Question

Find the equation of the perpendicular from the point $(1,6,3)$ to the line $\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$. Find also the coordinates of the foot of perpendicular.

Answer 1

Let the given equation be:

$\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}=\lambda$

therefore the direction cosines are $(1,2,3)$

therefore coordinates of any point on the line is given by,

$x=\lambda ,y=2\lambda +1,z=3\lambda +2$

Let $Q(\lambda ,2\lambda +1,3\lambda +2)$

and $P(1,6,3)$

direction cosines of PQ

$=(\lambda -1,2\lambda +1-6,3\lambda +2-3)$

$=(\lambda -1,2\lambda -5,3\lambda -1)$

sum of products of these direction cosines should be zero

$(\lambda -1)\left(1\right)+(2\lambda -5)\left(2\right)+(3\lambda -1)\left(3\right)=0$

$\lambda -1+4\lambda -10+9\lambda -3=0$

$14\lambda =14$

$\lambda =1$

substituting the value of $\lambda$ we get Q as

$Q(1,3,5)$ Cooordinates of foot of the perpendicular

direction ratios of PQ = (λ−1,2λ−5,3λ−1) = (0,-3,2)

equation of PQ

$\frac{x-1}{0}=\frac{y-3}{-3}=\frac{z-5}{2}$

=(λ−1,2λ−5,3λ−1)