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Rotation Concept

Distance from...

Question

Distance from point $(-^i + 2^j + 6^k)$ to straight line through point $(2, 3, - 4)$ and parallel to vector $6^i + 3^j - 4^k$ .

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Solution

$\begin{matrix} L_{1} : \frac{x - 2}{6} = \frac{y - 3}{3} = \frac{z + 4}{- 4} (- 1, 2, 6) - - \to P Q = (6 r + 3, 3 r + 1, - 4 r - 10) \end{matrix}$
is parallel to :
$\begin{matrix} 6 (6 r + 3) + 3 (3 r + 1) - 4 (- 4 r - 10) = 0 61 r + 61 = 0 r = - 1 P Q = \sqrt{{(3)}^{2} + {(2)}^{2} + {(6)}^{2}} = \sqrt{49} = 7 \end{matrix}$
Hence,
We get $7$ .

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