The distance of the point P(-2, 4, -5) from t | vedclass

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(B) Let the given point be $P(-2, 4, -5)$ and the line be $L: \frac{x+3}{3} = \frac{y-4}{5} = \frac{z+8}{6} = \lambda$.Any point $Q$ on the line is gi

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$\sqrt{\frac{37}{10}}$

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(B) Let the given point be $P(-2, 4, -5)$ and the line be $L: \frac{x+3}{3} = \frac{y-4}{5} = \frac{z+8}{6} = \lambda$.Any point $Q$ on the line is given by $Q(3\lambda-3, 5\lambda+4, 6\lambda-8)$.The vector $\vec{PQ} = (3\lambda-3 - (-2), 5\lambda+4-4, 6\lambda-8 - (-5)) = (3\lambda-1, 5\lambda, 6\lambda-3)$.Since $PQ$ is perpendicular to the line with direction ratios $(3, 5, 6)$,their dot product is zero:$3(3\lambda-1) + 5(5\lambda) + 6(6\lambda-3) = 0$$9\lambda - 3 + 25\lambda + 36\lambda - 18 = 0$$70\lambda - 21 = 0 \implies \lambda = \frac{21}{70} = \frac{3}{10}$.Substituting $\lambda = \frac{3}{10}$ in $\vec{PQ}$:$\vec{PQ} = (3(\frac{3}{10})-1, 5(\frac{3}{10}), 6(\frac{3}{10})-3) = (-\frac{1}{10}, \frac{15}{10}, -\frac{12}{10})$.The distance $PQ = \sqrt{(-\frac{1}{10})^2 + (\frac{15}{10})^2 + (-\frac{12}{10})^2} = \sqrt{\frac{1+225+144}{100}} = \sqrt{\frac{370}{100}} = \sqrt{\frac{37}{10}}$ units.

The distance of the point $P(-2, 4, -5)$ from the line $\frac{x+3}{3} = \frac{y-4}{5} = \frac{z+8}{6}$ is

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