Find the perpendicular distance of the point $P (4, - 5, 3)$ from the line $\to r = (5, - 2, 6) + k (3, - 4, 5), k \in R$ .

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Solution

$\begin{matrix} r = a + K (b - a) x i + y i + z k = (5^i - 2^j + 6^k) + K (3^i - 4^j + 5^k) x = (5 + 3 k) y = - 2 - 4 k z = 6 + 5 k \end{matrix}$
Thus, co-ordinate of L are $[(5 + 3 k), (- 2 - 4 k), (6 + 5 k)]$ So, the direction ratio of the line PL are
$\begin{matrix} x = (4 - 5 - 3 k) = (- 1 - 3 k) y = (- 5 + 2 + 4 k) = (- 3 + 4 k) z = (3 - 6 - 5 k) = (- 3 - 5 k) i - e [(- 1 + 3 k), (- 3 + 4 k), (- 3 - 5 k)] S i n c e P L i s ⊥ t o Q R, w e h a v e (- 1 - 3 k) (- 5 + 3) + (- 3 + 4 k) (+ 2 - 4) + (- 3 - 5 k) = 0 (- 1 - 3 k) (- 2) + (- 3 + 4 k) (- 2) + (- 3 - 5 k) (- 1) = 0 \Rightarrow 2 + 6 k + 6 - 8 k + 3 + 5 k = 0 \Rightarrow 3 k = - 11 \Rightarrow k = \frac{- 11}{3} ∴ r e q u i r e d c o - o r d i n a t e o f L i s (- 12; \frac{35}{3}; \frac{46}{3}) ∴ D i s t a n c e b e t w e e n P L i s = \sqrt{{(4 + 12)}^{2} + {(- 5 - \frac{35}{3})}^{2} + {(3 - \frac{46}{3})}^{2}} = \sqrt{256 + \frac{2500}{9} + \frac{1369}{9}} = \frac{1}{3} \sqrt{6173} = 26.2 \end{matrix}$

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