For three vectors $\vec{A}=(-x \hat{i}-6 \hat{j}-2 \hat{k}), \vec{B}=(-\hat{i}+4 \hat{j}+3 \hat{k})$ and…

For three vectors $\vec{A}=(-x \hat{i}-6 \hat{j}-2 \hat{k}), \vec{B}=(-\hat{i}+4 \hat{j}+3 \hat{k})$ and $\vec{C}=(-8 \hat{i}-\hat{j}+3 \hat{k})$, if $\vec{A} \cdot(\vec{B} \times \vec{C})=0$, then value of $x$ is _________

Solution

$\begin{aligned} & \vec{B} \times \vec{C}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ -1 & 4 & 3 \\ -8 & -1 & 3\end{array}\right|=15 \hat{i}-21 \hat{j}+33 \hat{k} \\ & \vec{A} \cdot(\vec{B} \times \vec{C})=(-x \hat{i}-6 \hat{j}-2 \hat{k}) \cdot(15 \hat{i}-21 \hat{j}+33 \hat{k}) \\ & 0=-15 x+126-66 \\ & 15 x=60 \\ & x=4\end{aligned}$

Asked in: JEE Main 2024 (06 Apr Shift 1)

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