The distance of the point having position vec | vedclass

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(D) Let the point be $P$ with position vector $\vec{\alpha} = \hat{i} - 2\hat{j} - 6\hat{k}$.Let the line pass through point $A$ with position vector

Vedclass · Accepted Answer

$\sqrt{\frac{341}{61}}$

Vedclass · Answer

(D) Let the point be $P$ with position vector $\vec{\alpha} = \hat{i} - 2\hat{j} - 6\hat{k}$.Let the line pass through point $A$ with position vector $\vec{a} = 2\hat{i} - 3\hat{j} - 4\hat{k}$ and be parallel to vector $\vec{b} = 6\hat{i} + 3\hat{j} - 4\hat{k}$.The distance $d$ of point $P$ from the line is given by $d = \frac{|(\vec{\alpha} - \vec{a}) 	imes \vec{b}|}{|\vec{b}|}$.First,calculate $\vec{\alpha} - \vec{a} = (1-2)\hat{i} + (-2 - (-3))\hat{j} + (-6 - (-4))\hat{k} = -\hat{i} + \hat{j} - 2\hat{k}$.Next,calculate the cross product $(\vec{\alpha} - \vec{a}) 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -1 & 1 & -2 \ 6 & 3 & -4 \end{vmatrix} = \hat{i}(-4 - (-6)) - \hat{j}(4 - (-12)) + \hat{k}(-3 - 6) = 2\hat{i} - 16\hat{j} - 9\hat{k}$.The magnitude is $|(\vec{\alpha} - \vec{a}) 	imes \vec{b}| = \sqrt{2^2 + (-16)^2 + (-9)^2} = \sqrt{4 + 256 + 81} = \sqrt{341}$.The magnitude of $\vec{b}$ is $|\vec{b}| = \sqrt{6^2 + 3^2 + (-4)^2} = \sqrt{36 + 9 + 16} = \sqrt{61}$.Therefore,the distance $d = \frac{\sqrt{341}}{\sqrt{61}} = \sqrt{\frac{341}{61}}$ units.

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 units.

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