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(A) Let $P(2\hat{i} - \hat{j} + 5\hat{k})$ be the given point and $L$ be the foot of the perpendicular drawn from $P$ to the line $\vec{r} = (11\hat{i

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$\sqrt{14}$

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(A) Let $P(2\hat{i} - \hat{j} + 5\hat{k})$ be the given point and $L$ be the foot of the perpendicular drawn from $P$ to the line $\vec{r} = (11\hat{i} - 2\hat{j} - 8\hat{k}) + \lambda(10\hat{i} - 4\hat{j} - 11\hat{k})$.The position vector of any point $L$ on the line is given by $\vec{L} = (11 + 10\lambda)\hat{i} + (-2 - 4\lambda)\hat{j} + (-8 - 11\lambda)\hat{k}$.The vector $\vec{PL} = \vec{L} - \vec{P} = [(11 + 10\lambda) - 2]\hat{i} + [(-2 - 4\lambda) - (-1)]\hat{j} + [(-8 - 11\lambda) - 5]\hat{k}$.$\vec{PL} = (9 + 10\lambda)\hat{i} + (-1 - 4\lambda)\hat{j} + (-13 - 11\lambda)\hat{k}$.Since $PL$ is perpendicular to the line,it must be perpendicular to the direction vector of the line $\vec{b} = 10\hat{i} - 4\hat{j} - 11\hat{k}$.Therefore,$\vec{PL} \cdot \vec{b} = 0$.$10(9 + 10\lambda) - 4(-1 - 4\lambda) - 11(-13 - 11\lambda) = 0$.$90 + 100\lambda + 4 + 16\lambda + 143 + 121\lambda = 0$.$237\lambda + 237 = 0 \Rightarrow \lambda = -1$.Substituting $\lambda = -1$ into the expression for $\vec{L}$,we get the coordinates of the foot of the perpendicular $L$ as $(1, 2, 3)$,i.e.,$\vec{L} = \hat{i} + 2\hat{j} + 3\hat{k}$.Now,$\vec{PL} = (1 - 2)\hat{i} + (2 - (-1))\hat{j} + (3 - 5)\hat{k} = -\hat{i} + 3\hat{j} - 2\hat{k}$.The length of the perpendicular is $|\vec{PL}| = \sqrt{(-1)^2 + 3^2 + (-2)^2} = \sqrt{1 + 9 + 4} = \sqrt{14}$.

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

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