JEE Main 2026 — 6 April, Morning Shift — Question 43

Using triple product identity with $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{b}}$ $\overrightarrow{\mathrm{a}} \times(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}})=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}$ $(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}) \overrightarrow{\mathrm{a}}-(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{a}}) \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}$ As $\vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k}\\ 1 & 1 & 1\\ 0 & 1 & -1\end{array}\right|=-2 \hat{i}+\hat{j}+\hat{k}$ Substituting given values $3 \overrightarrow{\mathrm{a}}-3 \overrightarrow{\mathrm{c}}=-2 \hat{\mathrm{i}}+\mathrm{j}+\mathrm{k}$ $3(\hat{i}+\hat{j}+\hat{k})+2 \hat{i}+\hat{j}+\hat{k}=3 \vec{c}$ $\overrightarrow{\mathrm{c}}=\frac{1}{3}(5 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})$ Now $\overrightarrow{\mathrm{c}} \cdot(\overrightarrow{\mathrm{a}}-2 \overrightarrow{\mathrm{~b}})=\frac{1}{3}(5 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}) \cdot(\hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}})$ $=\frac{1}{3}[5-2+6]=3$