\documentclass{article}
\usepackage{amsmath}
\usepackage{amssymb}
\begin{document}
Solution:
We are given:
- Point \( P(-1, 1, 0) \)
- Line passing through points \( A(0, 2, 4) \) and \( B(3, 0, 1) \)
We need to find the perpendicular distance from point \(P\) to the line \( \overline{AB} \).
Step 1: Direction Vector of the Line
The direction vector of the line joining points \(A\) and \(B\) is:
\[
\vec{AB} = (3 - 0)\hat{i} + (0 - 2)\hat{j} + (1 - 4)\hat{k}
= 3\hat{i} - 2\hat{j} - 3\hat{k}
\]
Step 2: Vector \( \vec{AP} \)
\[
\vec{AP} = (-1 - 0)\hat{i} + (1 - 2)\hat{j} + (0 - 4)\hat{k}
= -\hat{i} - \hat{j} - 4\hat{k}
\]
Step 3: Perpendicular Distance Formula
The perpendicular distance from point \( P \) to the line passing through \( A \) in the direction of \( \vec{AB} \) is given by:
\[
d = \frac{|\vec{AP} \times \vec{AB}|}{|\vec{AB}|}
\]
Step 4: Compute the Cross Product \( \vec{AP} \times \vec{AB} \)
\[
\vec{AP} \times \vec{AB} =
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k}
-1 & -1 & -4
3 & -2 & -3
\end{vmatrix}
\]
\[
= \hat{i} \left((-1)(-3) - (-4)(-2)\right)
- \hat{j} \left((-1)(-3) - (-4)(3)\right)
+ \hat{k} \left((-1)(-2) - (-1)(3)\right)
\]
\[
= \hat{i} (3 - 8) - \hat{j} (3 + 12) + \hat{k} (2 + 3)
\]
\[
= -5\hat{i} - 15\hat{j} + 5\hat{k}
\]
Step 5: Compute Magnitudes
\[
|\vec{AP} \times \vec{AB}| = \sqrt{(-5)^2 + (-15)^2 + 5^2}
= \sqrt{25 + 225 + 25}
= \sqrt{275} = 5\sqrt{11}
\]
\[
|\vec{AB}| = \sqrt{(3)^2 + (-2)^2 + (-3)^2}
= \sqrt{9 + 4 + 9} = \sqrt{22}
\]
Step 6: Compute the Distance
\[
d = \frac{5\sqrt{11}}{\sqrt{22}}
= \frac{5}{\sqrt{2}}
\]
Step 7: Final Answer
\[
\boxed{\frac{5}{\sqrt{2}}}
\]
Final Answer: (C) \( \frac{5}{\sqrt{2}} \)