If the position vectors of three points A, B, | vedclass

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Question

(B) Let the position vectors be $\vec{a} = \hat{i} + \hat{j} + \hat{k}$,$\vec{b} = 2\hat{i} + 3\hat{j} - 4\hat{k}$,and $\vec{c} = 7\hat{i} + 4\hat{j}

Vedclass · Accepted Answer

$\frac{31\hat{i} - 38\hat{j} - 9\hat{k}}{\sqrt{2486}}$

Vedclass · Answer

(B) Let the position vectors be $\vec{a} = \hat{i} + \hat{j} + \hat{k}$,$\vec{b} = 2\hat{i} + 3\hat{j} - 4\hat{k}$,and $\vec{c} = 7\hat{i} + 4\hat{j} + 9\hat{k}$.Two vectors in the plane are $\vec{AB} = \vec{b} - \vec{a} = (2-1)\hat{i} + (3-1)\hat{j} + (-4-1)\hat{k} = \hat{i} + 2\hat{j} - 5\hat{k}$ and $\vec{AC} = \vec{c} - \vec{a} = (7-1)\hat{i} + (4-1)\hat{j} + (9-1)\hat{k} = 6\hat{i} + 3\hat{j} + 8\hat{k}$.$A$ vector perpendicular to the plane is $\vec{n} = \vec{AB} 	imes \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & -5 \ 6 & 3 & 8 \end{vmatrix}$.$\vec{n} = \hat{i}(16 - (-15)) - \hat{j}(8 - (-30)) + \hat{k}(3 - 12) = 31\hat{i} - 38\hat{j} - 9\hat{k}$.The magnitude is $|\vec{n}| = \sqrt{31^2 + (-38)^2 + (-9)^2} = \sqrt{961 + 1444 + 81} = \sqrt{2486}$.The unit vector is $\hat{n} = \pm \frac{\vec{n}}{|\vec{n}|} = \pm \frac{31\hat{i} - 38\hat{j} - 9\hat{k}}{\sqrt{2486}}$.Comparing with the options,option $B$ is correct.

If the position vectors of three points $A, B, C$ are $\hat{i} + \hat{j} + \hat{k}$,$2\hat{i} + 3\hat{j} - 4\hat{k}$,and $7\hat{i} + 4\hat{j} + 9\hat{k}$ respectively,then find the unit vector perpendicular to the plane of triangle $ABC$.

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