Question 8 — JEE Main 23 Jan 2026 S Ii

Question

Let $\vec{a}=\hat{i}-2 \hat{j}+3 \hat{k}, \vec{b}=2 \hat{i}+\hat{j}-\hat{k}, \vec{c}=\lambda \hat{i}+\hat{j}+\hat{k}$ and $\vec{v}=\vec{a} \times \vec{b}$. If $\vec{v} \cdot \vec{c}=11$ and the length of the projection of $\vec{b}$ on $\vec{c}$ is $p$, then $9 p^{2}$ is equal to
[JEE MAIN 23 Jan 2026 S II]
(1) 9
(2) 6
(3) 4
(4) 12

Solution

$\vec{v}=\vec{a} \times \vec{b}, \quad \vec{v} \cdot \vec{c}=(\vec{a} \times \vec{b}) \cdot \vec{c}=11$
$\Rightarrow[\overrightarrow{\mathrm{a}} \overrightarrow{\mathrm{b}} \overrightarrow{\mathrm{c}}]=11$
$\Rightarrow\left|\begin{array}{ccc}1 & -2 & 3 \\ 2 & 1 & -1 \\ \lambda & 1 & 1\end{array}\right|=1(1+1)+2(2+\lambda)+3(2-\lambda)=11$
$\Rightarrow-\lambda+12=11 \Rightarrow \lambda=1$
Required projection's length of $\vec{b}$ on $\vec{c}=\left|\frac{\vec{b} \cdot \vec{c}}{|\vec{c}|}\right|$
$p=\left|\frac{(2 \hat{i}+\hat{j}-\hat{k}) \cdot(\hat{i}+\hat{j}+\hat{k})}{\sqrt{1+1+1}}\right|=\frac{2+1-1}{\sqrt{3}}=\frac{2}{\sqrt{3}}$
$\mathrm{p}^{2}=\frac{4}{3} \Rightarrow 9 \mathrm{p}^{2}=4 \times 3=12$ Answer(4)