<p>The length of perpendicular from <p>P(2, -3, 1)</p> to the line <p>\(\frac{x+1}{2} = \frac{y-3}{3} = \frac{z+2}{-1}\)</p> is</p>
<p>(a) \(\frac{\sqrt{531}}{15}\) units</p>
<p>(b) \(\frac{\sqrt{531}}{14}\) units</p>
<p>(c) \(\frac{15}{14}\) units</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: Use the formula for perpendicular distance from a point to a line: distance equals the magnitude of cross product of direction vector with vector joining the point to any point on the line, divided by the magnitude of direction vector.
Step 1: Given line is $\frac{x+1}{2} = \frac{y-3}{3} = \frac{z+2}{-1}$ The line passes through point $A(-1, 3, -2)$ with direction vector $\vec{b} = (2, 3, -1)$ Step 2: Vector from point on line to P: $\vec{AP} = (2-(-1), -3-3, 1-(-2)) = (3, -6, 3)$ Step 3: The perpendicular distance is given by $d = \frac{|\vec{AP} \times \vec{b}|}{|\vec{b}|}$ Step 4: Calculate cross product: $\vec{AP} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -6 & 3 \\ 2 & 3 & -1 \end{vmatrix} = (6-9)\hat{i} - (-3-6)\hat{j} + (9+12)\hat{k} = (-3, 9, 21)$ Step 5: $|\vec{AP} \times \vec{b}| = \sqrt{9 + 81 + 441} = \sqrt{531}$ Step 6: $|\vec{b}| = \sqrt{4 + 9 + 1} = \sqrt{14}$ Step 7: Distance $d = \frac{\sqrt{531}}{\sqrt{14}} = \frac{\sqrt{531}}{\sqrt{14}} \cdot \frac{\sqrt{14}}{\sqrt{14}} = \frac{\sqrt{531 \times 14}}{14} = \frac{\sqrt{531}}{\sqrt{14}} = \frac{\sqrt{531 \times 14}}{14}$ ∴ Answer is B: $\frac{\sqrt{531}}{14}$ units
Correct Answer: B