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The length of the perpendicular drawn from the point (2,1,4) to the plane

containing the lines $\vec{r}=(\hat{i}+\hat{j})+\lambda (\hat{i}+2\hat{j}-\hat{k})$  and

$\vec{r}=(\hat{i}+\hat{j})+\mu (-\hat{i}+\hat{j}-2\hat{k})$ is :

• Option 1)

3

• Option 2)

$\frac{1}{3}$

• Option 3)

$\sqrt{3}$

• Option 4)

$\frac{1}{\sqrt{3}}$

Perpendicula vector to the plane   Equation of the plane  Equation =>    Option 1) 3 Option 2) Option 3) Option 4)

A plane which bisects the angle between the two given planes $2x-y+2z-4=0$ and $x-2y+2z-2=0$, passes through the points:

• Option 1)

(1,-4,1)

• Option 2)

(1,4,-1)

• Option 3)

(2,4,1)

• Option 4)

(2,-4,1)

=> Required equation of planes are  and Now, substituting the options we will get , the point ( 2, -4 , 1 ).Option 1)(1,-4,1)Option 2)(1,4,-1)Option 3)(2,4,1)  Option 4)(2,-4,1)

If the line $\frac{x-2}{3}-\frac{y+1}{2}=\frac{z-1}{-1}$ intersects the plane $2x+3y-z+13=0$ at a point P and the plane $3x+y+4z=16$ at a point Q, then PQ is equal to :

• Option 1)

$2\sqrt{7}$

• Option 2)

$\sqrt{14}$

• Option 3)

$2\sqrt{14}$

• Option 4)

$14$

Intersection with plane  Intersection with plane                             Option 1)     Option 2)        Option 3)          Option 4)

If the plane $2x-y+2z+3=0$ has the distances $\frac{1}{3}$ and $\frac{2}{3}$ units

from the planes $4x-2y+4z+\lambda=0$ and $2x-y+2z+\mu=0$,

respectively, then the maximum value of $\lambda+\mu$ is equal to :

• Option 1)

9

• Option 2)

15

• Option 3)

5

• Option 4)

13

Given plane    unit distance from the plane  =>  Now ,   unit distance from the plane  =>   maximum value of  = 8 + 5 = 13 So, correct option is (4) Option 1) 9 Option 2) 15 Option 3) 5 Option 4) 13

A perpendicular is drawn from a point on the line $\frac{x-1}{2}=\frac{y+1}{-1}=\frac{z}{1}$

to the plane $x+y+z=3$ such that the foot of the perpendicular Q also

lies on the plane $x-y+z=3$. Then the co-ordinates of Q are :

• Option 1)

(1, 0, 2)

• Option 2)

(2, 0 , 1)

• Option 3)

( - 1, 0 , 4)

• Option 4)

(4, 0 , -1)

So, =>  =>  =>    point is  It lies on equation                                                     So, point Q                  Q ( 2 , 0 , 1 ) So, option (2) is correct. Option 1) (1, 0, 2) Option 2) (2, 0 , 1) Option 3) ( - 1, 0 , 4)   Option 4) (4, 0 , -1)

If the length of the perpendicular from the point $(\beta ,0,\beta )$

$(\beta \neq0)$ to the line, $\frac{x}{1}=\frac{y-1}{0}=\frac{z+1}{-1}$ is $\sqrt{\frac{3}{2}}$ ,

then $\beta$ is equal to :

• Option 1)

1

• Option 2)

2

• Option 3)

-1

• Option 4)

-2

Let point  given that length of perpendicular distance    from P to line is . Direction ratio of  PR is perpendicular to the line               Correct option is (3)        Option 1) 1 Option 2) 2 Option 3) -1 Option 4) -2

Let $A(3,0,-1),B(2,10,6)\: \: and\: \: C(1,2,1)$ be the

vertices of a triangle and M be the midpoint of AC. If G divides

BM in the ratio , 2:1, then $cos(\angle GOA)$ ( O being the origin)

is equal to :

• Option 1)

$\frac{1}{2\sqrt{15}}$

• Option 2)

$\frac{1}{\sqrt{15}}$

• Option 3)

$\frac{1}{6\sqrt{10}}$

• Option 4)

$\frac{1}{\sqrt{30}}$

Given that M is mid point of AC and G divides BM in the ratio 2:1  G is centroid of ABC So, correct option is (2) Option 1) Option 2) Option 3) Option 4)

If $Q(0,-1,-3)$ is the image of the point P in the plane $3x-y+4z=2$

and R is the point $(3,-1,-2)$, then the area ( in sq. units) of $\bigtriangleup PQR$  is :

• Option 1)

$2\sqrt{13}$

• Option 2)

$\frac{\sqrt{91}}{4}$

• Option 3)

$\frac{\sqrt{91}}{2}$

• Option 4)

$\frac{\sqrt{65}}{2}$

Given equation of plane is 3x-y+4z=2  Q ( 0 , -1, -3 ) is image of P  Point R is lie on plane   is right angled triangle So, Area of  correct option (3)  Option 1) Option 2) Option 3) Option 4)

If a unit vector  $\vec{a}$ make angles $\pi/3$    with  $\hat{i},\pi/4\:\:\:with\:\:\:\hat{j}\:\:and\:\:\theta\:\epsilon (0,\pi)\:\:with\:\:\hat{k}$ then a value of $\theta$  is :

• Option 1)

$\frac{5\pi}{6}$

• Option 2)

$\frac{\pi}{4}$

• Option 3)

• Option 4)

$\frac{2\pi}{3}$

Option 1) Option 2) Option 3) Option 4)

Let P be the plane , which contains the line of intersection of the planes ,$x+y+z-6=0\:\:and\:\:2x+3y+z+5=0$,  and it is perpendicular to the xy-plane.  Then the distance of the point $(0,0,256)$ from P is equal to :

• Option 1)

$17/\sqrt{5}$

• Option 2)

$63\sqrt{5}$

• Option 3)

$205 \sqrt{5}$

• Option 4)

$11/\sqrt{5}$

equation of plane which contains intersection of the planes  is  since it is   to xy plane so     distant of point  from this plane  =      Option 1) Option 2) Option 3) Option 4)

The vertices B and C of a $\Delta ABC$   lie on the line , $\frac{x+2}{3}=\frac{y-1}{0}=\frac{z}{4}$   such that $BC=5$  units. Then the area ( in sq. units ) of this triangle , given that the point $A(1,-1,2)$, is :

• Option 1)

$5\sqrt{17}$

• Option 2)

$2\sqrt{34}$

• Option 3)

$6$

• Option 4)

$\sqrt{34}$

Since Ad is perpendicular to BC       Option 1) Option 2) Option 3) Option 4)

A plane passing through the points $\left ( 0,-1,0 \right )$ and $\left ( 0,0,1 \right )$ and making an angle $\frac{\pi }{4}$ with the plane $y-z+5=0,$ also passes through the point :

• Option 1)

$\left ( -\sqrt{2},1,-4 \right )$

• Option 2)

$\left ( \sqrt{2},-1,4 \right )$

• Option 3)

$\left (- \sqrt{2},-1,-4 \right )$

• Option 4)

$\left ( \sqrt{2},1,4 \right )$

Let eq. of plane is       as given it passes through                equation of plane is Also given that this plane make  with the plane        Now eq.of plane become     Satisfies   Option 1)               Option 2)   Option 3)      Option 4)

If the line,$\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-2}{4}$ meets the plane,$x+2y+3z=15$ at a point $P,$ then the distance of $P$ from the origin is :

• Option 1)

$\sqrt{5}/2$

• Option 2)

$2\sqrt{5}$

• Option 3)

$9/2$

• Option 4)

$7/2$

Given Line     any  point on line is  This point will lie on plane                           Distance from origin=  Option 1)               Option 2) Option 3) Option 4)

If a point $R\left ( 4,y,z \right )$ lies on the line segment joining the points $P\left ( 2,-3,4 \right )$ and $Q\left ( 8,0,10 \right ),$ then distance of R from the origin is :

• Option 1)

$2\sqrt{14}$

• Option 2)

$2\sqrt{21}$

• Option 3)

$6$

• Option 4)

$\sqrt{53}$

Direction Ratio of PQ=  Direction ratio of PR=  P,R and Q lie on same line  Point  Distance from origin  Option 1)Option 2)Option 3)  Option 4)

The vector equation of the plane through the line of interectionb of the planes $x+y+z=1$ and $2x+3y+4z=5$ which is perpendicular to the plane $x-y+z=0$ is :

• Option 1)

$\vec{r}\times \left ( \vec{i}-\vec{k} \right )+2=0$

• Option 2)

$\vec{r}\cdot \left ( \vec{i}-\vec{k} \right )-2=0$

• Option 3)

$\vec{r}\times \left ( \vec{i}+\vec{k} \right )+2=0$

• Option 4)

$\vec{r}\cdot \left ( \vec{i}-\vec{k} \right )+2=0$

Equation of plane passing through line of interaction of plane  and  is  plane (ii) and  is perpendicular. so,  Equation of plane is  vector equation.                 Option 1) Option 2) Option 3)   Option 4)

The equation of a plane containg the line of intersection of the planes $2x-y-4=0$ and $y+2z-4=0$ and passing through the point $(1,1,0)$ is :

• Option 1)

$x+3y+z=4$

• Option 2)

$x-3y-2z=-2$

• Option 3)

$x-y-z=0$

• Option 4)

$2x-z=2$

Equation of plane passing through intersection of and if passes through Option 1) Option 2) Option 3)   Option 4)

The length of the perpendicular from the point $(2,-1,4)$ on the straight line, $\frac{x+3}{10}=\frac{y-2}{-7}=\frac{z}{1}$ is :

• Option 1)

greater than $2$ but less than $3$

• Option 2)

greater than $4$

• Option 3)

less than $2$

• Option 4)

greater than $3$ but less than $4$

Length of perpendicular from point to line ? Any point Q on line DR's of                        PQ is perpendicular to given line                                                 Option 1) greater than but less than   Option 2) greater than Option 3) less than   Option 4) greater than but less than

If the point $(2,\alpha ,\beta )$ lies on the plane which passes through the points (3,4,2) and (7,0,6) and is perpendicular to the plane $2x-5y=15$ ,

then $2\alpha -3\beta$ is equal to :

• Option 1)

17

• Option 2)

5

• Option 3)

7

• Option 4)

12

Cartesian equation of plane passing through a given point and normal to a given vector -   - wherein Putting in We get     Conversion of equation in normal form (vector form ) - The equation is converted in normal by (i)     (ii)    Making RHS position      (iii)    Dividing by (iv)     We get -   Normal Vector of plane  Equation of plane is  Option 1)17Option 2)5Option 3)7Option 4)12

The direction ratios of normal to the plane through the points (0,-1,0) and (0,0,1) and making an angle $\frac{\pi}{4}$  with the plane $y-z+5=0$ are :

• Option 1)

$\sqrt2,1,-1$

• Option 2)

$2,-1,1$

• Option 3)

$2,\sqrt2,-\sqrt2$

• Option 4)

$2\sqrt3,1,-1$

Cartesian equation of plane passing through a given point and normal to a given vector -   - wherein Putting in We get     Angle between two planes (Cartesian form) - Let the two planes be then the angle between them is defined as the angle between their normals -     Equation of line passing through  &   is    Now,  So direction ratios :  So answer are (1) and (3)Option...

Two lines $\frac{x-3}{1}=\frac{y+1}{3}=\frac{z-6}{-1}$ and  $\frac{x+5}{7}=\frac{y-2}{-6}=\frac{z-3}{4}$  intersect at the point R.

The reflection of R in the xy-plane has coordinates:

• Option 1)

(2,-4,7)

• Option 2)

(2,4,7)

• Option 3)

(-2,4,7)

• Option 4)

(2,-4,-7)

Image of a point - Let be the image of point in the plane will be given by the formula  - Point    on  line 1   Point    on  line 2   At point R     =  ....................(1) ..................(2) ....................(3) Solving any 2 equations out of (1),(2) and (3) point R (2 , -4 , 7 ) Reflection of R in xy plane is ( 2, -4, -7 ).  Option 1)(2,-4,7)Option 2)(2,4,7)Option...
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