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Let $\mathrm{d}$ be the distance between the foot of perpendiculars of the points $\mathrm{P(1,2,-1) \, and\: Q(2,-1,3)}$ on the plane $\mathrm{-x+y+z=1.}$ Then $\mathrm{d^{2}}$ is equal to______________.
Option: 1
26

Option: 2
-

Option: 3
-

Option: 4
-

Answers (1)

$\mathrm{P(1,2,-1)}$

Let $\mathrm{P'(a,b,c)}$ be its foot of perpendicular

$\mathrm{\frac{a-1}{-1} =\frac{b-2}{1}=\frac{c+1}{1}=\frac{-(-1+2-1-1)}{3}=\frac{1}{3}} \\$

$\mathrm{\Rightarrow a =-\frac{1}{3}+1=\frac{2}{3}} \\$

$\mathrm{b =2+\frac{1}{3}=\frac{7}{3}} \\$

$\mathrm{c =\frac{1}{3}-1=-\frac{2}{3}} \\$

$\mathrm{\therefore P^{\prime}\left(\frac{2}{3}, \frac{7}{3},-\frac{2}{3}\right) }$

Let $\mathrm{Q'(x,y,z)}$ be foot of perpendicular of Q

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

$\mathrm{x=2-\frac{1}{3}=\frac{5}{3}} \\$

$\mathrm{y=-1+\frac{1}{3}=-\frac{2}{3}}$

$\mathrm{Z=3+\frac{1}{3} =\frac{10}{3}} \\$

$\mathrm{Q^{\prime}\left(\frac{5}{3},-\frac{2}{3}, \frac{10}{3}\right) }\\$

$\mathrm{\therefore d^{2}=P^{\prime} Q^{\prime 2} =\left(\frac{5}{3}-\frac{2}{3}\right)^{2}+\left(-\frac{2}{3}-\frac{7}{3}\right)^{2}+\left(\frac{10}{3}+\frac{2}{3}\right)^{2}} \\$

$\mathrm{=1+9+16} \\$

$\mathrm{=26}$

Posted by

Ramraj Saini

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