3D Geometry — Page 18 | JEE Mathematics

Q171

Let

P Q R

be a triangle with

R(-1,4,2)

. Suppose

M(2,1,2)

is the mid point of

\mathrm{PQ}

. The distance of the centroid of

\triangle \mathrm{PQR}

from the point of intersection of the lines

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

and

\dfrac{x-1}{1}=\dfrac{y+3}{-3}=\dfrac{z+1}{1}

is

A 69

B

\sqrt{99}

C

\sqrt{69}

D 9

Correct Answer

Option C

Solution

Centroid

G

divides MR in

1: 2

\mathrm{G}(1,2,2)

Point of intersection

A

of given lines is

(2,-6,0)

\mathrm{AG}=\sqrt{69}

Q172

Let the plane P:

4 x-y+z=10

be rotated by an angle

\dfrac{\pi}{2}

about its line of intersection with the plane

x+y-z=4

. If

\alpha

is the distance of the point

(2,3,-4)

from the new position of the plane

\mathrm{P}

, then

35 \alpha

is equal to :

A 126

B 105

C 85

D 90

Correct Answer

Option A

Solution

Equation of plane after rotation :

\begin{aligned} & (4 x-y+z-10)+\lambda(x+y-z-y)=0 \\\\ \Rightarrow & (4+\lambda) x+y(\lambda-1)+z(1-\lambda)-4 \lambda-10=0 \\\\ & \overrightarrow{n_1} \cdot \overrightarrow{n_2}=0 \\\\ \Rightarrow & (4+\lambda) 4+(\lambda-1)(-1)+(1-\lambda) 1=0 \\\\ \Rightarrow & 16+4 \lambda-\lambda+1+1-\lambda=0 \\\\ \Rightarrow & 2 \lambda=-18 \\\\ \Rightarrow & \lambda=-9 \end{aligned}

$\therefore$ equation of plane : $-5 x-10 y+10 z+26=0$ Distance of plane from $(2,3,-4)$

\begin{aligned} & =\left|\frac{-10-30-40+26}{\sqrt{100+100+26}}\right|=\frac{54}{15}=\alpha \\\\ \therefore 35 \alpha & =35 \cdot \frac{54}{15}=7 \times \frac{54}{3}=7 \times 18=126 \end{aligned}

Q173

Let the line passing through the points

\mathrm{P}(2,-1,2)

and

\mathrm{Q}(5,3,4)

meet the plane

x-y+z=4

at the point

\mathrm{R}

. Then the distance of the point

\mathrm{R}

from the plane

x+2 y+3 z+2=0

measured parallel to the line

\dfrac{x-7}{2}=\dfrac{y+3}{2}=\dfrac{z-2}{1}

is equal to :

A

\sqrt{31}

B

\sqrt{189}

C

\sqrt{61}

D 3

Correct Answer

Option D

Solution

Equation of line $P Q$ :

\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{2}=\lambda

Let $R$ be $(3 \lambda+2,4 \lambda-1,2 \lambda+2)$ $\mathrm{R}$ lies on plane $x-y+z=4$

\begin{aligned} & \therefore \quad 3 \lambda+2-4 \lambda+1+2 \lambda+2=4 \\\\ & \Rightarrow \quad \lambda=-1 \\\\ & \therefore \quad R(-1,-5,0) \end{aligned}

\text{ Let } S R \text{ be }: \frac{x+1}{2}=\frac{y+5}{2}=\frac{z - 0}{1}=k

Point

S:(2 k-1,2 k-5, k)

$S$ lies on plane: $x+2 y+3 z+2=0$

\begin{aligned} & \Rightarrow(2 k-1)+(4 k-10)+3 k+2=0 \\\\ & \Rightarrow 9 k-9=0 \Rightarrow k=1 \end{aligned}

\therefore \mathrm{S}=(1,-3,1)

\therefore S R=\sqrt{4+4+1}=3

Q174

Let P be the plane passing through the points

(5,3,0),(13,3,-2)

and

(1,6,2)

. For

\alpha \in \mathbb{N}

, if the distances of the points

\mathrm{A}(3,4, \alpha)

and

\mathrm{B}(2, \alpha, a)

from the plane P are 2 and 3 respectively, then the positive value of a is :

A 6

B 4

C 5

D 3

Correct Answer

Option B

Solution

\begin{aligned} & \overrightarrow{A B}=8 \hat{i}-2 \hat{k} \\\\ & \overrightarrow{A C}=-4 \hat{i}+3 \hat{j}+2 \hat{k} \end{aligned}

\begin{aligned} & \overrightarrow{A B} \times \overrightarrow{A C}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 8 & 0 & -2 \\ -4 & 3 & 2 \end{array}\right| \\\\ & =6 \hat{i}-8 \hat{j}+24 \hat{k} \end{aligned}

Equation of plane : $6 x-8 y+24 z=d$ passes through $(5,3,0)$

\begin{aligned} & 6 \times 5-8 \times 3+24 \times 0=d \\\\ & \Rightarrow d=6 \\\\ & 6 x-8 y+24 z=6 \\\\ & \Rightarrow 3 x-4 y+12 z=3 \end{aligned}

Distance of point $(3,4, \alpha)$

\frac{9-16+12 \alpha-3}{\sqrt{9+16+144}}=2 \Rightarrow \alpha=3

Distance of point $(2, \alpha, a)$

\begin{aligned} & \frac{3 \times 2-4 \times 3+12 \times a-3}{13}=3 \\\\ & \Rightarrow 12 a-9=39 \\\\ & \Rightarrow 12 a=48 \\\\ & \Rightarrow a=4 \end{aligned}

Q175

Let

(\alpha, \beta, \gamma)

be the image of the point

\mathrm{P}(2,3,5)

in the plane

2 x+y-3 z=6

. Then

\alpha+\beta+\gamma

is equal to :

A 10

B 9

C 5

D 12

Correct Answer

Option A

Solution

Co-ordinate of image of point $P(2,3,5)$ in the plane $2 x+y-3 z=6$ is

\begin{aligned} \frac{\alpha-2}{2} & =\frac{\beta-3}{1}=\frac{\gamma-5}{-3}=\frac{-2(2 \times 2+3-3 \times 5-6)}{2^2+1^2+(-3)^2} \\\\ \frac{\alpha-2}{2} & =\frac{\beta-3}{1}=\frac{\gamma-5}{-3}=2 \\\\ \frac{\alpha-2}{2} & =2, \frac{\beta-3}{1}=2, \frac{\gamma-5}{-3}=2 \\\\ \alpha & =6, \beta=5, \gamma=-1 \end{aligned}

Hence, $\alpha+\beta+\gamma=6+5-1=10$

Q176

If equation of the plane that contains the point

(-2,3,5)

and is perpendicular to each of the planes

2 x+4 y+5 z=8

and

3 x-2 y+3 z=5

is

\alpha x+\beta y+\gamma z+97=0

then

\alpha+\beta+\gamma=

A 15

B 16

C 17

D 18

Correct Answer

Option A

Solution

The equation of plane that passes through the point $(-2,3,5)$ is

a(x+2)+b(y-3)+c(z-5)=0

..........(i) The plane is perpendicular to

\begin{array}{ll} 2 x+4 y+5 z =8 \text{ and } 3 x-2 y+3 z=5 \\\\ \end{array}

\begin{array}{ll} \therefore 2 a+4 b+5 c=0 ..........(ii)\\\\ \text{ and } 3 a-2 b+3 c=0 ..........(iii) \end{array}

\begin{aligned} & \therefore \quad \frac{\mathrm{a}}{\left|\begin{array}{cc} 4 & 5 \\ -2 & 3 \end{array}\right|}=\frac{-\mathrm{b}}{\left|\begin{array}{ll} 2 & 5 \\ 3 & 3 \end{array}\right|}=\frac{\mathrm{c}}{\left|\begin{array}{cc} 2 & 4 \\ 3 & -2 \end{array}\right|} \\\\ & \Rightarrow \frac{\mathrm{a}}{22}=\frac{\mathrm{b}}{9}=\frac{\mathrm{c}}{-16} \end{aligned}

$\therefore$ From Equation (i) equation of plane

\begin{aligned} &22(x+2)+9(y-3)-16(z-5)=0 \\\\ &\Rightarrow 22 x+9 y-16 z+97=0 \end{aligned}

Here, $\alpha=22, \beta=9, \gamma=-16$

\therefore \alpha+\beta+\gamma=22+9-16=15

Q177

Let the image of the point

\mathrm{P}(1,2,6)

in the plane passing through the points

\mathrm{A}(1,2,0), \mathrm{B}(1,4,1)

and

\mathrm{C}(0,5,1)

be

\mathrm{Q}(\alpha, \beta, \gamma)

. Then

\left(\alpha^{2}+\beta^{2}+\gamma^{2}\right)

is equal to :

A 76

B 62

C 70

D 65

Correct Answer

Option D

Solution

Equation of plane passing through the points $A(1,2$ , $0), B(1,4,1)$ and $C(0,5,1)$ is

\begin{aligned} & \left|\begin{array}{ccc} x-1 & y-2 & z-0 \\ 0 & 2 & 1 \\ -1 & 3 & 1 \end{array}\right|=0 \\\\ & \Rightarrow x+y-2 z=3 \end{aligned}

Now $Q(\alpha, \beta, \gamma)$ is the image of the point $P(1,2,6)$ in the plane $x+y-2 z-3=0$

\begin{array}{ll} &\therefore \frac{\alpha-1}{1}=\frac{\beta-2}{1}=\frac{\gamma-6}{-2}=\frac{-2[1+2-2(6)-3]}{1^2+1^2+(-2)^2} \\\\ &\Rightarrow \frac{\alpha-1}{1}=\frac{\beta-2}{1}=\frac{\gamma-6}{-2}=4 \\\\ &\Rightarrow \alpha=5, \beta=6, \gamma=-2 \end{array}

\text{ Hence, } \alpha^2+\beta^2+\gamma^2=5^2+6^2+(-2)^2=65

Q178

Let

\mathrm{P}(3,2,3), \mathrm{Q}(4,6,2)

and

\mathrm{R}(7,3,2)

be the vertices of

\triangle \mathrm{PQR}

. Then, the angle

\angle \mathrm{QPR}

is

A

\cos ^{-1}\left(\dfrac{7}{18}\right)

B

\dfrac{\pi}{6}

C

\cos ^{-1}\left(\dfrac{1}{18}\right)

D

\dfrac{\pi}{3}

Correct Answer

Option D

Solution

Direction ratio of

\mathrm{PR}=(4,1,-1)

Direction ratio of

\mathrm{PQ}=(1,4,-1)

Now,

\cos \theta=\left|\frac{4+4+1}{\sqrt{18} \cdot \sqrt{18}}\right|

\theta=\frac{\pi}{3}

Q179

Let the line

\dfrac{x}{1}=\dfrac{6-y}{2}=\dfrac{z+8}{5}

intersect the lines

\dfrac{x-5}{4}=\dfrac{y-7}{3}=\dfrac{z+2}{1}

and

\dfrac{x+3}{6}=\dfrac{3-y}{3}=\dfrac{z-6}{1}

at the points

\mathrm{A}

and

\mathrm{B}

respectively. Then the distance of the mid-point of the line segment

\mathrm{AB}

from the plane

2 x-2 y+z=14

is :

A 3

B

\dfrac{10}{3}

C 4

D

\dfrac{11}{3}

Correct Answer

Option C

Solution

We have, $\dfrac{x}{1}=\dfrac{6-y}{2}=\dfrac{z+8}{5}$ intersect the line $\dfrac{x-5}{4}=\dfrac{y-7}{3}=\dfrac{z+2}{1}$ and $\dfrac{x+3}{6}=\dfrac{3-y}{3}=\dfrac{z-6}{1}$

\begin{array}{ll} & \text{ Now, } \frac{x}{1}=\frac{6-y}{2}=\frac{z+8}{5}=\lambda ...........(i)\\\\ & \Rightarrow x=\lambda, y=6-2 \lambda, z=5 \lambda-8 \end{array}

\begin{array}{ll} &\text{ Also, } \frac{x-5}{4} =\frac{y-7}{3}=\frac{z+2}{1}=k ...........(ii)\\\\ &\Rightarrow x =4 k+5, y=3 k+7, z=k-2 \end{array}

\begin{array}{rlrl} & \frac{x+3}{6}=\frac{3-y}{3}=\frac{z-6}{1}=\mu ..........(iii)\\\\ &\Rightarrow x = 6 \mu-3, y=3-3 \mu, z=\mu+6 \end{array}

On solving Eqs. (i) and (ii), we get $\lambda=1, k=-1$ $\therefore$ Co-ordinate of $A$ is $(1,4,-3)$ On solving Eqs. (i) and (iii), we get $\lambda=3, \mu=1$ $\therefore$ Co-ordinate of $\beta$ is $(3,0,7)$ Co-ordinate of mid-point of $A B$ is $\left(\dfrac{1+3}{2}, \dfrac{4+0}{2}, \dfrac{-3+7}{2}\right)$ or $(2,2,2)$ Perpendicular distance of mid-point of $A B$ from the plane $2 x-2 y+z=14$ is

\frac{|2(2)-2(2)+2-14|}{\sqrt{2^2+(-2)^2+1^2}}=4

Q180

The shortest distance between the lines

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

and

{{x - 4} \over 1} = {{y - 1} \over 2} = {{z + 3} \over 0}

is :

A 8

B 7

C 6

D 9

Correct Answer

Option D

Solution

Given, the lines are

\begin{aligned} \frac{x+2}{1} & =\frac{y}{-2}=\frac{z-5}{2} ~~~~..........(i)\\\\ \text{ and } \frac{x-4}{1} & =\frac{y-1}{2}=\frac{z+3}{0} ~~~~..........(ii) \end{aligned}

Formula for shortest distance between two skew-lines,

\begin{aligned} S D & =\left|\frac{\left|\begin{array}{ccc} 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{array}\right|}{\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{array}\right|}\right|=\left|\frac{\left|\begin{array}{ccc} 6 & 1 & -8 \\ 1 & -2 & 2 \\ 1 & 2 & 0 \end{array}\right|}{\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & -2 & 2 \\ 1 & 2 & 0 \end{array}\right|}\right| \\\\ & =\left|\frac{6(-4)-1(-2)-8(4)}{|-4 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}|}\right| \\\\ & =\left|\frac{-24+2-32}{\sqrt{36}}\right| \\\\ & =\left|\frac{-54}{6}\right|=|-9|=9 \end{aligned}