Question

Find the foot of perpendicular from the point (2, 3, 4) to the line \[\frac{4 - x}{2} = \frac{y}{6} = \frac{1 - z}{3} .\] Also, find the perpendicular distance from the given point to the line.

Answer 1

Let L be the foot of the perpendicular drawn from the point P (2, 3, 4) to the given line.
The coordinates of a general point on the line

\[\frac{4 - x}{2} = \frac{y}{6} = \frac{1 - z}{3}\] are given by 

\[\frac{4 - x}{2} = \frac{y}{6} = \frac{1 - z}{3} = \lambda\]

\[\text{ They can be re - written as }  \]

\[\frac{x - 4}{- 2} = \frac{y}{6} = \frac{z - 1}{- 3} = \lambda\]

\[ \Rightarrow x = - 2\lambda + 4\]

\[ y = 6\lambda\]

\[ z = - 3\lambda + 1\]

Let the coordinates of L be

\[\left( - 2\lambda + 4, 6\lambda, - 3\lambda + 1 \right)\] 

The direction ratios of PL are proportional to 

\[- 2\lambda + 4 - 2, 6\lambda - 3, - 3\lambda + 1 - 4, i . e . - 2\lambda + 2, 6\lambda - 3, - 3\lambda - 3\]

The direction ratios of the given line are proportional to -2,6,-3,  but PL is perpendicular to the given line. 

\[\therefore - 2\left( - 2\lambda + 2 \right) + 6\left( 6\lambda - 3 \right) - 3\left( - 3\lambda - 3 \right) = 0\]

\[ \Rightarrow \lambda = \frac{13}{49}\] 

Substituting 

\[ \Rightarrow \lambda = \frac{13}{49}\]  in 

\[\left( - 2\lambda + 4, 6\lambda, - 3\lambda + 1 \right)\]  we get the coordinates of L as  

\[\left( \frac{170}{49}, \frac{78}{49}, \frac{10}{49} \right)\]

\[\therefore PL = \sqrt{\left( \frac{170}{49} - 2 \right)^2 + \left( \frac{78}{49} - 3 \right)^2 + \left( \frac{10}{49} - 4 \right)^2}\]

\[ = \sqrt{\frac{44541}{2401}}\]

\[ = \sqrt{\frac{909}{49}}\]

\[ = \frac{3}{7}\sqrt{101}\]

Hence, the length of the perpendicular from P on PL is 

\[\frac{3}{7}\sqrt{101} \text{ units } \] .