The length of the perpendicular drawn from th | vedclass

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Question

(A) Let the point be $P = (2, -1, 5)$. The equation of the line is $\vec{r} = (11, -2, -8) + \lambda(10, -4, -11)$.Any point $Q$ on the line is given

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$\sqrt{14}$ units

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(A) Let the point be $P = (2, -1, 5)$. The equation of the line is $\vec{r} = (11, -2, -8) + \lambda(10, -4, -11)$.Any point $Q$ on the line is given by $Q = (10\lambda + 11, -4\lambda - 2, -11\lambda - 8)$.The direction ratios of the line $PQ$ are $(10\lambda + 11 - 2, -4\lambda - 2 + 1, -11\lambda - 8 - 5) = (10\lambda + 9, -4\lambda - 1, -11\lambda - 13)$.Since $PQ$ is perpendicular to the line with direction ratios $(10, -4, -11)$,their dot product is zero:$10(10\lambda + 9) - 4(-4\lambda - 1) - 11(-11\lambda - 13) = 0$.$100\lambda + 90 + 16\lambda + 4 + 121\lambda + 143 = 0$.$237\lambda + 237 = 0 \Rightarrow \lambda = -1$.Substituting $\lambda = -1$ into $Q$,we get $Q = (10(-1) + 11, -4(-1) - 2, -11(-1) - 8) = (1, 2, 3)$.The length of the perpendicular $PQ$ is $\sqrt{(1 - 2)^2 + (2 - (-1))^2 + (3 - 5)^2} = \sqrt{(-1)^2 + 3^2 + (-2)^2} = \sqrt{1 + 9 + 4} = \sqrt{14}$ units.

The length of the perpendicular drawn from the point $2 \hat{i} - \hat{j} + 5 \hat{k}$ to the line $\vec{r} = (11 \hat{i} - 2 \hat{j} - 8 \hat{k}) + \lambda(10 \hat{i} - 4 \hat{j} - 11 \hat{k})$ is

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