JEE Main 2025 Jan 29 Shift 2, Mathematics Q19: Vector (or Cross) Product of Two Vectors

Compute $\vec{b}\times \vec{c}$:

$\vec{b}\times \vec{c}=((-2)(-1)-3\cdot 3,3\cdot 2-1\cdot (-1),1\cdot 3-(-2)\cdot 2)=(-7,7,7)=7(-1,1,1)$.

So $\hat{a}=\pm \frac{1}{\sqrt{3}}(-1,1,1)$. Testing each sign with $\vec{v}=(1,1,1)$:

$\hat{a}\cdot \frac{\vec{v}}{|\vec{v}|}=\pm \frac{-1+1+1}{3}=\pm \frac{1}{3}$.

Matching the prescribed cosine $-\frac{1}{3}$ forces the minus sign, giving $\hat{a}=\frac{1}{\sqrt{3}}(1,-1,-1)$.

Now impose $\angle (\hat{a},(1,\alpha ,1))=\frac{\pi }{3}$:

$\cos \frac{\pi }{3}=\frac{1}{2}=\frac{1\cdot 1+(-1)\alpha +(-1)(1)}{\sqrt{3}\cdot \sqrt{1+{\alpha }^2+1}}=\frac{-\alpha }{\sqrt{3}\sqrt{{\alpha }^2+2}}$.

Since the right side must be positive, $\alpha <0$. Squaring: $\frac{{\alpha }^2}{3({\alpha }^2+2)}=\frac{1}{4}\Rightarrow 4{\alpha }^2=3{\alpha }^2+6\Rightarrow {\alpha }^2=6$.

$\therefore \alpha =-\sqrt{6}$.