The distance of the point P(-3, 2, 3) from th | vedclass

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(C) The line passes through $A(4, 6, -2)$ with direction vector $\vec{v} = -\hat{i} + 2\hat{j} + 3\hat{k}$.Let $P = (-3, 2, 3)$. The vector $\vec{AP}

Vedclass · Accepted Answer

$2 \sqrt{19}$

Vedclass · Answer

(C) The line passes through $A(4, 6, -2)$ with direction vector $\vec{v} = -\hat{i} + 2\hat{j} + 3\hat{k}$.Let $P = (-3, 2, 3)$. The vector $\vec{AP} = (-3-4)\hat{i} + (2-6)\hat{j} + (3-(-2))\hat{k} = -7\hat{i} - 4\hat{j} + 5\hat{k}$.The distance $d$ of point $P$ from the line is given by $d = \frac{|\vec{AP} 	imes \vec{v}|}{|\vec{v}|}$.First,calculate the cross product $\vec{AP} 	imes \vec{v}$:$\vec{AP} 	imes \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -7 & -4 & 5 \ -1 & 2 & 3 \end{vmatrix} = \hat{i}(-12-10) - \hat{j}(-21+5) + \hat{k}(-14-4) = -22\hat{i} + 16\hat{j} - 18\hat{k}$.The magnitude $|\vec{AP} 	imes \vec{v}| = \sqrt{(-22)^2 + 16^2 + (-18)^2} = \sqrt{484 + 256 + 324} = \sqrt{1064} = 2\sqrt{266}$.The magnitude $|\vec{v}| = \sqrt{(-1)^2 + 2^2 + 3^2} = \sqrt{1+4+9} = \sqrt{14}$.Thus,$d = \frac{2\sqrt{266}}{\sqrt{14}} = 2\sqrt{\frac{266}{14}} = 2\sqrt{19}$ units.

The distance of the point $P(-3, 2, 3)$ from the line passing through $A(4, 6, -2)$ with direction ratios $\langle -1, 2, 3 \rangle$ is . . . . . . units.

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