Find the distance of the point P(-hat{i} + 2h | vedclass

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(A) Let the point $A$ be $(2, 3, -4)$ and the point $P$ be $( -1, 2, 6)$.The vector $\vec{AP} = (-1 - 2)\hat{i} + (2 - 3)\hat{j} + (6 - (-4))\hat{k} =

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(A) Let the point $A$ be $(2, 3, -4)$ and the point $P$ be $( -1, 2, 6)$.The vector $\vec{AP} = (-1 - 2)\hat{i} + (2 - 3)\hat{j} + (6 - (-4))\hat{k} = -3\hat{i} - \hat{j} + 10\hat{k}$.The magnitude $|\vec{AP}| = \sqrt{(-3)^2 + (-1)^2 + (10)^2} = \sqrt{9 + 1 + 100} = \sqrt{110}$.The line is parallel to $\vec{v} = 6\hat{i} + 3\hat{j} - 4\hat{k}$.The unit vector along the line is $\hat{u} = \frac{6\hat{i} + 3\hat{j} - 4\hat{k}}{\sqrt{6^2 + 3^2 + (-4)^2}} = \frac{6\hat{i} + 3\hat{j} - 4\hat{k}}{\sqrt{36 + 9 + 16}} = \frac{6\hat{i} + 3\hat{j} - 4\hat{k}}{\sqrt{61}}$.The projection of $\vec{AP}$ on the line is $AN = |\vec{AP} \cdot \hat{u}| = \left| \frac{(-3)(6) + (-1)(3) + (10)(-4)}{\sqrt{61}} \right| = \left| \frac{-18 - 3 - 40}{\sqrt{61}} \right| = \frac{61}{\sqrt{61}} = \sqrt{61}$.The distance $PN$ from the point to the line is given by $PN = \sqrt{|\vec{AP}|^2 - AN^2}$.$PN = \sqrt{110 - 61} = \sqrt{49} = 7$.

Find the distance of the point $P(-\hat{i} + 2\hat{j} + 6\hat{k})$ from the line passing through the point $A(2, 3, -4)$ and parallel to the vector $\vec{v} = 6\hat{i} + 3\hat{j} - 4\hat{k}$.

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