Let $\vec{a}=2 \hat{i}-3 \hat{j}+4 \hat{k}, \vec{b}=3 \hat{i}+4 \hat{j}-5 \hat{k}$ and a vector $\vec{c}$ be…
Solution
$\begin{aligned} & (\vec{a} \cdot \vec{b}) \vec{a}-a^2 \vec{b}+(\vec{a} \cdot \vec{c}) \vec{a}-a^2 \vec{c}+(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{c}=\vec{a} \times(\hat{i}+8 \hat{j}+13 \hat{k}) \\ & \Rightarrow-26 \vec{a}-29 \vec{b}+13 \vec{a}-29 \vec{c}+13 \vec{b}+26 \vec{c}=\vec{a} \times(\hat{i}+8 \hat{j}+13 \hat{k})\end{aligned}$ $\begin{aligned} & \Rightarrow \quad-13 \overrightarrow{\mathrm{a}}-16 \overrightarrow{\mathrm{b}}-3 \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{a}} \times(\hat{\mathrm{i}}+8 \hat{\mathrm{j}}+13 \hat{\mathrm{k}}) \\ & \Rightarrow \quad-13 \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}-16 \mathrm{~b}^2-3 \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=\{\overrightarrow{\mathrm{a}} \times(\hat{\mathrm{i}}+8 \hat{\mathrm{j}}+13 \hat{\mathrm{k}})\} \cdot \overrightarrow{\mathrm{b}} \\ & \Rightarrow \quad(-13)(-26)-16(50)-3 \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=\left|\begin{array}{ccc} 2 & -3 & 4 \\ 1 & 8 & 13 \\ 3 & 4 & -5 \end{array}\right| \\ & \Rightarrow \quad-462-3 \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=-396 \\ & \Rightarrow \quad \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=-22 \end{aligned}$
Hence $24-\vec{b} \cdot \vec{c}=46$
Asked in: JEE Main 2024 (06 Apr Shift 1)