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 find the co-ordinates 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 (x₁,y₁,z₁) and (x₂,y₂,z₂) is

${\displaystyle \frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z1}}$

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

${\displaystyle \frac{x-1}{4}=\frac{y-4}{0}=\frac{z-6}{-2}=\lambda }$

${\displaystyle x=4\lambda +1;y=4;z=-2\lambda +6}$

 ∴ Coordinates of M are${\displaystyle (4\lambda +1,4,-2\lambda +6)}$      ..........(1)

The direction ratios of AM are

${\displaystyle 4\lambda +1-1,4-2,-2\lambda +6-1}$

i.e ${\displaystyle 4\lambda ,2,-2\lambda +5}$

The direction ratios of given line are 4,0,-2.
Since AM is perpendicular to the given line

${\displaystyle \therefore 4\left(4\lambda \right)+0\left(2\right)+(-2)(-2\lambda +5)=0}$

${\displaystyle \therefore \lambda =\frac{1}{2}}$

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

Length of perpendicular from A on the given line

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