If overrightarrow{OA}=2 hat{i}-hat{j}+hat{k}, | vedclass

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(D) First,we find the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ which lie in the plane containing points $A, B$,and $C$. $\overrightarro

Vedclass · Accepted Answer

$\frac{8 \hat{i}+2 \hat{j}+5 \hat{k}}{\sqrt{93}}$

Vedclass · Answer

(D) First,we find the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ which lie in the plane containing points $A, B$,and $C$. $\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (3\hat{i} - \hat{k}) - (2\hat{i} - \hat{j} + \hat{k}) = \hat{i} + \hat{j} - 2\hat{k}$. $\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = (2\hat{j} + 3\hat{k}) - (2\hat{i} - \hat{j} + \hat{k}) = -2\hat{i} + 3\hat{j} + 2\hat{k}$. $A$ vector perpendicular to the plane is given by the cross product $\vec{n} = \overrightarrow{AB} 	imes \overrightarrow{AC}$. $\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & -2 \ -2 & 3 & 2 \end{vmatrix} = \hat{i}(2 - (-6)) - \hat{j}(2 - 4) + \hat{k}(3 - (-2)) = 8\hat{i} + 2\hat{j} + 5\hat{k}$. The magnitude of this vector is $|\vec{n}| = \sqrt{8^2 + 2^2 + 5^2} = \sqrt{64 + 4 + 25} = \sqrt{93}$. The unit vector perpendicular to the plane is $\hat{n} = \frac{\vec{n}}{|\vec{n}|} = \frac{8\hat{i} + 2\hat{j} + 5\hat{k}}{\sqrt{93}}$.

If $\overrightarrow{OA}=2 \hat{i}-\hat{j}+\hat{k}$,$\overrightarrow{OB}=3 \hat{i}-\hat{k}$ and $\overrightarrow{OC}=2 \hat{j}+3 \hat{k}$ are the position vectors of the points $A, B$ and $C$,then a unit vector perpendicular to the plane containing $A, B$ and $C$ is

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