3\widehat{k})

= (-9\times3)+(3\times3)+(9\times3) = 9

Now, after substituting the value in the above formula we get,

d= \left | \frac{9}{3\sqrt{19}} \right | = \frac{3}{\sqrt{19}}

Therefore,\frac{3}{\sqrt{19}} is the shortest distance between the two given lines. 

Question:17 Find the shortest distance between the lines whose vector equations are 

  \overrightarrow{r}=(1-t)\widehat{i}+(t-2)\widehat{j}+(3-2t)\widehat{k} and \overrightarrow{r}=(s+1)\widehat{i}+(2s-1)\widehat{j}-(2s+1)\widehat{k}

Answer:

Given two equations of the line 

\overrightarrow{r}=(1-t)\widehat{i}+(t-2)\widehat{j}+(3-2t)\widehat{k}  \overrightarrow{r}=(s+1)\widehat{i}+(2s-1)\widehat{j}-(2s+1)\widehat{k}  in the vector form.

So, we will apply the distance formula  for knowing the distance between two lines \vec{r} =\vec{a_{1}}+\lambda{b_{1}}   and   \vec{r} =\vec{a_{2}}+\lambda{b_{2}}

d= \left | \frac{\left ( \vec{b_{1}}\times\vec{b_{2}} \right ).\left ( \vec{a_{2}}-\vec{a_{1}} \right )}{\left | \vec{b_{1}}\times \vec{b_{2}} \right |} \right |

After comparing the given equations, we obtain

\vec{a_{1}} = \widehat{i}-2\widehat{j}+3\widehat{k}                \vec{b_{1}} = -\widehat{i}+\widehat{j}-2\widehat{k}

\vec{a_{2}} = \widehat{i}-\widehat{j}-\widehat{k}             \vec{b_{2}} = \widehat{i}+2\widehat{j}-2\widehat{k}

\vec{a_{2}}-\vec{a_{1}} = (\widehat{i}-\widehat{j}-\widehat{k}) - (\widehat{i}-2\widehat{j}+3\widehat{k}) = \widehat{j}-4\widehat{k} 

Then calculating the determinant value numerator.

\vec{b_{1}}\times\vec{b_{2}} = \begin{vmatrix} \widehat{i} &\widehat{j} &\widehat{k} \\ -1& 1 &-2 \\ 1& 2& -2 \end{vmatrix}

= (-2+4)\widehat{i}-(2+2)\widehat{j}+(-2-1)\widehat{k} = 2\widehat{i}-4\widehat{j}-3\widehat{k}

That implies,

\left | \vec{b_{1}}\times\vec{b_{2}} \right | = \sqrt{(2)^2+(-4)^2+(-3)^2}

= \sqrt{4+16+9} = \sqrt{29}

\left ( \vec{b_{1}}\times\vec{b_{2}} \right ).\left ( \vec{a_{2}}-\vec{a_{1}} \right )=(2\widehat{i}-4\widehat{j}-3\widehat{k})(\widehat{j}-4\widehat{k}) = -4+12 = 8

Now, after substituting the value in the above formula we get,

d= \left | \frac{8}{\sqrt{29}} \right | = \frac{8}{\sqrt{29}}

Therefore,\frac{8}{\sqrt{29}} units are the shortest distance between the two given lines. 

NCERT solutions for class 12 maths chapter 11 three dimensional geometry-Exercise: 11.3

Question:1(a) In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

    z = 2 

Answer:

Equation of plane Z=2, i.e.  0x+0y+z=2

The direction ratio of normal is 0,0,1

\therefore \, \, \, \sqrt{0^2+0^2+1^2}=1

Divide equation 0x+0y+z=2 by 1 from both side 

We get,      0x+0y+z=2

Hence, direction cosins are 0,0,1.

The distance of the plane from the origin is 2.

Question:1(b) In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

   x + y + z = 1 

Answer:

Given the equation of the plane is x+y+z=1 or we can write 1x+1y+1z=1

So, the direction ratios of normal from the above equation are, 1,\1,\ and\ 1.

Therefore \sqrt{1^2+1^2+1^2} =\sqrt{3}

Then dividing both sides of the plane equation by \sqrt{3}, we get

\frac{x}{\sqrt3}+\frac{y}{\sqrt3}+\frac{z}{\sqrt3}=\frac{1}{\sqrt3}

So, this is the form of  lx+my+nz = d the plane, where l,\ m,\ n are the direction cosines of normal to the plane and d is the distance of the perpendicular drawn from the origin.

\therefore The direction cosines of the given line are \frac{1}{\sqrt3},\ \frac{1}{\sqrt3},\ \frac{1}{\sqrt3} and the distance of the plane from the origin is \frac{1}{\sqrt3} units.

Question:1(c) In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

  2x + 3y - z = 5

Answer:

Given the equation of plane is 2x+3y-z=5  

So, the direction ratios of normal from the above equation are, 2,\3,\ and\ -1.

Therefore \sqrt{2^2+3^2+(-1)^2} =\sqrt{14}

Then dividing both sides of the plane equation by \sqrt{14}, we get

\frac{2x}{\sqrt{14}}+\frac{3y}{\sqrt{14}}-\frac{z}{\sqrt{14}}=\frac{5}{\sqrt{14}}

So, this is the form of  lx+my+nz = d the plane, where l,\ m,\ n are the direction cosines of normal to the plane and d is the distance of the perpendicular drawn from the origin.

\therefore The direction cosines of the given line are \frac{2}{\sqrt{14}},\ \frac{3}{\sqrt{14}},\ \frac{-1}{\sqrt{14}} and the distance of the plane from the origin is \frac{5}{\sqrt{14}} units.

Question:1(d) In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

    5y + 8 = 0

Answer:

Given the equation of plane is 5y+8=0  or we can write 0x-5y+0z=8

So, the direction ratios of normal from the above equation are, 0,\ -5,\ and\ 0.

Therefore \sqrt{0^2+(-5)^2+0^2} =5

Then dividing both sides of the plane equation by 5, we get

-y = \frac{8}{5}

So, this is the form of  lx+my+nz = d the plane, where l,\ m,\ n are the direction cosines of normal to the plane and d is the distance of the perpendicular drawn from the origin.

\therefore The direction cosines of the given line are 0,\ -1,\ and\ 0 and the distance of the plane from the origin is \frac{8}{5} units.

Question:2 Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector 3\widehat{i}+5\widehat{j}-6\widehat{k}.

Answer:

We have given the distance between the plane and origin equal to 7 units and normal to the vector 3\widehat{i}+5\widehat{j}-6\widehat{k}.

So, it is known that the equation of the plane with position vector \vec{r} is given by, the relation,

\vec{r}.\widehat{n} =d , where d is the distance of the plane from the origin.

Calculating \widehat{n};

\widehat{n} = \frac{\vec{n}}{|\vec{n}|} = \frac{3\widehat{i}+5\widehat{j}-6\widehat{k}}{\sqrt{(3)^2+(5)^2+(6)^2}} = \frac{3\widehat{i}+5\widehat{j}-6\widehat{k}}{\sqrt{70}}

\vec{r}.\left ( \frac{3\widehat{i}+5\widehat{j}-6\widehat{k}}{\sqrt{70}} \right ) = 7  is the vector equation of the required plane.

Question:3(a) Find the Cartesian equation of the following planes:

  \overrightarrow{r}.(\widehat{i}+\widehat{j}-\widehat{k})=2

Answer:

Given the equation of the plane \overrightarrow{r}.(\widehat{i}+\widehat{j}-\widehat{k})=2

So we have to find the Cartesian equation,

Any point A (x,y,z) on this plane will satisfy the equation and its position vector given by,

\vec{r}=x\widehat{i}+y\widehat{j}-z\widehat{k}

Hence we have,

 (x\widehat{i}+y\widehat{j}+z\widehat{k}).(\widehat{i}+\widehat{j}-\widehat{k}) =2

Or,  x+y-z=2

Therefore this is the required Cartesian equation of the plane.

Question:3(b) Find the Cartesian equation of the following planes:

   \overrightarrow{r}.(2\widehat{i}+3\widehat{i}-4\widehat{k})=1

Answer:

Given the equation of plane \overrightarrow{r}.(2\widehat{i}+3\widehat{i}-4\widehat{k})=1

So we have to find the Cartesian equation,

Any point A (x,y,z) on this plane will satisfy the equation and its position vector given by,

\vec{r}=x\widehat{i}+y\widehat{j}-z\widehat{k}

Hence we have,

 (x\widehat{i}+y\widehat{j}+z\widehat{k}).(2\widehat{i}+3\widehat{j}-4\widehat{k}) =1

Or,  2x+3y-4z=1

Therefore this is the required Cartesian equation of the plane.

Question:3(c) Find the Cartesian equation of the following planes:

     \overrightarrow{r}.\left [ \left ( s-2t \right )\widehat{i}+(3-t) \widehat{j}+(2s+t)\widehat{k}\right ]=15 

Answer:

Given the equation of plane \overrightarrow{r}.\left [ \left ( s-2t \right )\widehat{i}+(3-t) \widehat{j}+(2s+t)\widehat{k}\right ]=15

So we have to find the Cartesian equation,

Any point A (x,y,z) on this plane will satisfy the equation and its position vector given by,  \vec{r}=x\widehat{i}+y\widehat{j}-z\widehat{k}

Hence we have,

 (x\widehat{i}+y\widehat{j}+z\widehat{k}).\left [ \left ( s-2t \right )\widehat{i}+(3-t) \widehat{j}+(2s+t)\widehat{k}\right] =15

Or,  (s-2t)x+(3-t)y+(2s+t)z=15

Therefore this is the required Cartesian equation of the plane.

Question:4(a) In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

   2 x + 3y + 4 z - 12 = 0

Answer:

Let the coordinates of the foot of perpendicular P from the origin to the plane be (x_{1},y_{1},z_{1})

Given a plane equation 2x+3y+4z-12=0,

Or, 2x+3y+4z=12

The direction ratios of the normal of the plane are 2,\ 3,\and\ 4.

Therefore \sqrt{(2)^2+(3)^2+(4)^2} = \sqrt{29}

So, now dividing both sides of the equation by \sqrt{29} we will obtain,

\frac{2}{\sqrt{29}}x+\frac{3}{\sqrt{29}}y+\frac{4}{\sqrt{29}}z = \frac{12}{\sqrt{29}}

This equation is similar to lx+my+nz = d where, l,\ m,\ n are the directions cosines of normal to the plane and d is the distance of normal from the origin.

Then finding the coordinates of the foot of the perpendicular are given by (ld,md,nd)

\therefore The coordinates of the foot of the perpendicular are;

\left [ \frac{2}{\sqrt{29}}.\frac{12}{\sqrt{29}},\frac{3}{\sqrt{29}}.\frac{12}{\sqrt{29}},\frac{4}{\sqrt{29}}.\frac{12}{\sqrt{29}} \right ]  or  \left [ \frac{24}{29}, \frac{36}{49}, \frac{48}{29} \right ]

Question:4(b) In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

 3y + 4z - 6 =  0

Answer:

Let the coordinates of the foot of perpendicular P from the origin to the plane be (x_{1},y_{1},z_{1})

Given a plane equation 3y+4z-6=0,

Or, 0x+3y+4z=6

The direction ratios of the normal of the plane are 0,\ 3,\and\ 4.

Therefore \sqrt{(0)^2+(3)^2+(4)^2} = 5

So, now dividing both sides of the equation by 5 we will obtain,

0x+\frac{3}{5}y+\frac{4}{5}z = \frac{6}{5}

This equation is similar to lx+my+nz = d where, l,\ m,\ n are the directions cosines of normal to the plane and d is the distance of normal from the origin.

Then finding the coordinates of the foot of the perpendicular are given by (ld,md,nd)

\therefore The coordinates of the foot of the perpendicular are;

\left (0,\frac{3}{5}.\frac{6}{5},\frac{4}{5}.\frac{6}{5} \right )  or  \left ( 0, \frac{18}{25}, \frac{24}{25} \right )

Question:4(c) In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

  x + y + z = 1 

Answer:

Let the coordinates of the foot of perpendicular P from the origin to the plane be (x_{1},y_{1},z_{1})

Given plane equation x+y+z=1.

The direction ratios of the normal of the plane are 1,\ 1,\and\ 1.

Therefore \sqrt{(1)^2+(1)^2+(1)^2} = \sqrt3

So, now dividing both sides of the equation by \sqrt3 we will obtain,

\frac{x}{\sqrt3}+\frac{y}{\sqrt3}+\frac{z}{\sqrt3} = \frac{1}{\sqrt3}

This equation is similar to lx+my+nz = d where, l,\ m,\ n are the directions cosines of normal to the plane and d is the distance of normal from the origin.

Then finding the coordinates of the foot of the perpendicular are given by (ld,md,nd)

\therefore The coordinates of the foot of the perpendicular are;

\left ( \frac{1}{\sqrt3}.\frac{1}{\sqrt3},\frac{1}{\sqrt3}.\frac{1}{\sqrt3},\frac{1}{\sqrt3}.\frac{1}{\sqrt3} \right )  or  \left ( \frac{1}{3},\frac{1}{3},\frac{1}{3} \right )..

Question:4(d) In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

 5y + 8 = 0   

Answer:

Let the coordinates of the foot of perpendicular P from the origin to the plane be (x_{1},y_{1},z_{1})

Given plane equation 5y+8=0.

or written as 0x-5y+0z=8

The direction ratios of the normal of the plane are 0,\ -5,\and\ 0.

Therefore \sqrt{(0)^2+(-5)^2+(0)^2} = 5

So, now dividing both sides of the equation by 5 we will obtain,

-y=\frac{8}{5}

This equation is similar to lx+my+nz = d where, l,\ m,\ n are the directions cosines of normal to the plane and d is the distance of normal from the origin.

Then finding the coordinates of the foot of the perpendicular are given by (ld,md,nd)

\therefore The coordinates of the foot of the perpendicular are;

\left ( 0,-1(\frac{8}{5}),0 \right )  or  \left ( 0,\frac{-8}{5},0 \right ).

Question:5(a)  Find the vector and cartesian equations of the planes (a) that passes through the point (1, 0, – 2) and the normal to the plane is \widehat{i}+\widehat{j}-\widehat{k}.

Answer:

Given the point A (1,0,-2) and the normal vector \widehat{n} which is perpendicular to the plane is \widehat{n} = \widehat{i}+\widehat{j}-\widehat{k}

The position vector of point A is \vec {a} = \widehat{i}-2\widehat{k}

So, the vector equation of the plane would be given by,

(\vec{r}-\vec{a}).\widehat{n} = 0

Or \left [ \vec{r}-(\widehat{i}-2\widehat{k}) \right ].(\widehat{i}+\widehat{j}-\widehat{k}) = 0

where \vec{r} is the position vector of any arbitrary point A(x,y,z) in the plane.

\therefore \vec{r} = x\widehat{i}+y\widehat{j}+z\widehat{k}

Therefore, the equation we get,

\left [(x\widehat{i}+y\widehat{j}+z\widehat{k})-(\widehat{i}-2\widehat{k}) \right ].(\widehat{i}+\widehat{j}-\widehat{k}) = 0

\Rightarrow \left [(x-1)\widehat{i}+y\widehat{j}+(z+2)\widehat{k}\right ].(\widehat{i}+\widehat{j}-\widehat{k}) = 0

\Rightarrow(x-1)+y-(z+2) = 0

\Rightarrow x+y-z-3=0    or   x+y-z=3

So, this is the required Cartesian equation of the plane.

Question:5(b) Find the vector and cartesian equations of the planes 

   that passes through the point (1,4, 6) and the normal vector to the plane is \widehat{i}-2\widehat{j}+\widehat{k}.

Answer:

Given the point A (1,4,6) and the normal vector \widehat{n} which is perpendicular to the plane is \widehat{n} = \widehat{i}-2\widehat{j}+\widehat{k}

The position vector of point A is \vec {a} = \widehat{i}+4\widehat{j}+6\widehat{k}

So, the vector equation of the plane would be given by,

(\vec{r}-\vec{a}).\widehat{n} = 0

Or \left [ \vec{r}-(\widehat{i}+4\widehat{j}+6\widehat{k}) \right ].(\widehat{i}-2\widehat{j}+\widehat{k}) = 0

where \vec{r} is the position vector of any arbitrary point A(x,y,z) in the plane.

\therefore \vec{r} = x\widehat{i}+y\widehat{j}+z\widehat{k}

Therefore, the equation we get,

\left [ (x\widehat{i}+y\widehat{j}+z\widehat{k})-(\widehat{i}+4\widehat{j}+6\widehat{k}) \right ].(\widehat{i}-2\widehat{j}+\widehat{k}) = 0

\Rightarrow \left [(x-1)\widehat{i}+(y-4)\widehat{j}+(z-6)\widehat{k}\right ].(\widehat{i}-2\widehat{j}+\widehat{k}) = 0

(x-1)-2(y-4)+(z-6)=0

\Rightarrow x-2y+z+1=0    

So, this is the required Cartesian equation of the plane.

Question:6(a) Find the equations of the planes that passes through three points.

     (1, 1, – 1), (6, 4, – 5), (– 4, – 2, 3) 

Answer:

The equation of the plane which passes through the three points A(1,1,-1),\ B(6,4,-5),\ and\ C(-4,-2,3)  is given by;

Determinant method,

\begin{vmatrix} 1 &1 &-1 \\ 6& 4 & -5\\ -4& -2 &3 \end{vmatrix} = (12-10)-(18-20)-(-12+16)

Or, = 2+2-4 = 0

Here, these three points A, B, C are collinear points.

Hence there will be an infinite number of planes possible which passing through the given points.

Question:6(b)  Find the equations of the planes that passes through three points.

  (1, 1, 0), (1, 2, 1), (– 2, 2, – 1)

Answer:

The equation of the plane which passes through the three points A(1,1,0),\ B(1,2,1),\ and\ C(-2,2,-1)  is given by;

Determinant method,

\begin{vmatrix} 1 &1 &0 \\ 1& 2 & 1\\ -2& 2 &-1 \end{vmatrix} = (-2-2)-(2+2)= -8 \neq 0

As determinant value is not equal to zero hence there must be a plane that passes through the points A, B, and C.

Finding the equation of the plane through the points, (x_{1},y_{1},z_{1}), (x_{2},y_{2},z_{2})\ and\ (x_{3},y_{3},z_{3})

\begin{vmatrix} x-x_{1} &y-y_{1} &z-z_{1} \\ x_{2}-x_{1} &y_{2}-y_{1} &z_{2}-z_{1} \\ x_{3}-x_{1}&y_{3}-y_{1} &z_{3}-z_{1} \end{vmatrix} = 0

After substituting the values in the determinant we get,

\begin{vmatrix} x-1 &y-1 &z \\ 0& 1 &1 \\ -3& 1&-1 \end{vmatrix} = 0

\Rightarrow(x-1)(-1-1)-(y-1)(0+3)+z(0+3) = 0

\Rightarrow-2x+2-3y+3+3z = 0

2x+3y-3z = 5

So, this is the required Cartesian equation of the plane.

Question:7 Find the intercepts cut off by the plane 2x + y – z = 5.

Answer:

Given plane  2x + y-z = 5

We have to find the intercepts that this plane would make so,

Making it look like intercept form first:

By dividing both sides of the equation by 5 (as we have to make the R.H.S =1) , we get then,

\frac{2}{5}x+\frac{y}{5}-\frac{z}{5} =1

\Rightarrow \frac{x}{\frac{5}{2}}+\frac{y}{5}+\frac{z}{-5} =1

So, as we know that from the equation of a plane in intercept form,\frac{x}{a}+\frac{y}{b}+\frac{z}{c} = 1 where a,b,c are the intercepts cut off by the plane at x,y, and z-axes respectively.

Therefore after comparison, we get the values of a,b, and c.

a = \frac{5}{2},\ b=5,\ and\ c=-5.

Hence the intercepts are \frac{5}{2},\ 5,\ and\ -5.

Question:8 Find the equation of the plane with intercept 3 on the y-axis and parallel to ZOX plane.

Answer:

Given that the plane is parallel to the ZOX plane.

So, we have the equation of plane ZOX as  y = 0.

And an intercept of 3 on the y-axis \Rightarrow b =3

Intercept form of a plane given by;

\frac{x}{a}+\frac{y}{b}+\frac{z}{c} = 1

So, here the plane would be parallel to the x and z-axes both.

we have any plane parallel to it is of the form, y=a.

Equation of the plane required is y=3.

Question:9 Find the equation of the plane through the intersection of the planes 3x – y + 2z – 4 = 0 and x + y + z – 2 = 0 and the point (2, 2, 1).

Answer:

The equation of any plane through the intersection of the planes,

3x-y+2z-4=0\ and\ x+y+z-2=0

Can be written in the form of; (3x-y+2z-4)\ +\alpha( x+y+z-2)= 0, where \alpha \epsilon R

So, the plane passes through the point (2,2,1), will satisfy the above equation.

(3\times2-2+2\times1-4)+\alpha(2+2+1-2) = 0

That implies 2+3\alpha= 0

\alpha = \frac{-2}{3}

Now, substituting the value of \alpha in the equation above we get the final equation of the plane;

(3x-y+2z-4)\ +\alpha( x+y+z-2)= 0

(3x-y+2z-4)\ +\frac{-2}{3}( x+y+z-2)= 0

\Rightarrow 9x-3y+6z-12\ -2 x-2y-2z+4= 0

\Rightarrow 7x-5y+4z-8= 0 is the required equation of the plane.

Question:10 Find the vector equation of the plane passing through the intersection of the planes \overrightarrow{r}.(2\widehat{i}+2\widehat{j}-3\widehat{k})=7   ,  \overrightarrow{r}(2\widehat{i}+5\widehat{j}+3\widehat{k})=9 and   through the point (2, 1, 3).

Answer:

Here \vec{n_{1}} =2 \widehat{i}+2\widehat{j}-3\widehat{k} and \vec{n_{2}} = 2\widehat{i}+5\widehat{j}+3\widehat{k} 

and d_{1} = 7  and  d_{2} = 9

Hence, using the relation \vec{r}.(\vec{n_{1}}+\lambda\vec{n_{2}}) = d_{1}+\lambda d_{2}, we get

\vec{r}.[2\widehat{i}+2\widehat{j}-3\widehat{k}+\lambda(2\widehat{i}+5\widehat{j}+3\widehat{k})] = 7+9\lambda

or     \vec{r}.[(2+2\lambda)\widehat{i}+(2+5\lambda)\widehat{j}+(3\lambda-3)\widehat{k}] = 7+9\lambda             ..............(1)

where, \lambda is some real number.

Taking \vec{r} = x\widehat{i}+y\widehat{j}+z\widehat{k}, we get

(\vec{x\widehat{i}+y\widehat{j}+z\widehat{k}}).[(2+2\lambda)\widehat{i}+(2+5\lambda)\widehat{j}+(3\lambda-3)\widehat{k}] = 7+9\lambda

or   x(2+2\lambda) + y(2+5\lambda) +z(3\lambda-3) = 7+9\lambda

or   2x+2y-3z-7 + \lambda(2x+5y+3z-9) = 0                       .............(2)

Given that the plane passes through the point (2,1,3), it must satisfy (2), i.e.,

                 (4+2-9-7) + \lambda(4+5+9-9) = 0

or                \lambda = \frac{10}{9}

Putting the values of \lambda in (1), we get

\vec{r}\left [\left ( 2+\frac{20}{9} \right )\widehat{i}+\left ( 2+\frac{50}{9} \right )\widehat{j}+\left ( \frac{10}{3}-3 \right )\widehat{k} \right ] = 7+10

or       \vec{r}\left ( \frac{38}{9}\widehat{i}+\frac{68}{9}\widehat{j}+\frac{1}{3}\widehat{k} \right ) = 17

or      \vec{r}.\left ( 38\widehat{i}+68\widehat{j}+3\widehat{k} \right ) = 153

which is the required vector equation of the plane.

Question:11  Find the equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x – y + z = 0.

Answer:

The equation of the plane through the intersection of the given two planes, x+y+z =1  and  2x+3y+4z =5 is given in Cartesian form as;

(x+y+z-1) +\lambda(2x+3y+4z -5) = 0

or (1+2\lambda)x(1+3\lambda)y+(1+4\lambda)z-(1+5\lambda) = 0             ..................(1)

So, the direction ratios of (1) plane are a_{1},b_{1},c_{1} which are (1+2\lambda),(1+3\lambda),\ and\ (1+4\lambda).

Then, the plane in equation (1) is perpendicular to x-y+z= 0 whose direction ratios a_{2},b_{2},c_{2} are 1,-1,\ and\ 1.

As planes are perpendicular then,

a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2} = 0

we get,

(1+2\lambda) -(1+3\lambda)+(1+4\lambda) = 0

or  1+3\lambda = 0

or  \lambda = -\frac{1}{3}

Then we will substitute the values of \lambda in the equation (1), we get

\frac{1}{3}x-\frac{1}{3}z+\frac{2}{3} = 0

or   x-z+2=0

This is the required equation of the plane.

Question:12 Find the angle between the planes whose vector equations are \overrightarrow{r}.(2\widehat{i}+2\widehat{j}-3\widehat{k})= 5 and \overrightarrow{r}.(3\widehat{i}-3\widehat{j}+5\widehat{k})= 3.          

Answer:

Given two vector equations of plane

\overrightarrow{r}.(2\widehat{i}+2\widehat{j}-3\widehat{k})= 5 and \overrightarrow{r}.(3\widehat{i}-3\widehat{j}+5\widehat{k})= 3.

Here, \vec{n_{1}} = 2\widehat{i}+2\widehat{j}-3\widehat{k}  and  \vec{n_{2}} = 3\widehat{i}-3\widehat{j}+5\widehat{k}

The formula for finding the angle between two planes,

\cos A = \left | \frac{\vec{n_{1}}.\vec{n_{2}}}{|\vec{n_{1}}||\vec{n_{2}}|} \right |                                         .............................(1)

\vec{n_{1}}.\vec{n_{2}} = (2\widehat{i}+2\widehat{j}-3\widehat{k})(3\widehat{i}-3\widehat{j}+5\widehat{k}) = 2(3)+2(-3)-3(5) = -15

|\vec{n_{1}}| =\sqrt{(2)^2+(2)^2+(-3)^2} =\sqrt{17}

and       |\vec{n_{2}}| =\sqrt{(3)^2+(-3)^2+(5)^2} =\sqrt{43}

Now, we can substitute the values in the angle formula (1) to get,

\cos A = \left | \frac{-15}{\sqrt{17}\sqrt{43}} \right |

or  \cos A =\frac{15}{\sqrt{731}}

or  A = \cos^{-1}\left ( \frac{15}{\sqrt{731}} \right )

Question:13(a) In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them. 

7x + 5y + 6z + 30 = 0 and 3x – y – 10z + 4 = 0

Answer:

Two planes

L_{1}:a_{1}x+b_{1}y+c_{1}z = 0 whose direction ratios are a_{1},b_{1},c_{1} and L_{2}:a_{2}x+b_{2}y+c_{2}z = 0 whose direction ratios are a_{2},b_{2},c_{2},

are said to Parallel: 

If, \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}

and Perpendicular:

If, a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2} = 0

And the angle between L_{1}\ and\ L_{2} is given by the relation,

A = \cos^{-1}\left | \frac{a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}}{\sqrt{a_{1}^2+b_{1}^2+c_{1}^2}.\sqrt{a_{2}^2+b_{2}^2+c_{2}^2}} \right |
So, given two planes  7x + 5y + 6z + 30 = 0\ and\ 3x -y - 10z + 4 = 0

Here,

a_{1} = 7,b_{1} = 5, c_{1} = 6   and   a_{2} = 3,b_{2} = -1, c_{2} = -10

So, applying each condition to check:

Parallel check:   \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}

\Rightarrow \frac{a_{1}}{a_{2}} =\frac{7}{3}, \frac{b_{1}}{b_{2}}=\frac{5}{-1},\frac{c_{1}}{c_{2}} = \frac{6}{-10}

Clearly, the given planes are NOT parallel.\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}

Perpendicular check: a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2} = 0

\Rightarrow 7(3)+5(-1)+6(-10) = 21-5-60 = -44 \neq 0.

Clearly, the given planes are NOT perpendicular.

Then find the angle between them,

A = \cos^{-1}\left | \frac{a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}}{\sqrt{a_{1}^2+b_{1}^2+c_{1}^2}.\sqrt{a_{2}^2+b_{2}^2+c_{2}^2}} \right |

     = \cos^{-1}\left | \frac{-44}{\sqrt{7^2+5^2+6^2}.\sqrt{3^2+(-1)^2+(-10)^2}} \right |

     = \cos^{-1}\left | \frac{-44}{\sqrt{110}.\sqrt{110}} \right |

    = \cos^{-1}\left ( \frac{44}{110} \right )

    = \cos^{-1}\left ( \frac{2}{5} \right )

Question:13(b) In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.

 2x + y + 3z – 2 = 0 and x – 2y + 5 = 0

Answer:

Two planes

L_{1}:a_{1}x+b_{1}y+c_{1}z = 0 whose direction ratios are a_{1},b_{1},c_{1} and L_{2}:a_{2}x+b_{2}y+c_{2}z = 0 whose direction ratios are a_{2},b_{2},c_{2},

are said to Parallel: 

If, \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}

and Perpendicular:

If, a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2} = 0

And the angle between L_{1}\ and\ L_{2} is given by the relation,

A = \cos^{-1}\left | \frac{a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}}{\sqrt{a_{1}^2+b_{1}^2+c_{1}^2}.\sqrt{a_{2}^2+b_{2}^2+c_{2}^2}} \right |
So, given two planes  2x + y + 3z -2 = 0\ and\ x -2y + 5 = 0

Here,

a_{1} = 2,b_{1} = 1, c_{1} = 3   and   a_{2} = 1,b_{2} = -2, c_{2} = 0

So, applying each condition to check:

Perpendicular check: a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2} = 0

\Rightarrow 2(1)+1(-2)+3(0) = 2-2+0 = 0.

Thus, the given planes are perpendicular to each other.

Question:13(c)  In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.

 2x – 2y + 4z + 5 = 0 and 3x – 3y + 6z – 1 = 0

Answer:

Two planes

L_{1}:a_{1}x+b_{1}y+c_{1}z = 0 whose direction ratios are a_{1},b_{1},c_{1} and L_{2}:a_{2}x+b_{2}y+c_{2}z = 0 whose direction ratios are a_{2},b_{2},c_{2},

are said to Parallel: 

If, \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}

and Perpendicular:

If, a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2} = 0

And the angle between L_{1}\ and\ L_{2} is given by the relation,

A = \cos^{-1}\left | \frac{a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}}{\sqrt{a_{1}^2+b_{1}^2+c_{1}^2}.\sqrt{a_{2}^2+b_{2}^2+c_{2}^2}} \right |
So, given two planes  2x - 2y + 4z + 5 = 0\ and\ 3x -3y +6z -1 = 0

Here,

a_{1} = 2,b_{1} = -2, c_{1} = 4   and   a_{2} = 3,b_{2} = -3, c_{2} = 6

So, applying each condition to check:

Parallel check:   \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}

\Rightarrow \frac{a_{1}}{a_{2}} =\frac{2}{3}, \frac{b_{1}}{b_{2}}=\frac{-2}{-3}=\frac{2}{3},\ and\ \frac{c_{1}}{c_{2}} = \frac{4}{6}=\frac{2}{3}

Thus, the given planes are parallel as  \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}

Question:13(d) In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.

 2x – y + 3z – 1 = 0 and 2x – y + 3z + 3 = 0

Answer:

Two planes

L_{1}:a_{1}x+b_{1}y+c_{1}z = 0 whose direction ratios are a_{1},b_{1},c_{1} and L_{2}:a_{2}x+b_{2}y+c_{2}z = 0 whose direction ratios are a_{2},b_{2},c_{2},

are said to Parallel: 

If, \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}

and Perpendicular:

If, a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2} = 0

And the angle between L_{1}\ and\ L_{2} is given by the relation,

A = \cos^{-1}\left | \frac{a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}}{\sqrt{a_{1}^2+b_{1}^2+c_{1}^2}.\sqrt{a_{2}^2+b_{2}^2+c_{2}^2}} \right |
So, given two planes  2x - y + 3z -1 = 0\ and\ 2x -y +3z + 3 = 0

Here,

a_{1} = 2,b_{1} = -1, c_{1} = 3   and   a_{2} = 2,b_{2} = -1, c_{2} = 3

So, applying each condition to check:

Parallel check:   \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}

\Rightarrow \frac{a_{1}}{a_{2}} =\frac{2}{2}=1, \frac{b_{1}}{b_{2}}=\frac{-1}{-1} =1,\frac{c_{1}}{c_{2}} = \frac{3}{3} = 1

Therefore \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}

Thus, the given planes are parallel to each other.

Question:13(e) In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.

  4x + 8y + z – 8 = 0 and y + z – 4 = 0

Answer:

Two planes

L_{1}:a_{1}x+b_{1}y+c_{1}z = 0 whose direction ratios are a_{1},b_{1},c_{1} and L_{2}:a_{2}x+b_{2}y+c_{2}z = 0 whose direction ratios are a_{2},b_{2},c_{2},

are said to Parallel: 

If, \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}

and Perpendicular:

If, a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2} = 0

And the angle between L_{1}\ and\ L_{2} is given by the relation,

A = \cos^{-1}\left | \frac{a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}}{\sqrt{a_{1}^2+b_{1}^2+c_{1}^2}.\sqrt{a_{2}^2+b_{2}^2+c_{2}^2}} \right |
So, given two planes  4x + 8y + z -8 = 0\ and\ y + z - 4 = 0

Here,

a_{1} = 4,b_{1} = 8, c_{1} = 1   and   a_{2} = 0,b_{2} = 1, c_{2} = 1

So, applying each condition to check:

Parallel check:   \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}

\Rightarrow \frac{a_{1}}{a_{2}} =\frac{4}{0}, \frac{b_{1}}{b_{2}}=\frac{8}{1},\frac{c_{1}}{c_{2}} = \frac{1}{1}

Clearly, the given planes are NOT parallel as  \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}.

Perpendicular check: a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2} = 0

\Rightarrow 4(0)+8(1)+1(1) =0+8+1 = 9 \neq 0.

Clearly, the given planes are NOT perpendicular.

Then finding the angle between them,

A = \cos^{-1}\left | \frac{a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}}{\sqrt{a_{1}^2+b_{1}^2+c_{1}^2}.\sqrt{a_{2}^2+b_{2}^2+c_{2}^2}} \right |

     = \cos^{-1}\left | \frac{9}{\sqrt{4^2+8^2+1^2}.\sqrt{0^2+1^2+1^2}} \right |

     = \cos^{-1}\left | \frac{9}{\sqrt{81}.\sqrt{2}} \right |

    = \cos^{-1}\left ( \frac{9}{9\sqrt{2}} \right ) = \cos^{-1}\left ( \frac{1}{\sqrt{2}} \right )

   = 45^{\circ}

Question:14 In the following cases, find the distance of each of the given points from the corresponding given plane

POINT

PLANE

a. (0, 0, 0)

3x – 4y + 12 z = 3

b. (3, – 2, 1)

2x – y + 2z + 3 = 0

c. (2, 3, – 5)

x + 2y – 2z = 9

d. (– 6, 0, 0)

2x – 3y + 6z – 2 = 0

 

Answer:

We know that the distance between a point P (x_{1},y_{1},z_{1}) and a plane Ax+By+Cz =D is given by,

d =\left | \frac{Ax_{1}+By_{1}+Cz_{1}-D}{\sqrt{A^2+B^2+C^2}} \right |                                   .......................(1)

So, calculating for each case;

(a) Point (0,0,0) and Plane 3x-4y+12z = 3

Therefore, d =\left | \frac{3(0)-4(0)+12(0)-3}{\sqrt{3^2+(-4)^2+12^2}} \right | = \frac{3}{\sqrt{169}} = \frac{3}{13}

(b) Point (3,-2,1) and Plane 2x-y+2z +3= 0

Therefore, d =\left | \frac{2(3)-(-2)+2(1)+3}{\sqrt{2^2+(-1)^2+2^2}} \right | = \frac{13}{3}

(c) Point (2,3,-5) and Plane x+2y-2z =9

Therefore, d =\left | \frac{2+2(3)-2(-5)-9}{\sqrt{1^2+2^2+(-2)^2}} \right | = \frac{9}{3} = 3

(d) Point (-6,0,0) and Plane 2x-3y+6z -2= 0

Therefore, d =\left | \frac{2(-6)-3(0)+6(0)-2}{\sqrt{2^2+(-3)^2+6^2}} \right | = \frac{-14}{\sqrt{49}} = \frac{14}{7} =2

CBSE NCERT solutions for class 12 maths chapter 11 three dimensional geometry-Miscellaneous Exercise

Question:1 Show that the line joining the origin to the point (2, 1, 1) is perpendicular to the line determined by the points (3, 5, – 1), (4, 3, – 1).

Answer:

We can assume the line joining the origin, be OA where O(0,0,0) and the point A(2,1,1) and PQ be the line joining the points P(3,5,-1) and Q(4,3,-1).

Then the direction ratios of the line OA will be (2-0),(1-0),\ and\ (1-0) = 2,1,1  and that of line PQ will be

(4-3),(3-5),\ and\ (-1+1) = 1,-2,0

So to check whether line OA is perpendicular to line PQ then,

Applying the relation we know,

a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2} = 0

\Rightarrow 2(1)+1(-2)+1(0) = 2-2+0 = 0

Therefore OA is perpendicular to line PQ.

Question:2 If l1 , m1 , n1 and l 2 , m2 , n2 are the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of these are m_{1}n_{2}-m_{2}n_{1}, n_{1}l_{2}-n_{2}l_{1},l_{1}m_{2}-l_{2}m_{1}.

Answer:

Given that l_1,m_1,n_1\ and\ l_2,m_2,n_2 are the direction cosines of two mutually perpendicular lines.

Therefore, we have the relation:

l_{1}l_{2}+m_{1}m_{2}+n_{1}n_{2} = 0                              .........................(1)

l_{1}^2+m_{1}^2+n_{1}^2 =1\ and\ l_{2}^2+m_{2}^2+n_{2}^2 =1         .............(2)

Now, let us assume l,m,n be the new direction cosines of the lines which are perpendicular to the line with direction cosines.l_1,m_1,n_1\ and\ l_2,m_2,n_2

Therefore we have, ll_{1}+mm_{1}+nn_{1} = 0 \and\ ll_{2}+mm_{2}+nn_{2} = 0

Or, \frac{l}{m_{1}n_{2}-m_{2}n_{1}} = \frac{m}{n_{1}l_{2}-n_{2}l_{1}} = \frac{n}{l_{1}m_{2}-l_{2}m_{1}}

\Rightarrow \frac{l^2}{(m_{1}n_{2}-m_{2}n_{1})^2} = \frac{m^2}{(n_{1}l_{2}-n_{2}l_{1})^2} = \frac{n^2}{(l_{1}m_{2}-l_{2}m_{1})^2}

\Rightarrow \frac{l^2+m^2+n^2}{(m_{1}n_{2}-m_{2}n_{1})^2 +(n_{1}l_{2}-n_{2}l_{1})^2+(l_{1}m_{2}-l_{2}m_{1})^2}    ......(3)

So, l,m,n are the direction cosines of the line.

where, l^2+m^2+n^2 =1                                   ........................(4)

Then we know that,

\Rightarrow (l_{1}^2+m_{1}^2+n_{1}^2)(l_{2}^2+m_{2}^2+n_{2}^2) - (l_{1}l_{2}+m_{1}m_{2}+n_{1}n_{2})^2

= (m_{1}n_{2}-m_{2}n_{1})^2 + (n_{1}l_{2}-n_{2}l_{1})^2+(l_{1}m_{2}-l_{2}m_{1})^2

So, from the equation (1) and (2) we have,

(1)(1) -(0) =(m_{1}n_{2}-m_{2}n_{1})^2 + (n_{1}l_{2}-n_{2}l_{1})^2+(l_{1}m_{2}-l_{2}m_{1})^2

Therefore, (m_{1}n_{2}-m_{2}n_{1})^2 + (n_{1}l_{2}-n_{2}l_{1})^2+(l_{1}m_{2}-l_{2}m_{1})^2 =1     ..(5)

Now, we will substitute the values from the equation (4) and (5) in equation (3), to get

\frac{l^2}{(m_{1}n_{2}-m_{2}n_{1})^2} = \frac{m^2}{(n_{1}l_{2}-n_{2}l_{1})^2} = \frac{n^2}{(l_{1}m_{2}-l_{2}m_{1})^2} =1

Therefore we have the direction cosines of the required line as;

l =m_{1}n_{2} - m_{2}n_{1}   

m =n_{1}l_{2} - n_{2}l_{1}

n =l_{1}m_{2} - l_{2}m_{1}

Question:3 Find the angle between the lines whose direction ratios are a, b, c and b – c, c – a, a – b.

Answer:

Given direction ratios a,b,c  and  b-c,\ c-a,\ a-b.

Thus the angle between the lines A is given by;

A = \left | \frac{a(b-c)+b(c-a)+c(a-b)}{\sqrt{a^2+b^2+c^2}.\sqrt{(b-c)^2+(c-a)^2+(a-b)^2}} \right |

\Rightarrow \cos A = 0

\Rightarrow A = \cos^{-1}(0) = 90^{\circ}a

Thus, the angle between the lines is 90^{\circ}

Question:4 Find the equation of a line parallel to x-axis and passing through the origin.

Answer:

Equation of a line parallel to the x-axis and passing through the origin (0,0,0) is itself x-axis.

So, let A be a point on the x-axis.

Therefore, the coordinates of A are given by (a,0,0), where a\epsilon R.

Now, the direction ratios of OA are (a-0) =a,0 , 0

So, the equation of OA is given by,

\frac{x-0}{a} = \frac{y-0}{0} = \frac{z-0}{0} 

or  \Rightarrow \frac{x}{1} = \frac{y}{0} = \frac{z}{0} = a

Thus, the equation of the line parallel to the x-axis and passing through origin is 

\frac{x}{1} = \frac{y}{0} = \frac{z}{0}

Question:5 If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (– 4, 3, – 6) and (2, 9, 2) respectively, then find the angle between the lines AB and CD.

Answer:

Direction ratios of  AB are (4-1),(5-2),(7-3) = 3,3,4

and Direction ratios of CD are (2-(-4)), (9-3), (2-(-6)) = 6,6,8

So, it can be noticed that, \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} =\frac{c_{1}}{c_{2}} = \frac{1}{2}

Therefore, AB is parallel to CD.

Thus, we can easily say the angle between AB and CD which is either 0^{\circ}\ or\ 180^{\circ}.

Question:6 If the lines \frac{x-1}{-3}=\frac{y-2}{k}=\frac{z-3}{2} and \frac{x-1}{3k}=\frac{y-1}{1}=\frac{z-6}{-5}are perpendicular, find the value of k.

Answer:

Given both lines are perpendicular so we have the relation; a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2} = 0

For the two lines whose direction ratios are known,

a_{1},b_{1},c_{1}\ and\ a_{2},b_{2},c_{2} 

We have the direction ratios of the lines, \frac{x-1}{-3}=\frac{y-2}{k}=\frac{z-3}{2}  and  \frac{x-1}{3k}=\frac{y-1}{1}=\frac{z-6}{-5}  are -3,2k,2  and 3k,1,-5 respectively.

Therefore applying the formula,

-3(3k)+2k(1)+2(-5) = 0

\Rightarrow -9k +2k -10 = 0

\Rightarrow7k=-10  or  k= \frac{-10}{7}

\therefore For, k= \frac{-10}{7} the lines are perpendicular.

Question:7 Find the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane \overrightarrow{r}.(\widehat{i}+2\widehat{j}-5\widehat{k})+9=0

Answer:

Given that the plane is passing through the point A (1,2,3) so, the position vector of the point A is \vec{r_{A}} = \widehat{i}+2\widehat{j}+3\widehat{k}  and perpendicular to the plane \overrightarrow{r}.(\widehat{i}+2\widehat{j}-5\widehat{k})+9=0 whose direction ratios are  1,2,\ and\ -5 and the normal vector is \vec{n} = \widehat{i}+2\widehat{j}-5\widehat{k}

So, the equation of a line passing through a point and perpendicular to the given plane is given by,

\vec{l} = \vec{r} + \lambda\vec{n}, where \lambda \epsilon R

\Rightarrow \vec{l} = (\widehat{i}+2\widehat{j}+3\widehat{k}) + \lambda(\widehat{i}+2\widehat{j}-5\widehat{k})

Question:8 Find the equation of the plane passing through (a, b, c) and parallel to the plane \overrightarrow{r}.(\widehat{i}+\widehat{j}+\widehat{k})=2.

Answer:

Given that the plane is passing through (a,b,c) and is parallel to the plane \overrightarrow{r}.(\widehat{i}+\widehat{j}+\widehat{k})=2

So, we have

The position vector of the point A(a,b,c) is, \vec{r_{A}} = a\widehat{i}+b\widehat{j}+c\widehat{k}

and any plane which is parallel to the plane, \overrightarrow{r}.(\widehat{i}+\widehat{j}+\widehat{k})=2 is of the form,

\overrightarrow{r}.(\widehat{i}+\widehat{j}+\widehat{k})=\lambda.                      .......................(1)

Therefore the equation we get,

( a\widehat{i}+b\widehat{j}+c\widehat{k}).(\widehat{i}+\widehat{j}+\widehat{k})=\lambda

Or, a+b+c = \lambda

So, now substituting the value of \lambda = a+b+c in equation (1), we get

\overrightarrow{r}.(\widehat{i}+\widehat{j}+\widehat{k})=a+b+c            .................(2)

So, this is the required equation of the plane .

Now, substituting \vec{r} = x\widehat{i}+y\widehat{j}+z\widehat{k} in equation (2), we get

(x\widehat{i}+y\widehat{j}+z\widehat{k}).(\widehat{i}+\widehat{j}+\widehat{k})=a+b+c

Or, x+y+z = a+b+c

Question:9 Find the shortest distance between lines \overrightarrow{r}=6\widehat{i}+2\widehat{j}+2\widehat{k}+\lambda (\widehat{i}-2\widehat{j}-2\widehat{k}) and \overrightarrow{r}=-4\widehat{i}-\widehat{k}+\mu(3\widehat{i}-2\widehat{j}-2\widehat{k}).

Answer:

Given lines are; 

\overrightarrow{r}=6\widehat{i}+2\widehat{j}+2\widehat{k}+\lambda (\widehat{i}-2\widehat{j}-2\widehat{k})  and

\overrightarrow{r}=-4\widehat{i}-\widehat{k}+\mu(3\widehat{i}-2\widehat{j}-2\widehat{k})

So, we can find the shortest distance between two lines \vec{r} = \vec{a_{1}}+\lambda \vec{b_{1}} and \vec{r} = \vec{a_{1}}+\lambda \vec{b_{1}} by the formula,

d = \left | \frac{(\vec{b_{1}}\times\vec{b_{2}}).(\vec{a_{2}}-\vec{a_{1}})}{\left | \vec{b_{1}}\times\vec{b_{2}} \right |} \right |                                               ...........................(1)

Now, we have from the comparisons of the given equations of lines.

\vec{a_{1}} = 6\widehat{i}+2\widehat{j}+2\widehat{k}     \vec{b_{1}} = \widehat{i}-2\widehat{j}+2\widehat{k}

\vec{a_{2}} = -4\widehat{i}-\widehat{k}             \vec{b_{2}} = 3\widehat{i}-2\widehat{j}-2\widehat{k}

So, \vec{a_{2}} -\vec{a_{1}} = (-4\widehat{i}-\widehat{k}) -(6\widehat{i}+2\widehat{j}+2\widehat{k}) = -10\widehat{i}-2\widehat{j}-3\widehat{k}

and \Rightarrow \vec{b_{1}}\times \vec{b_{2}} = \begin{vmatrix} \widehat{i} &\widehat{j} &\widehat{k} \\ 1 &-2 &2 \\ 3& -2 &-2 \end{vmatrix} = (4+4)\widehat{i}-(-2-6)\widehat{j}+(-2+6)\widehat{k}

=8\widehat{i}+8\widehat{j}+4\widehat{k}

\therefore \left | \vec{b_{1}}\times \vec{b_{2}} \right | = \sqrt{8^2+8^2+4^2} =12

(\vec{b_{1}}\times \vec{b_{2}}).(\vec{a_{2}}-\vec{a_{1}}) = (8\widehat{i}+8\widehat{j}+4\widehat{k}).(-10\widehat{i}-2\widehat{j}-3\widehat{k}) = -80-16-12 =-108Now, substituting all values in equation (3) we get,

d = | \frac{-108}{12}| = 9

Hence the shortest distance between the two given lines is 9 units.

Question:10 Find the coordinates of the point where the line through (5, 1, 6) and (3, 4,1) crosses the YZ-plane.

Answer:

We know that the equation of the line that passes through the points (x_{1},y_{1},z_{1})and (x_{2},y_{2},z_{2}) is given by the relation;

\frac{x-x_{1}}{x_{2}-x_{1}} = \frac{y-y_{1}}{y_{2}-y_{1}} = \frac{z-z_{1}}{z_{2}-z_{1}}

and the line passing through the points, \frac{x-5}{3-5} =\frac{y-1}{4-1} = \frac{z-6}{1-6}

\implies \frac{x-5}{-2} = \frac{y-1}{3} = \frac{z-6}{-5} = k\ (say)

\implies x=5-2k,\ y=3k+1,\ z=6-5k  

And any point on the line is of the form (5-2k,3k+1,6-5k).

So, the equation of the YZ plane is x=0

Since the line passes through YZ- plane,

we have then,

5-2k = 0

\Rightarrow k =\frac{5}{2}

or  3k+1 = 3(\frac{5}{2})+1 = \frac{17}{2}   and  6-5k= 6-5(\frac{5}{2}) = \frac{-13}{2}

So, therefore the required point is \left ( 0,\frac{17}{2},\frac{-13}{2} \right )

Question:11 Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the ZX-plane.

Answer:

We know that the equation of the line that passes through the points (x_{1},y_{1},z_{1})and (x_{2},y_{2},z_{2}) is given by the relation;

\frac{x-x_{1}}{x_{2}-x_{1}} = \frac{y-y_{1}}{y_{2}-y_{1}} = \frac{z-z_{1}}{z_{2}-z_{1}}

and the line passing through the points, \frac{x-5}{3-5} =\frac{y-1}{4-1} = \frac{z-6}{1-6}

\implies \frac{x-5}{-2} = \frac{y-1}{3} = \frac{z-6}{-5} = k\ (say)

\implies x=5-2k,\ y=3k+1,\ z=6-5k  

And any point on the line is of the form (5-2k,3k+1,6-5k).

So, the equation of ZX plane is y=0

Since the line passes through YZ- plane,

we have then,

3k+1 = 0

\Rightarrow k =-\frac{1}{3}

or  5-2k = 5-2\left ( -\frac{1}{3} \right ) = \frac{17}{3}   and  6-5k= 6-5(\frac{-1}{3}) = \frac{23}{3}

So, therefore the required point is \left ( \frac{17}{3},0,\frac{23}{3} \right )