Let $\vec{a}=9 \hat{i}-13 \hat{j}+25 \hat{k}, \vec{b}=3 \hat{i}+7 \hat{j}-13 \hat{k}$ and $\vec{c}=17…

Let $\vec{a}=9 \hat{i}-13 \hat{j}+25 \hat{k}, \vec{b}=3 \hat{i}+7 \hat{j}-13 \hat{k}$ and $\vec{c}=17 \hat{i}-2 \hat{j}+\hat{k}$ be three given vectors. If $\vec{r}$ is a vector such that $\vec{r} \times \vec{a}=(\vec{b}+\vec{c}) \times \vec{a}$ and $\vec{r} \cdot(\vec{b}-\vec{c})=0$, then $\frac{|593 \vec{r}+67 \vec{a}|^2}{(593)^2}$ is equal to___________

Solution

$\begin{aligned} & \overrightarrow{\mathrm{a}}=9 \hat{\mathrm{i}}-13 \hat{\mathrm{j}}+25 \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{b}}=3 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}-13 \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{c}}=17 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=20 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}-12 \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{c}}=-14 \hat{\mathrm{i}}+9 \hat{\mathrm{j}}-14 \hat{\mathrm{k}} \\ & (\overrightarrow{\mathrm{r}}-(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}))_{\times \mathrm{a}}=0 \\ & \mathrm{r}-(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})=\lambda \overrightarrow{\mathrm{a}} \\ & \overrightarrow{\mathrm{r}}=\lambda \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}} \\ & \text { But } \overrightarrow{\mathrm{r}} \cdot(\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{c}})=0 \\ & \Rightarrow(\lambda \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}) \cdot(\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{c}})=0 \\ & \Rightarrow \lambda \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{b}}-\lambda \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{c}}=0 \\ & \lambda=\frac{\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{b}}}{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}}=\frac{294-227}{-389=204}=\frac{-67}{593} \\ & \therefore \overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}-\frac{67}{593} \overrightarrow{\mathrm{a}} \\ & \Rightarrow 593 \overrightarrow{\mathrm{r}}+67 \overrightarrow{\mathrm{a}}=593(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}) \\ & \Rightarrow|\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}|^2=569\end{aligned}$

Asked in: JEE Main 2024 (08 Apr Shift 1)

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