Question ID: #551
Let a straight line $L$ pass through the point $P(2, -1, 3)$ and be perpendicular to the lines $\frac{x-1}{2} = \frac{y+1}{1} = \frac{z-3}{-2}$ and $\frac{x-3}{1} = \frac{y-2}{3} = \frac{z+2}{4}$. If the line $L$ intersects the $yz$-plane at the point $Q$, then the distance between the points $P$ and $Q$ is:
- (1) 2
- (2) $\sqrt{10}$
- (3) 3
- (4) $2\sqrt{3}$
Solution:
The direction ratios of the given lines are $\vec{b_1} = 2\hat{i} + \hat{j} – 2\hat{k}$ and $\vec{b_2} = \hat{i} + 3\hat{j} + 4\hat{k}$.
Since line $L$ is perpendicular to both, its direction vector $\vec{d}$ is parallel to $\vec{b_1} \times \vec{b_2}$.
$\vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 1 & 3 & 4 \end{vmatrix}$
$= \hat{i}(4 – (-6)) – \hat{j}(8 – (-2)) + \hat{k}(6 – 1)$
$= 10\hat{i} – 10\hat{j} + 5\hat{k}$
Direction ratios can be simplified to $\langle 2, -2, 1 \rangle$.
The equation of line $L$ passing through $P(2, -1, 3)$ is:
$\frac{x-2}{2} = \frac{y+1}{-2} = \frac{z-3}{1} = \lambda$
The line intersects the $yz$-plane where $x = 0$.
$\frac{0-2}{2} = \lambda \Rightarrow \lambda = -1$.
Substitute $\lambda = -1$ to find $y$ and $z$:
$y = -2(-1) – 1 = 1$
$z = 1(-1) + 3 = 2$
So, the point $Q$ is $(0, 1, 2)$.
Distance $PQ = \sqrt{(2-0)^2 + (-1-1)^2 + (3-2)^2}$
$= \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.
Ans. (3)
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