The distance between a point P whose position | vedclass

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(D) Let the point $P$ be $(5, 1, 3)$.The line passes through point $Q(3, 7, 1)$ and is parallel to the vector $\vec{v} = \hat{j} + \hat{k}$.The vector

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$6$

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(D) Let the point $P$ be $(5, 1, 3)$.The line passes through point $Q(3, 7, 1)$ and is parallel to the vector $\vec{v} = \hat{j} + \hat{k}$.The vector $\vec{PQ} = (3-5)\hat{i} + (7-1)\hat{j} + (1-3)\hat{k} = -2\hat{i} + 6\hat{j} - 2\hat{k}$.The distance $d$ from a point to a line is given by $d = \frac{|\vec{PQ} 	imes \vec{v}|}{|\vec{v}|}$.First,calculate the cross product $\vec{PQ} 	imes \vec{v}$:$\vec{PQ} 	imes \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -2 & 6 & -2 \ 0 & 1 & 1 \end{vmatrix} = \hat{i}(6 - (-2)) - \hat{j}(-2 - 0) + \hat{k}(-2 - 0) = 8\hat{i} + 2\hat{j} - 2\hat{k}$.The magnitude $|\vec{PQ} 	imes \vec{v}| = \sqrt{8^2 + 2^2 + (-2)^2} = \sqrt{64 + 4 + 4} = \sqrt{72} = 6\sqrt{2}$.The magnitude $|\vec{v}| = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{2}$.Therefore,$d = \frac{6\sqrt{2}}{\sqrt{2}} = 6$.

The distance between a point $P$ whose position vector is $5 \hat{i}+\hat{j}+3 \hat{k}$ and the line $r=(3 \hat{i}+7 \hat{j}+\hat{k})+t(\hat{j}+\hat{k})$ is

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