The perpendicular distance from the point P(3 | vedclass

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(D) Let the point $P$ be $(3, -2, 1)$. The line passes through $A(1, -3, 5)$ and $B(2, 1, -4)$.Vector $\vec{a} = \vec{i} - 3\vec{j} + 5\vec{k}$ and th

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$\sqrt{3}$

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(D) Let the point $P$ be $(3, -2, 1)$. The line passes through $A(1, -3, 5)$ and $B(2, 1, -4)$.Vector $\vec{a} = \vec{i} - 3\vec{j} + 5\vec{k}$ and the direction vector of the line is $\vec{v} = \vec{AB} = (2-1)\vec{i} + (1-(-3))\vec{j} + (-4-5)\vec{k} = \vec{i} + 4\vec{j} - 9\vec{k}$.Let $\vec{p} = 3\vec{i} - 2\vec{j} + \vec{k}$. The vector $\vec{AP} = \vec{p} - \vec{a} = (3-1)\vec{i} + (-2-(-3))\vec{j} + (1-5)\vec{k} = 2\vec{i} + \vec{j} - 4\vec{k}$.The perpendicular distance $d$ is given by $d = \frac{|\vec{AP} 	imes \vec{v}|}{|\vec{v}|}$.$\vec{AP} 	imes \vec{v} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \ 2 & 1 & -4 \ 1 & 4 & -9 \end{vmatrix} = \vec{i}(-9 - (-16)) - \vec{j}(-18 - (-4)) + \vec{k}(8 - 1) = 7\vec{i} + 14\vec{j} + 7\vec{k}$.Magnitude $|\vec{AP} 	imes \vec{v}| = \sqrt{7^2 + 14^2 + 7^2} = \sqrt{49 + 196 + 49} = \sqrt{294} = 7\sqrt{6}$.Magnitude $|\vec{v}| = \sqrt{1^2 + 4^2 + (-9)^2} = \sqrt{1 + 16 + 81} = \sqrt{98} = 7\sqrt{2}$.Distance $d = \frac{7\sqrt{6}}{7\sqrt{2}} = \sqrt{3}$.

The perpendicular distance from the point $P(3, -2, 1)$ to the line joining the points $A(1, -3, 5)$ and $B(2, 1, -4)$ is:

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