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
- A$\frac{8 \hat{i}-4 \hat{j}+2 \hat{k}}{2 \sqrt{21}}$
- B$\frac{6 \hat{i}+2 \hat{j}+3 \hat{k}}{7}$
- C$\frac{9 \hat{i}+2 \hat{j}+6 \hat{k}}{11}$
- D$\frac{8 \hat{i}+2 \hat{j}+5 \hat{k}}{\sqrt{93}}$
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The unit vector perpendicular to each of the vectors $\bar{a}+\bar{b}$ and $\bar{a}-\bar{b}$,where $\bar{a}=\hat{i}+\hat{j}+\hat{k}$ and $\bar{b}=3 \hat{i}-2 \hat{j}+5 \hat{k}$ is
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DifficultJEE MAIN 2024
View SolutionLet $\vec{a}, \vec{b}, \vec{c}$ be three vectors such that $|\vec{a}|=\sqrt{31}$,$4|\vec{b}|=|\vec{c}|=2$ and $2(\vec{a} \times \vec{b})=3(\vec{c} \times \vec{a})$. If the angle between $\vec{b}$ and $\vec{c}$ is $\frac{2\pi}{3}$,then $\left(\frac{\vec{a} \times \vec{c}}{\vec{a} \cdot \vec{b}}\right)^2$ is equal to $............$.
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