Find the perpendicular distance of the point $Q(5,-2,1)$ from the line passing through $A(1,0,2)$ and having direction ratios $2,-1,3$.

Step-by-step solution

1. Form the direction vector of the line: $\mathbf{d}=(2,-1,3)$. Its magnitude $|\mathbf{d}|=\sqrt{2^2+(-1{)}^2+3^2}=\sqrt{14}$.
2. Find the vector $\mathbf{AQ}=Q-A=(5-1,-2-0,1-2)=(4,-2,-1)$.
3. Compute the cross product $\mathbf{AQ}\times \mathbf{d}$:
4. $\begin{array}{rl}& (4,-2,-1)\times (2,-1,3)\\ =& ((-2)\cdot 3-(-1)(-1),-[4\cdot 3-(-1)\cdot 2],4(-1)-(-2)\cdot 2)\\ =& (-6-1,-[12+2],-4+4)\\ =& (-7,-14,0).\end{array}$
5. Magnitude of the cross product: $|\mathbf{AQ}\times \mathbf{d}|=\sqrt{(-7{)}^2+(-14{)}^2+0^2}=\sqrt{245}=7\sqrt{5}$.
6. Distance formula $d=\frac{|\mathbf{AQ}\times \mathbf{d}|}{|\mathbf{d}|}=\frac{7\sqrt{5}}{\sqrt{14}}=\frac{\sqrt{70}}{2}$.

Concept tested: Distance of a point from a line

Published 5 September 2026 · How this solution was prepared and checked