SATHEE: 3D Geometry 3 Question 7

3D Geometry 3 Question 7

####7. If $Q (0, - 1, - 3)$ is the image of the point $P$ in the plane $3 x - y + 4 z = 2$ and $R$ is the point $(3, - 1, - 2)$ , then the area (in sq units) of $△ P Q R$ is

(a) $\frac{\sqrt{91}}{2}$

(b) $2 \sqrt{13}$

(d) $\frac{\sqrt{65}}{2}$

(2019 Main, 10 April I)

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Answwer:

(a)

Solution:

Given, equation of plane $3 x - y + 4 z = 2$ …(i) and the point $Q (0, - 1, - 3)$ is the image of point $P$ in the plane (i), so point $P$ is also image of point $Q$ w.r.t. plane (i).

Let the coordinates of point $P$ is $(x_{1}, y_{1}, z_{1})$ , then

$\begin{aligned} \frac{x_{1} - 0}{3} & = \frac{y_{1} + 1}{- 1} = \frac{z_{1} + 3}{4} \\ = - 2 \frac{[3 (0) - 1 (- 1) + 4 (- 3) - 2]}{3^{2} + (- 1)^{2} + 4^{2}} \end{aligned}$

$[∵$ image of the point $(x_{1}, y_{1}, z_{1})$ in the plane $a x + b y + c z + d = 0$ is $(x, y, z)$ , where

$\begin{aligned} \frac{x - x_{1}}{a} = \frac{y - y_{1}}{b} = \frac{z - z_{1}}{c} = \frac{- 2 (a x_{1} + b y_{1} + c z_{1} + d)}{a^{2} + b^{2} + c^{2}}] \\ \Rightarrow \frac{x_{1} - 0}{3} = \frac{y_{1} + 1}{- 1} = \frac{z_{1} + 3}{4} \\ = 2 \frac{(1 - 12 - 2)}{26} = \frac{26}{26} = 1 \\ \Rightarrow P (x_{1}, y_{1}, z_{1}) = (3, - 2, 1) \end{aligned}$

Now, area of $△ P Q R$ , where point $R (3, - 1, - 2)$

$\begin{aligned} = \frac{1}{2} | \vec{P \vec{Q}} \times P \vec{R} | & = \frac{1}{2} | (- 3 \hat{i} + \hat{j} - 4 \hat{k}) \times (0 \hat{i} + \hat{j} - 3 \hat{k}) | \\ = \frac{1}{2} | \begin{array}{rrr} \hat{i} & \hat{j} & \hat{k} \\ - 3 & 1 & - 4 \\ 0 & 1 & - 3 \end{array} | = \frac{1}{2} | \hat{i} - 9 \hat{j} - 3 \hat{k} | \\ = \frac{1}{2} \sqrt{1 + 81 + 9} = \frac{\sqrt{91}}{2} squnits \end{aligned}$

3D Geometry 3 Question 7

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3D Geometry 3 Question 7

Answwer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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