Question

The coordinate of foot of perpendicular drawn from the point p(2,2,-1) to the join of the point Q(1,2,-1) and R(2,-1,0) are

Answer 1

Let foot of perpendicular is A(\(\alpha , \beta, \Gamma\))

∴ PA is perpendicular to QR.

\(\therefore a_1a_2 + b_1b_2 + c_1 c_2 = 0 \) where

\(a_1,b_1,c_1\) are direction ratios of QR.

\(a_2,b_2,c_2\) are direction ratios of PR.

⇒ \((2 -1) (\alpha -2) + (-1 - 2)(\beta -2) + (0 - (-1))(\Gamma - (-1) )=0\) 

⇒ \(\alpha - 2 - 3\beta + 6 + \Gamma + 1 = 0\)

⇒ \(\alpha - 3\beta + \Gamma + 5 = 0\)    .....(1)

Equation of line QR is \(\frac{x-1}{2-1} = \frac{y-2}{-1-2} = \frac{z-(-1)}{0-(-1)}\)

or \(\frac{x-1}{1} = \frac{y-2}{-3} = \frac{z+1}{1}\)

∵ A(\(\alpha , \beta, \Gamma\)) will lie on line QR.

\(\therefore \frac{\alpha -1}1 = \frac{\beta - 2}{-3}= \frac{\Gamma + 1}1 = \gamma \) (Let)

⇒ \(\alpha = \gamma + 1, \beta = -3\gamma + 2, \Gamma = \gamma -1\)

 From (1), we get

\((\gamma + 1)-(-9\gamma + 6)+ \gamma - 1+5 = 0\)

\(11\gamma - 1 = 0\)

\(\gamma = \frac1{11}\)

\(\therefore \alpha = \frac 1{11}+ 1 = \frac{12}{11}\)

\(\beta = \frac{-3}{11} + 2 = \frac{19}{11}\)

\(\Gamma= \frac1{11 } - 1 = \frac{-10}{11}\)

Hence, foot of perpendicular is \(\left(\frac{12}{11}, \frac{19}{11}, \frac{-10}{11}\right).\)