# 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 . and Direction cosines of a line joining two points are , where If l, m, n are the direction cosines and a, b, c are the direction of a line then-

## 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

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

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

So, we have

Therefore the direction cosines of the lines are .

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

Let the common angle made is  with each coordinate axes.

Therefore, we can write;

And as we know the relation;

or

Thus the direction cosines of the line are

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

Line having direction ratio -18 has direction cosine:

Line having direction ratio 12 has direction cosine:

Line having direction ratio -4 has direction cosine:

Thus, the direction cosines are .

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  is given by

The direction ratios of AB are  i.e.,

The direction ratios of BC are  i.e., .

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

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.

Given vertices of the triangle   (3, 5, – 4), (– 1, 1, 2) and (– 5, – 5, – 2). Finding each side direction ratios;

Direction ratios of side AB are  i.e.,

Therefore its direction cosines values are;

SImilarly for side BC;

Direction ratios of side BC are  i.e.,

Therefore its direction cosines values are;

Direction ratios of side CA are  i.e.,

Therefore its direction cosines values are;

## Question:1 Show that the three lines with direction cosines

are mutually perpendicular.

GIven direction cosines of the three lines;

And we know that two lines with direction cosines    and   are perpendicular to each other, if

Hence we will check each pair of lines:

Lines ;

the lines  are perpendicular.

Lines ;

the lines  are perpendicular.

Lines ;

the lines  are perpendicular.

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

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

or

Direction ratios of CD are

or  .

Now, lines AB and CD will be perpendicular to each other if

Therefore, AB and CD are perpendicular to each other.

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

or

Direction ratios of CD are

or  .

Now, lines AB and CD will be parallel to each other if

Therefore we have now;

Hence we can say that AB is parallel to CD.

It is given that the line is passing through A (1, 2, 3) and is parallel to the vector

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

, where  is a constant.

So, we have now,

Thus the required equation of the line.

Given that the line is passing through the point with position vector  and is in the direction of the line .

And we know the equation of the line which passes through the point with the position vector and parallel to the vector  is given by the equation,

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

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

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

Given a line which passes through the point (– 2, 4, – 5) and is parallel to the line given by the ;

The direction ratios of the line,  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  and with direction ratios a, b, c is written by: .

Therefore we have the equation of the required line:

or

The required line equation.

Given the Cartesian equation of the line;

Here the given line is passing through the point

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

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, .

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

As we know that the line passing through the position vector  and in the direction of the vector  is given by the relation,

So, we get the equation.

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

GIven that the line is passing through the  and

Thus the required line passes through the origin.

its position vector is given by,

So, the direction ratios of the line through   and  are,

The line is parallel to the vector given by the equation,

Therefore the equation of the line passing through the point with position vector  and parallel to  is given by;

Now, the equation of the line through the point  and the direction ratios a, b, c is given by;

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

OR

Let the line passing through the points  and  is AB;

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

Then direction ratios of PQ are given by,

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

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

So, the equation of AB in Cartesian form is;

or

(i)      and

To find the angle A between the pair of lines  we have the formula;

We have two lines :

and

The given lines are parallel to the vectors ;

where    and    respectively,

Then we have

and

Therefore we have;

or

(ii)       and

To find the angle A between the pair of lines  we have the formula;

We have two lines :

and

The given lines are parallel to the vectors ;

where    and    respectively,

Then we have

and

Therefore we have;

or

(i)      and

Given lines are;

and

So, we two vectors  which are parallel to the pair of above lines respectively.

and

To find the angle A between the pair of lines  we have the formula;

Then we have

and

Therefore we have;

or

(ii)      and

Given lines are;

and

So, we two vectors  which are parallel to the pair of above lines respectively.

and

To find the angle A between the pair of lines  we have the formula;

Then we have

and

Therefore we have;

or

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

and

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

and      respectively..

Two lines with direction ratios  and   are perpendicular to each other if,

Thus, the value of p is .

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

and

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

and      respectively..

Two lines with direction ratios  and   are perpendicular to each other if,

Therefore the two lines are perpendicular to each other.

Question:14 Find the shortest distance between the lines

So given equation of lines;

and   in the vector form.

Now, we can find the shortest distance between the lines  and , is given by the formula,

Now comparing the values from the equation, we obtain

Then calculating

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

Therefore, the shortest distance between the two lines is  units.

Question:15 Find the shortest distance between the lines

and

We have given two lines:

and

Calculating the shortest distance between the two lines,

and

by the formula

Now, comparing the given equations, we obtain

Then calculating determinant

Now calculating the denominator,

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

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

units.

Given two equations of line

in the vector form.

So, we will apply the distance formula  for knowing the distance between two lines    and

After comparing the given equations, we obtain

Then calculating the determinant value numerator.

That implies,

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

Therefore, is the shortest distance between the two given lines.

and

Given two equations of the line

in the vector form.

So, we will apply the distance formula  for knowing the distance between two lines    and

After comparing the given equations, we obtain

Then calculating the determinant value numerator.

That implies,

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

Therefore, units are the shortest distance between the two given lines.

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

z = 2

Equation of plane Z=2, i.e.

The direction ratio of normal is 0,0,1

Divide equation  by 1 from both side

We get,

Hence, direction cosins are 0,0,1.

The distance of the plane from the origin is 2.

x + y + z = 1

Given the equation of the plane is  or we can write

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

Therefore

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

So, this is the form of   the plane, where  are the direction cosines of normal to the plane and d is the distance of the perpendicular drawn from the origin.

The direction cosines of the given line are  and the distance of the plane from the origin is  units.

2x + 3y - z = 5

Given the equation of plane is

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

Therefore

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

So, this is the form of   the plane, where  are the direction cosines of normal to the plane and d is the distance of the perpendicular drawn from the origin.

The direction cosines of the given line are  and the distance of the plane from the origin is  units.

5y + 8 = 0

Given the equation of plane is   or we can write

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

Therefore

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

So, this is the form of   the plane, where  are the direction cosines of normal to the plane and d is the distance of the perpendicular drawn from the origin.

The direction cosines of the given line are  and the distance of the plane from the origin is  units.

We have given the distance between the plane and origin equal to 7 units and normal to the vector .

So, it is known that the equation of the plane with position vector  is given by, the relation,

, where d is the distance of the plane from the origin.

Calculating ;

is the vector equation of the required plane.

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

Given the equation of the plane

So we have to find the Cartesian equation,

Any point  on this plane will satisfy the equation and its position vector given by,

Hence we have,

Or,

Therefore this is the required Cartesian equation of the plane.

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

Given the equation of plane

So we have to find the Cartesian equation,

Any point  on this plane will satisfy the equation and its position vector given by,

Hence we have,

Or,

Therefore this is the required Cartesian equation of the plane.

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

Given the equation of plane

So we have to find the Cartesian equation,

Any point  on this plane will satisfy the equation and its position vector given by,

Hence we have,

Or,

Therefore this is the required Cartesian equation of the plane.

2 x + 3y + 4 z - 12 = 0

Let the coordinates of the foot of perpendicular P from the origin to the plane be

Given a plane equation ,

Or,

The direction ratios of the normal of the plane are .

Therefore

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

This equation is similar to  where,  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

The coordinates of the foot of the perpendicular are;

or

3y + 4z - 6 =  0

Let the coordinates of the foot of perpendicular P from the origin to the plane be

Given a plane equation ,

Or,

The direction ratios of the normal of the plane are .

Therefore

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

This equation is similar to  where,  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

The coordinates of the foot of the perpendicular are;

or

x + y + z = 1

Let the coordinates of the foot of perpendicular P from the origin to the plane be

Given plane equation .

The direction ratios of the normal of the plane are .

Therefore

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

This equation is similar to  where,  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

The coordinates of the foot of the perpendicular are;

or  ..

5y + 8 = 0

Let the coordinates of the foot of perpendicular P from the origin to the plane be

Given plane equation .

or written as

The direction ratios of the normal of the plane are .

Therefore

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

This equation is similar to  where,  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

The coordinates of the foot of the perpendicular are;

or  .

Given the point  and the normal vector  which is perpendicular to the plane is

The position vector of point A is

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

Or

where  is the position vector of any arbitrary point  in the plane.

Therefore, the equation we get,

or

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

that passes through the point (1,4, 6) and the normal vector to the plane is .

Given the point  and the normal vector  which is perpendicular to the plane is

The position vector of point A is

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

Or

where  is the position vector of any arbitrary point  in the plane.

Therefore, the equation we get,

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

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

The equation of the plane which passes through the three points   is given by;

Determinant method,

Or,

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.

(1, 1, 0), (1, 2, 1), (– 2, 2, – 1)

The equation of the plane which passes through the three points   is given by;

Determinant method,

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,

After substituting the values in the determinant we get,

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

Given plane

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,

So, as we know that from the equation of a plane in intercept form, 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.

.

Hence the intercepts are .

Given that the plane is parallel to the ZOX plane.

So, we have the equation of plane ZOX as  .

And an intercept of 3 on the y-axis

Intercept form of a plane given by;

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, .

Equation of the plane required is .

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

Can be written in the form of; , where

So, the plane passes through the point , will satisfy the above equation.

That implies

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

is the required equation of the plane.

Here  and

and   and

Hence, using the relation , we get

or                  ..............(1)

where,  is some real number.

Taking , we get

or

or                          .............(2)

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

or

Putting the values of  in (1), we get

or

or

which is the required vector equation of the plane.

The equation of the plane through the intersection of the given two planes,   and   is given in Cartesian form as;

or              ..................(1)

So, the direction ratios of (1) plane are  which are .

Then, the plane in equation (1) is perpendicular to  whose direction ratios  are .

As planes are perpendicular then,

we get,

or

or

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

or

This is the required equation of the plane.

Given two vector equations of plane

and .

Here,   and

The formula for finding the angle between two planes,

.............................(1)

and

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

or

or

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

Two planes

whose direction ratios are  and  whose direction ratios are ,

are said to Parallel:

If,

and Perpendicular:

If,

And the angle between  is given by the relation,

So, given two planes

Here,

and

So, applying each condition to check:

Parallel check:

Clearly, the given planes are NOT parallel.

Perpendicular check:

.

Clearly, the given planes are NOT perpendicular.

Then find the angle between them,

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

Two planes

whose direction ratios are  and  whose direction ratios are ,

are said to Parallel:

If,

and Perpendicular:

If,

And the angle between  is given by the relation,

So, given two planes

Here,

and

So, applying each condition to check:

Perpendicular check:

.

Thus, the given planes are perpendicular to each other.

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

Two planes

whose direction ratios are  and  whose direction ratios are ,

are said to Parallel:

If,

and Perpendicular:

If,

And the angle between  is given by the relation,

So, given two planes

Here,

and

So, applying each condition to check:

Parallel check:

Thus, the given planes are parallel as

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

Two planes

whose direction ratios are  and  whose direction ratios are ,

are said to Parallel:

If,

and Perpendicular:

If,

And the angle between  is given by the relation,

So, given two planes

Here,

and

So, applying each condition to check:

Parallel check:

Therefore

Thus, the given planes are parallel to each other.

4x + 8y + z – 8 = 0 and y + z – 4 = 0

Two planes

whose direction ratios are  and  whose direction ratios are ,

are said to Parallel:

If,

and Perpendicular:

If,

And the angle between  is given by the relation,

So, given two planes

Here,

and

So, applying each condition to check:

Parallel check:

Clearly, the given planes are NOT parallel as  .

Perpendicular check:

.

Clearly, the given planes are NOT perpendicular.

Then finding the angle between them,

 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

We know that the distance between a point  and a plane  is given by,

.......................(1)

So, calculating for each case;

(a) Point  and Plane

Therefore,

(b) Point  and Plane

Therefore,

(c) Point  and Plane

Therefore,

(d) Point  and Plane

Therefore,

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

We can assume the line joining the origin, be OA where  and the point  and PQ be the line joining the points  and .

Then the direction ratios of the line OA will be   and that of line PQ will be

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

Applying the relation we know,

Therefore OA is perpendicular to line PQ.

Given that  are the direction cosines of two mutually perpendicular lines.

Therefore, we have the relation:

.........................(1)

.............(2)

Now, let us assume  be the new direction cosines of the lines which are perpendicular to the line with direction cosines.

Therefore we have,

Or,

......(3)

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

where,                                    ........................(4)

Then we know that,

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

Therefore,      ..(5)

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

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

Given direction ratios   and  .

Thus the angle between the lines A is given by;

a

Thus, the angle between the lines is

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

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

Therefore, the coordinates of A are given by , where .

Now, the direction ratios of OA are

So, the equation of OA is given by,

or

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

Direction ratios of  AB are

and Direction ratios of CD are

So, it can be noticed that,

Therefore, AB is parallel to CD.

Thus, we can easily say the angle between AB and CD which is either .

Given both lines are perpendicular so we have the relation;

For the two lines whose direction ratios are known,

We have the direction ratios of the lines,   and    are   and  respectively.

Therefore applying the formula,

or

For,  the lines are perpendicular.

Given that the plane is passing through the point  so, the position vector of the point A is   and perpendicular to the plane  whose direction ratios are   and the normal vector is

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

, where

Given that the plane is passing through  and is parallel to the plane

So, we have

The position vector of the point  is,

and any plane which is parallel to the plane,  is of the form,

.                      .......................(1)

Therefore the equation we get,

Or,

So, now substituting the value of  in equation (1), we get

.................(2)

So, this is the required equation of the plane .

Now, substituting  in equation (2), we get

Or,

Given lines are;

and

So, we can find the shortest distance between two lines  and  by the formula,

...........................(1)

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

So,

and

Now, substituting all values in equation (3) we get,

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

We know that the equation of the line that passes through the points and  is given by the relation;

and the line passing through the points,

And any point on the line is of the form .

So, the equation of the YZ plane is

Since the line passes through YZ- plane,

we have then,

or     and

So, therefore the required point is

We know that the equation of the line that passes through the points and  is given by the relation;

and the line passing through the points,

And any point on the line is of the form .

So, the equation of ZX plane is

Since the line passes through YZ- plane,

we have then,

or     and

So, therefore the required point is

We know that the equation of the line that passes through the points and  is given by the relation;

and the line passing through the points, .

And any point on the line is of the form.

This point lies on the plane,

or  .

Hence, the coordinates of the required point are   or .

Given

two planes x + 2y + 3z = 5 and 3x + 3y + z = 0.

the normal vectors  of these plane are

Since the normal vector of the required plane is perpendicular to the normal vector of given planes, the required plane's normal vector will be :

Now, as we know

the equation of a plane in vector form is :

Now Since this plane passes through the point (-1,3,2)

Hence the equation of the plane is

Given that the points  and  are equidistant from the plane

So we can write the position vector through the point  is

Similarly, the position vector through the point  is

The equation of the given plane is

and We know that the perpendicular distance between a point whose position vector is   and the plane,   and

Therefore, the distance between the point  and the given plane is

nbsp;                      .........................(1)

Similarly, the distance between the point , and the given plane is

.........................(2)

And it is given that the distance between the required plane and the points,   and     is equal.

therefore we have,

or   or

So, the given planes are:

and

The equation of any plane passing through the line of intersection of these planes is

..............(1)

Its direction ratios are   and   = 0

The required plane is parallel to the x-axis.

Therefore, its normal is perpendicular to the x-axis.

The direction ratios of the x-axis are 1,0, and 0.

Substituting  in equation (1), we obtain

So, the Cartesian equation is

We have the coordinates of the points   and   respectively.

Therefore, the direction ratios of OP are

And we know that the equation of the plane passing through the point  is

where a,b,c are the direction ratios of normal.

Here, the direction ratios of normal are  and  and the point P is .

Thus, the equation of the required plane is

The equation of the plane passing through the line of intersection of the given plane in

,,,,,,,,,,,,,(1)

The plane in equation (1) is perpendicular to the plane,

Therefore

Substituting  in equation (1), we obtain

.......................(4)

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

The Cartesian equation of this plane can be obtained by substituting  in equation (1).

Given,

Equation of a line :

Equation of the plane

Let's first find out the point of intersection of line and plane.

putting the value of  into the equation of a plane from the equation from line

Now, from the equation, any  point p  in line is

So the point of intersection is

SO, Now,

The distance between the points (-1,-5,-10) and (2,-1,2) is

Hence the required distance is 13.

Given

A point through which line passes

two plane

And

it can be seen that normals of the planes are

since the line is parallel to both planes, its parallel vector will be perpendicular to normals of both planes.

So, a vector perpendicular to both these normal vector is

Now a line which passes through  and parallels to  is

So the required line is

and