If $\overrightarrow{\mathrm{OA}}=2 \hat{i}-\hat{j}+\hat{k}, \overrightarrow{O B}=3 \hat{i}-\hat{k}$ and…

If $\overrightarrow{\mathrm{OA}}=2 \hat{i}-\hat{j}+\hat{k}, \overrightarrow{O B}=3 \hat{i}-\hat{k}$ and $\overrightarrow{O C}=2 \hat{j}+3 \hat{k}$ are the position vectors of the points $A, B$ and $C$, then a unit vector perpendicular to the plane containing $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ is
  1. $\frac{8 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}}{2 \sqrt{21}}$
  2. $\frac{6 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}}{7}$
  3. $\frac{9 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}}{11}$
  4. $\frac{8 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}}{\sqrt{93}}$

Solution

$\begin{aligned} & \text {} \overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A}=\hat{i}+\hat{j}-2 \hat{k} \\ & \overrightarrow{A C}=-2 \hat{i}+3 \hat{j}+2 \hat{k} \end{aligned}$ $\because \overrightarrow{A B}$ and $\overrightarrow{A C}$ lies on the plane. Then $\overrightarrow{A B} \times \overrightarrow{A C}$ will be perpendicular to the plane. Now, $\vec{n}=\overrightarrow{A B} \times \overrightarrow{A C}$ $=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -2 \\ -2 & 3 & 2 \end{array}\right|=8 \hat{i}+2 \hat{j}+5 \hat{k}$ The unit vector perpendicular to the plane is $\hat{n}=\frac{\vec{n}}{|\vec{n}|}=\frac{8 \hat{i}+2 \hat{j}+5 \hat{k}}{\sqrt{64+4+25}}=\frac{8 \hat{i}+2 \hat{j}+5 \hat{k}}{\sqrt{93}}$

Asked in: AP EAMCET 2023 (17 May Shift 1)

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