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

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(B) The distance $d$ of a point $P$ with position vector $\vec{p}$ from a line passing through point $A$ with position vector $\vec{a}$ and parallel t

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$7$

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(B) The distance $d$ of a point $P$ with position vector $\vec{p}$ from a line passing through point $A$ with position vector $\vec{a}$ and parallel to vector $\vec{b}$ is given by the formula: $d = \frac{|(\vec{p} - \vec{a}) 	imes \vec{b}|}{|\vec{b}|}$.Here,$\vec{a} = 2\hat{i} + 3\hat{j} - 4\hat{k}$,$\vec{p} = -\hat{i} + 2\hat{j} + 6\hat{k}$,and $\vec{b} = 6\hat{i} + 3\hat{j} - 4\hat{k}$.First,calculate $\vec{p} - \vec{a} = (-1 - 2)\hat{i} + (2 - 3)\hat{j} + (6 - (-4))\hat{k} = -3\hat{i} - \hat{j} + 10\hat{k}$.Next,calculate the cross product $(\vec{p} - \vec{a}) 	imes \vec{b}$:$(\vec{p} - \vec{a}) 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -3 & -1 & 10 \ 6 & 3 & -4 \end{vmatrix} = \hat{i}(4 - 30) - \hat{j}(12 - 60) + \hat{k}(-9 - (-6)) = -26\hat{i} + 48\hat{j} - 3\hat{k}$.The magnitude $|(\vec{p} - \vec{a}) 	imes \vec{b}| = \sqrt{(-26)^2 + 48^2 + (-3)^2} = \sqrt{676 + 2304 + 9} = \sqrt{2989} = \sqrt{49 	imes 61} = 7\sqrt{61}$ (Wait,let's recheck the calculation).Re-calculating: $\vec{p}-\vec{a} = -3\hat{i}-\hat{j}+10\hat{k}$. Cross product: $\hat{i}(4-30) - \hat{j}(12-60) + \hat{k}(-9+6) = -26\hat{i} + 48\hat{j} - 3\hat{k}$. Magnitude is $\sqrt{676+2304+9} = \sqrt{2989}$.Let's re-verify the vector $\vec{b} = 6\hat{i} + 3\hat{j} - 4\hat{k}$. Magnitude $|\vec{b}| = \sqrt{36+9+16} = \sqrt{61}$.Distance $d = \frac{\sqrt{2989}}{\sqrt{61}} = \sqrt{\frac{2989}{61}} = \sqrt{49} = 7$.

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

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