Let d be the distance between the foot of perpendiculars of the points P(1, 2, $-$1) and Q(2, $-$1, 3) on the plane $-$x + y + z = 1. Then d2 is equal to ___________.
Answer (integer)
26
Solution
<p>Foot of perpendicular from P</p>
<p>$${{x - 1} \over { - 1}} = {{y - 2} \over 1} = {{z + 1} \over 1} = {{ - ( - 1 + 2 - 1 - 1)} \over 3}$$</p>
<p>$\Rightarrow p' \equiv \left( {{2 \over 3},{7 \over 3},{{ - 2} \over 3}} \right)$</p>
<p>and foot of perpendicular from Q</p>
<p>$${{x - 2} \over { - 1}} = {{y + 1} \over 1} = {{z - 3} \over 1} = {{ - ( - 2 - 1 + 3 - 1)} \over 3}$$</p>
<p>$$ \Rightarrow Q' \equiv \left( {{5 \over 3},{{ - 2} \over 3},{{10} \over 3}} \right)$$</p>
<p>$P'Q' = \sqrt {{{(1)}^2} + {{(3)}^2} + {{(4)}^2}} = d = \sqrt {26}$</p>
<p>$\Rightarrow {d^2} = 26$</p>
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
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