SATHEE: 3D Geometry 3 Question 53

3D Geometry 3 Question 53

53. Let $P_{1} : 2 x + y - z = 3$ and $P_{2} : x + 2 y + z = 2$ be two planes. Then, which of the following statement(s) is (are) TRUE?

(2018 Adv.)

(a) The line of intersection of $P_{1}$ and $P_{2}$ has direction ratios $1, 2, - 1$

(b) The line $\frac{3 x - 4}{9} = \frac{1 - 3 y}{9} = \frac{z}{3}$ is perpendicular to the line of intersection of $P_{1}$ and $P_{2}$

(d) If $P_{3}$ is the plane passing through the point $(4, 2, - 2)$ and perpendicular to the line of intersection of $P_{1}$ and $P_{2}$ , then the distance of the point $(2, 1, 1)$ from the plane $P_{3}$ is $\frac{2}{\sqrt{3}}$

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Answer:

Correct Answer: 53. (a, b)

Solution:

We have,

	$P_{1} : 2 x + y - z = 3$
and	$P_{2} : x + 2 y + z = 2$
Here,	${\vec{n}}_{1} = 2 \hat{i} + \hat{j} - \hat{k}$
and	${\vec{n}}_{2} = \hat{i} + 2 \hat{j} + \hat{k}$

(a) Direction ratio of the line of intersection of $P_{1}$

$\begin{aligned} and P_{2} is θ {\vec{n}}_{1} \times {\vec{n}}_{2} \end{aligned}$

Hence, statement $a$ is false.

(b) We have, $\frac{3 x - 4}{9} = \frac{1 - 3 y}{9} = \frac{z}{3}$

$\Rightarrow \frac{x - \frac{4}{3}}{3} = \frac{y - \frac{1}{3}}{- 3} = \frac{z}{3}$

This line is parallel to the line of intersection of $P_{1}$ and $P_{2}$ .

Hence, statement (b) is false.

We know that,

$\begin{aligned} \cos θ = \frac{{\vec{n}}_{1} \cdot {\vec{n}}_{2}}{| {\vec{n}}_{1} | | {\vec{n}}_{2} |} & = \frac{(2 \hat{i} + \hat{j} - \hat{k}) \cdot (\hat{i} + 2 \hat{j} + \hat{k})}{| 2 \hat{i} + \hat{j} - \hat{k} | | \hat{i} + 2 \hat{j} + \hat{k} |} \\ = \frac{2 + 2 - 1}{\sqrt{6} \times \sqrt{6}} = \frac{1}{2} \\ θ & = 60^{\circ} \end{aligned}$

Hence, statement (c) is true.

(d) Equation of plane passing through the point $(4, 2, - 2)$ and perpendicular to the line of intersection of $P_{1}$ and $P_{2}$ is

$\begin{aligned} 3 (x - 4) - 3 (y - 2) + 3 (z + 2) & = 0 \\ \Rightarrow & 3 x - 3 y + 3 z - 12 + 6 + 6 & = 0 \\ \Rightarrow & x - y + z & = 0 \end{aligned}$