Question

Let a line $L$ pass through the point $P(2,3,1)$ and be parallel to the line $x+3 y-2 z-2=0=x-y+2 z$ If the distance of $L$ from the point $(5,3,8)$ is $\alpha$, then $3 \alpha^2$ is equal to_______

Answer 1

Correct Answer: 158

To solve this problem, we need to determine the distance \( \alpha \) of the line \( L \) from a given point and compute \( 3 \alpha^2 \). The line \( L \) is parallel to the intersection of the planes given by equations \( x + 3y - 2z - 2 = 0 \) and \( x - y + 2z = 0 \).
The direction vector of \( L \) is obtained from the cross-product of the normal vectors of these planes: 
Normal of first plane: \( \mathbf{n_1} = (1, 3, -2) \)
Normal of second plane: \( \mathbf{n_2} = (1, -1, 2) \)
Direction vector of \( L: \mathbf{d} = \mathbf{n_1} \times \mathbf{n_2} \) 
\(\Rightarrow \mathbf{d} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\1 & 3 & -2\\1 & -1 & 2\end{vmatrix} = (4, -4, -4)\)
Simplifying: \(\mathbf{d} = (1, -1, -1)\)
The line \( L \) passes through \( P(2, 3, 1) \) and is parallel to \( \mathbf{d} \).
We need the distance from \( Q(5, 3, 8) \) to line \( L \). Using the formula for the shortest distance from a point to a line, \( d = \frac{\|\mathbf{PQ} \times \mathbf{d}\|}{\|\mathbf{d}\|} \), where \( \mathbf{PQ} = Q - P = (3, 0, 7) \).
\(\mathbf{PQ} \times \mathbf{d} = \begin{vmatrix}\mathbf{i} & \mathbf{j} & \mathbf{k}\\3 & 0 & 7\\1 & -1 & -1\end{vmatrix} = (7, 10, -3)\)
\(\|\mathbf{PQ} \times \mathbf{d}\| = \sqrt{7^2 + 10^2 + (-3)^2} = \sqrt{158}\)
\(\|\mathbf{d}\| = \sqrt{1^2 + (-1)^2 + (-1)^2} = \sqrt{3}\)
\(\alpha = \frac{\sqrt{158}}{\sqrt{3}} = \sqrt{\frac{158}{3}}\)
\(3 \alpha^2 = 3 \times \frac{158}{3} = 158\)
The computed value of \( 3 \alpha^2 \) is 158, which is within the given range (158, 158).