Let $\overrightarrow{\mathrm{a}}=\hat{i}-3 \hat{j}+7 \hat{k}, \overrightarrow{\mathrm{b}}=2…

Let $\overrightarrow{\mathrm{a}}=\hat{i}-3 \hat{j}+7 \hat{k}, \overrightarrow{\mathrm{b}}=2 \hat{i}-\hat{j}+\hat{k}$ and $\overrightarrow{\mathrm{c}}$ be a vector such that $(\overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}=3(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}})$. If $\vec{a} \cdot \vec{c}=130$, then $\vec{b} \cdot \vec{c}$ is equal to _______

Solution

$\begin{aligned} & (\overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{c}}=3(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}) \\ & (2 \overrightarrow{\mathrm{b}}+4 \overrightarrow{\mathrm{a}}) \times \overrightarrow{\mathrm{c}}=0 \\ & \overrightarrow{\mathrm{c}}=\lambda(4 \overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{b}})=\lambda(8 \hat{\mathrm{i}}-14 \hat{\mathrm{j}}+30 \hat{\mathrm{k}}) \\ & \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}=130 \\ & 8 \lambda+42 \lambda+210 \lambda=130 \\ & \lambda=\frac{1}{2} \\ & \overrightarrow{\mathrm{c}}=4 \hat{\mathrm{i}}-7 \hat{\mathrm{j}}+15 \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=8+7+15=30\end{aligned}$

Asked in: JEE Main 2024 (05 Apr Shift 1)

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