The distance of the point Q(0, 2, -2) from th | vedclass

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(D) The direction vector of the required line is the cross product of the direction vectors of the two given lines:$\vec{v} = \begin{vmatrix} \hat{i}

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$\sqrt{74}$

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(D) The direction vector of the required line is the cross product of the direction vectors of the two given lines:$\vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 3 & 5 \ -1 & 3 & 2 \end{vmatrix} = \hat{i}(6 - 15) - \hat{j}(4 + 5) + \hat{k}(6 + 3) = -9\hat{i} - 9\hat{j} + 9\hat{k}$.We can take the direction vector as $\vec{d} = \hat{i} + \hat{j} - \hat{k}$.The equation of the line passing through $P(5, -4, 3)$ is $\vec{r} = (5\hat{i} - 4\hat{j} + 3\hat{k}) + \lambda(\hat{i} + \hat{j} - \hat{k})$.Any point $M$ on this line is $(5+\lambda, -4+\lambda, 3-\lambda)$.Let $Q$ be $(0, 2, -2)$. The vector $\vec{QM} = (5+\lambda - 0)\hat{i} + (-4+\lambda - 2)\hat{j} + (3-\lambda + 2)\hat{k} = (5+\lambda)\hat{i} + (\lambda-6)\hat{j} + (5-\lambda)\hat{k}$.Since $QM \perp$ line,$\vec{QM} \cdot (\hat{i} + \hat{j} - \hat{k}) = 0$.$(5+\lambda)(1) + (\lambda-6)(1) + (5-\lambda)(-1) = 0 \implies 5+\lambda + \lambda-6 - 5+\lambda = 0 \implies 3\lambda - 6 = 0 \implies \lambda = 2$.The point $M$ is $(5+2, -4+2, 3-2) = (7, -2, 1)$.The distance $QM = \sqrt{(7-0)^2 + (-2-2)^2 + (1 - (-2))^2} = \sqrt{49 + 16 + 9} = \sqrt{74}$.

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