Question 7 — JEE Main 23 Jan 2026 (S I)

Question

Let $\vec{a}=-\hat{i}+\hat{j}+2 \hat{k}, \vec{b}=\hat{i}-\hat{j}-3 \hat{k}, \vec{c}=\vec{a} \times \vec{b}$ and $\vec{d}=\vec{c} \times \vec{a}$. Then $(\vec{a}-\vec{b}) \cdot \vec{d}$ is equal to
[JEE MAIN 23 Jan 2026 S I]
(1) 4
(2) -4
(3) -2
(4) 2

Solution

$\vec{d}=\vec{c} \times \vec{a}=(\vec{a} \times \vec{b}) \times \vec{a}$
$\overrightarrow{\mathrm{d}}=-\overrightarrow{\mathrm{a}} \times(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}})=-[(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}) \overrightarrow{\mathrm{a}}-(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{a}}) \overrightarrow{\mathrm{b}}]$
$\overrightarrow{\mathrm{d}}=-[(-1-1-6) \overrightarrow{\mathrm{a}}-(1+1+4) \overrightarrow{\mathrm{b}}]$
$\overrightarrow{\mathrm{d}}=8 \overrightarrow{\mathrm{a}}+6 \overrightarrow{\mathrm{~b}}$
Now $(\vec{a}-\vec{b}) \cdot \vec{d}=(\vec{a}-\vec{b}) \cdot(8 \vec{a}+6 \vec{b})$
$=8 \vec{a} \cdot \vec{a}+6 \vec{a} \cdot \vec{b}-8 \vec{b} \cdot \vec{a}-6 \vec{b} \cdot \vec{b}=8|\vec{a}|^{2}-2 \vec{a} \cdot \vec{b}-6|\vec{b}|^{2}$
$=8(1+1+4)-2(-1-1-6)-6(1+1+9)$
$=48+16-66=-2$ Answer(3)