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(B) The length of the altitude $h$ from vertex $A$ to the side $BC$ is given by $h = \frac{2 	imes 	ext{Area}(	riangle ABC)}{|BC|} = \frac{|\vec{AB

Vedclass · Accepted Answer

$\frac{3}{\sqrt{2}}$

Vedclass · Answer

(B) The length of the altitude $h$ from vertex $A$ to the side $BC$ is given by $h = \frac{2 	imes 	ext{Area}(	riangle ABC)}{|BC|} = \frac{|\vec{AB} 	imes \vec{AC}|}{|BC|}$.First,we find the vectors $\vec{AB}$,$\vec{AC}$,and $\vec{BC}$:$\vec{AB} = \vec{OB} - \vec{OA} = (1-3)\hat{i} + (2-1)\hat{j} + (3-2)\hat{k} = -2\hat{i} + \hat{j} + \hat{k}$.$\vec{AC} = \vec{OC} - \vec{OA} = (2-3)\hat{i} + (3-1)\hat{j} + (1-2)\hat{k} = -\hat{i} + 2\hat{j} - \hat{k}$.$\vec{BC} = \vec{OC} - \vec{OB} = (2-1)\hat{i} + (3-2)\hat{j} + (1-3)\hat{k} = \hat{i} + \hat{j} - 2\hat{k}$.Now,calculate the cross product $\vec{AB} 	imes \vec{AC}$:$\vec{AB} 	imes \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -2 & 1 & 1 \ -1 & 2 & -1 \end{vmatrix} = \hat{i}(-1-2) - \hat{j}(2+1) + \hat{k}(-4+1) = -3\hat{i} - 3\hat{j} - 3\hat{k}$.The magnitude is $|\vec{AB} 	imes \vec{AC}| = \sqrt{(-3)^2 + (-3)^2 + (-3)^2} = \sqrt{9+9+9} = \sqrt{27} = 3\sqrt{3}$.The magnitude of the base $BC$ is $|\vec{BC}| = \sqrt{1^2 + 1^2 + (-2)^2} = \sqrt{1+1+4} = \sqrt{6}$.Thus,$h = \frac{3\sqrt{3}}{\sqrt{6}} = \frac{3\sqrt{3}}{\sqrt{2}\sqrt{3}} = \frac{3}{\sqrt{2}}$.Therefore,option $B$ is correct.

If the position vectors of the vertices of a $\triangle ABC$ are $\vec{OA} = 3\hat{i} + \hat{j} + 2\hat{k}$,$\vec{OB} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{OC} = 2\hat{i} + 3\hat{j} + \hat{k}$,then the length of the altitude of $\triangle ABC$ drawn from $A$ is

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