JEE Main 2024 — 27 January, Shift 1 — Question 15

$\vec{a} \cdot[(\vec{c} \times \vec{b})-\vec{b}-\vec{c}]$

$\overrightarrow{\mathrm{a}} \cdot(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{b}})-\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}$

given $\vec{a} \times \vec{c}=\vec{b}$

$\Rightarrow(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}) \cdot \overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{b}}=|\overrightarrow{\mathrm{b}}|^{2}=27$

$\Rightarrow \overrightarrow{\mathrm{a}} \cdot(\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{b}})=\left[\begin{array}{lll}\overrightarrow{\mathrm{a}} & \overrightarrow{\mathrm{c}} & \vec{b}\end{array}\right]=(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}}) \cdot \overrightarrow{\mathrm{b}}=27$

Now $\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=3-6+3=0$

$$\overrightarrow{a}.\overrightarrow{c}=3$$

By (i),(ii),(iii)& (iv)

27-0-3-=24