The distance of the point having position vec | vedclass

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(A) Let the given point be $P(-1, 2, 6)$ and the point on the line be $A(2, 3, -4)$. The vector $\vec{AP}$ is given by:$\vec{AP} = (-1 - 2)\hat{i} + (

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(A) Let the given point be $P(-1, 2, 6)$ and the point on the line be $A(2, 3, -4)$. The vector $\vec{AP}$ is given by:$\vec{AP} = (-1 - 2)\hat{i} + (2 - 3)\hat{j} + (6 - (-4))\hat{k} = -3\hat{i} - \hat{j} + 10\hat{k}$.The line is parallel to the vector $\vec{b} = 6\hat{i} + 3\hat{j} - 4\hat{k}$.The distance $d$ of point $P$ from the line is given by the formula:$d = \frac{|\vec{AP} 	imes \vec{b}|}{|\vec{b}|}$.First,calculate the cross product $\vec{AP} 	imes \vec{b}$:$\vec{AP} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -3 & -1 & 10 \ 6 & 3 & -4 \end{vmatrix} = \hat{i}(4 - 30) - \hat{j}(12 - 60) + \hat{k}(-9 - (-6)) = -26\hat{i} + 48\hat{j} - 3\hat{k}$.Now,calculate the magnitude $|\vec{AP} 	imes \vec{b}| = \sqrt{(-26)^2 + 48^2 + (-3)^2} = \sqrt{676 + 2304 + 9} = \sqrt{2989}$.Calculate the magnitude $|\vec{b}| = \sqrt{6^2 + 3^2 + (-4)^2} = \sqrt{36 + 9 + 16} = \sqrt{61}$.Thus,$d = \sqrt{\frac{2989}{61}} = \sqrt{49} = 7$.

The distance of the point having position vector $-\hat{i} + 2\hat{j} + 6\hat{k}$ from the straight line passing through the point $(2, 3, -4)$ and parallel to the vector $6\hat{i} + 3\hat{j} - 4\hat{k}$ is

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