The distance of the point -hat{i} + 2hat{j} + | vedclass

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(C) Let the given point be $P(-1, 2, 6)$ and the line $L$ pass through $A(2, 3, -4)$ with direction vector $\vec{v} = 6\hat{i} + 3\hat{j} - 4\hat{k}$.

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(C) Let the given point be $P(-1, 2, 6)$ and the line $L$ pass through $A(2, 3, -4)$ with direction vector $\vec{v} = 6\hat{i} + 3\hat{j} - 4\hat{k}$.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 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 $\vec{AP} 	imes \vec{v}$:$\vec{AP} 	imes \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -3 & -1 & 10 \ 6 & 3 & -4 \end{vmatrix} = \hat{i}(4 - 30) - \hat{j}(12 - 60) + \hat{k}(-9 + 6) = -26\hat{i} + 48\hat{j} - 3\hat{k}$.The magnitude $|\vec{AP} 	imes \vec{v}| = \sqrt{(-26)^2 + 48^2 + (-3)^2} = \sqrt{676 + 2304 + 9} = \sqrt{2989}$.The magnitude $|\vec{v}| = \sqrt{6^2 + 3^2 + (-4)^2} = \sqrt{36 + 9 + 16} = \sqrt{61}$.Thus,$d = \sqrt{\frac{2989}{61}} = \sqrt{49} = 7$.

The distance of the point $-\hat{i} + 2\hat{j} + 6\hat{k}$ from the straight line that passes through the point $2\hat{i} + 3\hat{j} - 4\hat{k}$ and is parallel to the vector $6\hat{i} + 3\hat{j} - 4\hat{k}$ is

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