Line $L_{1}$ passes through the point $(1, 2, 3)$ and is parallel to the z-axis. Line $L_{2}$ passes through the point $(\lambda, 5, 6)$ and is parallel to the y-axis. Let for $\lambda \in \{\lambda_{1}, \lambda_{2}\}$ with $\lambda_{2} < \lambda_{1}$, the shortest distance between the two lines be $3$. Then the square of the distance of the point $(\lambda_{1}, \lambda_{2}, 7)$ from the line $L_{1}$ is
- (1) 40
- (2) 32
- (3) 25
- (4) 37
Solution:
$$L_{1}: \frac{x-1}{0} = \frac{y-2}{0} = \frac{z-3}{1}$$
$$L_{2}: \frac{x-\lambda}{0} = \frac{y-5}{1} = \frac{z-6}{0}$$
Direction vector of $L_1$, $\vec{b_1} = 0\hat{i} + 0\hat{j} + 1\hat{k} = \hat{k}$
Direction vector of $L_2$, $\vec{b_2} = 0\hat{i} + 1\hat{j} + 0\hat{k} = \hat{j}$
Point on $L_1$, $\vec{a_1} = \hat{i} + 2\hat{j} + 3\hat{k}$
Point on $L_2$, $\vec{a_2} = \lambda\hat{i} + 5\hat{j} + 6\hat{k}$
Vector joining the points, $\vec{a_2} – \vec{a_1} = (\lambda – 1)\hat{i} + 3\hat{j} + 3\hat{k}$
$$ \vec{b_1} \times \vec{b_2} = \hat{k} \times \hat{j} = -\hat{i} $$
Shortest Distance (SD) = $ \frac{|(\vec{a_2} – \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{|\vec{b_1} \times \vec{b_2}|} $
$$ SD = \frac{|((\lambda – 1)\hat{i} + 3\hat{j} + 3\hat{k}) \cdot (-\hat{i})|}{|-\hat{i}|} $$
$$ SD = \frac{|-(\lambda – 1)|}{1} = |\lambda – 1| $$
Given SD = 3
$$ |\lambda – 1| = 3 $$
$$ \lambda – 1 = 3 \Rightarrow \lambda = 4 $$
$$ \lambda – 1 = -3 \Rightarrow \lambda = -2 $$
Since $\lambda_{2} < \lambda_{1}$, we have $\lambda_{1} = 4$ and $\lambda_{2} = -2$.
The given point is $P(\lambda_{1}, \lambda_{2}, 7) = P(4, -2, 7)$.
Let the foot of the perpendicular from $P(4, -2, 7)$ to line $L_{1}$ be $Q$.
Any generic point on line $L_{1}$ can be taken as $(1, 2, t+3)$. Let this be $Q$.
Direction ratios of $PQ$ = $(1-4, 2-(-2), t+3-7) = (-3, 4, t-4)$
Since $PQ$ is perpendicular to $L_{1}$ (whose direction ratios are $0, 0, 1$):
$$ -3(0) + 4(0) + (t-4)(1) = 0 $$
$$ t – 4 = 0 \Rightarrow t = 4 $$
So, the foot of the perpendicular is $Q(1, 2, 7)$.
Square of the distance from $L_1$ is $PQ^2$:
$$ PQ^2 = (1-4)^2 + (2-(-2))^2 + (7-7)^2 $$
$$ PQ^2 = (-3)^2 + (4)^2 + 0^2 $$
$$ PQ^2 = 9 + 16 = 25 $$
Ans. (3)