SATHEE: 3D Geometry 3 Question 26

3D Geometry 3 Question 26

26. The equation of the plane containing the straight line $\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$ and perpendicular to the plane containing the straight lines $\frac{x}{3} = \frac{y}{4} = \frac{z}{2}$ and $\frac{x}{4} = \frac{y}{2} = \frac{z}{3}$ is

(a) $5 x + 2 y - 4 z = 0$

(b) $x + 2 y - 2 z = 0$

(d) $x - 2 y + z = 0$

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Answer:

(d)

Solution:

Let $P_{1}$ be the plane containing the lines

$\frac{x}{3} = \frac{y}{4} = \frac{z}{2} and \frac{x}{4} = \frac{y}{2} = \frac{z}{3}$

For these two lines, direction vectors are

$b_{1} = 3 \hat{i} + 4 \hat{j} + 2 \hat{k} and b_{2} = 4 \hat{i} + 2 \hat{j} + 3 \hat{k}$

A vector along the normal to the plane $P_{1}$ is given by

$\begin{aligned} n_{1} = b_{1} \times b_{2} & = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 2 \\ 4 & 2 & 3 \end{array} | \\ = \hat{i} (12 - 4) - \hat{j} (9 - 8) + \hat{k} (6 - 16) = 8 \hat{i} - \hat{j} - 10 \hat{k} \end{aligned}$

Let $P_{2}$ be the plane containing the line $\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$ and perpendicular to plane $P_{1}$ .

For the line $\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$ , the direction vector is $b = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}$ and it passes through the point with position vector $a = 0 \hat{i} + 0 \hat{j} + 0 \hat{k}$ .

$∵ P_{2}$ is perpendicular to $P_{1}$ , therefore $n_{1}$ and $b$ lies along the plane.

Also, $P_{2}$ also passes through the point with position vector $a$ .

$∴$ Equation of plane $P_{2}$ is given by

$\begin{aligned} (r - a) \cdot (n_{1} \times b) = 0 \Rightarrow | \begin{array}{ccc} x - 0 & y - 0 & z - 0 \\ 8 & - 1 & - 10 \\ 2 & 3 & 4 \end{array} | = 0 \\ \Rightarrow x (- 4 + 30) - y (32 + 20) + z (24 + 2) = 0 \\ \Rightarrow 26 x - 52 y + 26 z = 0 \\ \Rightarrow x - 2 y + z = 0 \end{aligned}$