The perpendicular distance of the point (2, 4 | vedclass

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(C) Let the given point be $P(2, 4, -1)$ and the line be $L: \frac{x + 5}{1} = \frac{y + 3}{4} = \frac{z - 6}{-9} = \lambda$.Any point $Q$ on the line

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(C) Let the given point be $P(2, 4, -1)$ and the line be $L: \frac{x + 5}{1} = \frac{y + 3}{4} = \frac{z - 6}{-9} = \lambda$.Any point $Q$ on the line is given by $Q(\lambda - 5, 4\lambda - 3, -9\lambda + 6)$.The vector $\vec{PQ} = (\lambda - 5 - 2, 4\lambda - 3 - 4, -9\lambda + 6 - (-1)) = (\lambda - 7, 4\lambda - 7, -9\lambda + 7)$.The direction vector of the line is $\vec{v} = (1, 4, -9)$.Since $PQ$ is perpendicular to the line,$\vec{PQ} \cdot \vec{v} = 0$.$1(\lambda - 7) + 4(4\lambda - 7) - 9(-9\lambda + 7) = 0$.$\lambda - 7 + 16\lambda - 28 + 81\lambda - 63 = 0$.$98\lambda - 98 = 0 \implies \lambda = 1$.Substituting $\lambda = 1$ in $Q$,we get $Q(1 - 5, 4(1) - 3, -9(1) + 6) = Q(-4, 1, -3)$.The distance $PQ = \sqrt{(-4 - 2)^2 + (1 - 4)^2 + (-3 - (-1))^2} = \sqrt{(-6)^2 + (-3)^2 + (-2)^2} = \sqrt{36 + 9 + 4} = \sqrt{49} = 7$.

The perpendicular distance of the point $(2, 4, -1)$ from the line $\frac{x + 5}{1} = \frac{y + 3}{4} = \frac{z - 6}{-9}$ is

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