Question

If a line drawn from the point $A(1,2,1)$ is perpendicular to the line joining $P(1,4,6)$ and $Q(5,4,4)$, then the find the cordinates of the foot of the perpendicular.

Answer 1

Let M be the foot of the perpendicular drawn from the point A ($1,2,1$) to the line joining P ($1,4,6$) and Q ($5,4,4$) .

Equation of a line passing through the points $\left(x_1,y_1,z_1\right)and\left(x_2,y_2,z_2\right)$ is

$\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$

Equation of the required line passing through P (1, 4, 6) and Q( 5, 4, 4) is

$\begin{array}{c}\frac{x-1}{4}=\frac{y-4}{0}=\frac{z-6}{-2}=\lambda \\ x=4\lambda +1;y=4;z=-2\lambda +6\end{array}$

Coordinates of M are $\left(4\lambda +1,4,-2\lambda +6\right).........\left(1\right)$

The direction ratios of AM are

$\begin{array}{c}4\lambda +1-1,4-2,-2\lambda +6-1\\ i.e.4\lambda ,2,-2\lambda +5\end{array}$

The direction ratios of given line are 4,0,-2

Since AM is perpendicular to the given line

$\begin{array}{c}\therefore 4\left(4\lambda \right)+0\left(2\right)+\left(-2\right)\left(-2\lambda +5\right)=0\\ \therefore \lambda =\frac{1}{2}\end{array}$

Putting $\lambda =\frac{1}{2}$ in (i), the coordinates of M are (3,4,5)

Length of perpendicular from A on the given line

$AM=\sqrt{{\left(3-1\right)}^2+{\left(4-2\right)}^2+{\left(5-1\right)}^2}=\sqrt{24}units$

The coordinates of M i.e the foot of the perpendicular $=(3,4,5)$