The length and foot of the perpendicular from | vedclass

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(C) Let the point be $P(2, -1, 5)$ and the line be $L: \frac{x - 11}{10} = \frac{y + 2}{-4} = \frac{z + 8}{-11} = k$. Any point $Q$ on the line is giv

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$\sqrt{14}, (1, 2, 3)$

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(C) Let the point be $P(2, -1, 5)$ and the line be $L: \frac{x - 11}{10} = \frac{y + 2}{-4} = \frac{z + 8}{-11} = k$. Any point $Q$ on the line is given by $(10k + 11, -4k - 2, -11k - 8)$. The direction ratios of the line $PQ$ are $(10k + 11 - 2, -4k - 2 + 1, -11k - 8 - 5) = (10k + 9, -4k - 1, -11k - 13)$. Since $PQ$ is perpendicular to the line $L$ with direction ratios $(10, -4, -11)$,their dot product must be zero: $10(10k + 9) - 4(-4k - 1) - 11(-11k - 13) = 0$. $100k + 90 + 16k + 4 + 121k + 143 = 0$. $237k + 237 = 0 \implies k = -1$. Substituting $k = -1$ into the coordinates of $Q$: $x = 10(-1) + 11 = 1$,$y = -4(-1) - 2 = 2$,$z = -11(-1) - 8 = 3$. So,the foot of the perpendicular is $(1, 2, 3)$. The length of the perpendicular $PQ$ is $\sqrt{(1 - 2)^2 + (2 - (-1))^2 + (3 - 5)^2} = \sqrt{(-1)^2 + 3^2 + (-2)^2} = \sqrt{1 + 9 + 4} = \sqrt{14}$.

The length and foot of the perpendicular from the point $(2, -1, 5)$ to the line $\frac{x - 11}{10} = \frac{y + 2}{-4} = \frac{z + 8}{-11}$ are

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