Title: Vector Algebra – Dot Product Properties – JEE Main 29 Jan 2025 Shift 2

Solution:

First, we find a vector perpendicular to both $\vec{b}$ and $\vec{c}$ by calculating their cross product:
$$ \vec{v} = \vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 2 & 3 & -1 \end{vmatrix} $$
$$ \vec{v} = \hat{i}(2 – 9) – \hat{j}(-1 – 6) + \hat{k}(3 – (-4)) $$
$$ \vec{v} = -7\hat{i} + 7\hat{j} + 7\hat{k} $$

The unit vector $\hat{a}$ is parallel to $\vec{v}$. Normalizing $\vec{v}$, we get two possibilities:
$$ \hat{a} = \pm \frac{1}{\sqrt{3}}(-\hat{i} + \hat{j} + \hat{k}) $$

It is given that the angle between $\hat{a}$ and $\vec{d} = \hat{i} + \hat{j} + \hat{k}$ is $\theta = \cos^{-1}\left(-\frac{1}{3}\right)$, so $\cos \theta = -\frac{1}{3}$.
We check the dot product with $\vec{d}$ for both cases.
Case 1: $\hat{a} = \frac{1}{\sqrt{3}}(-\hat{i} + \hat{j} + \hat{k})$
$$ \hat{a} \cdot \vec{d} = \frac{1}{\sqrt{3}} (-1 + 1 + 1) \frac{1}{|\vec{d}|} = \frac{1}{\sqrt{3}}(1) \frac{1}{\sqrt{3}} = \frac{1}{3} \neq -\frac{1}{3} $$
Case 2: $\hat{a} = -\frac{1}{\sqrt{3}}(-\hat{i} + \hat{j} + \hat{k}) = \frac{1}{\sqrt{3}}(\hat{i} – \hat{j} – \hat{k})$
$$ \hat{a} \cdot \vec{d} = \frac{1}{\sqrt{3}} (1 – 1 – 1) \frac{1}{\sqrt{3}} = -\frac{1}{3} $$
Thus, $\hat{a} = \frac{1}{\sqrt{3}}(\hat{i} – \hat{j} – \hat{k})$.

Now, it is given that $\hat{a}$ makes an angle of $\frac{\pi}{3}$ with $\vec{w} = \hat{i} + \alpha\hat{j} + \hat{k}$.
$$ \cos\left(\frac{\pi}{3}\right) = \frac{\hat{a} \cdot \vec{w}}{|\hat{a}||\vec{w}|} $$
$$ \frac{1}{2} = \frac{\frac{1}{\sqrt{3}}(1 – \alpha – 1)}{1 \cdot \sqrt{1^2 + \alpha^2 + 1^2}} $$
$$ \frac{1}{2} = \frac{-\alpha}{\sqrt{3}\sqrt{\alpha^2 + 2}} $$

Since the LHS is positive, the RHS must be positive, which implies $-\alpha > 0$, so $\alpha < 0$. Squaring both sides: $$ \frac{1}{4} = \frac{\alpha^2}{3(\alpha^2 + 2)} $$ $$ 3(\alpha^2 + 2) = 4\alpha^2 $$ $$ 3\alpha^2 + 6 = 4\alpha^2 $$ $$ \alpha^2 = 6 $$
Since $\alpha < 0$, we have $\alpha = -\sqrt{6}$.Ans. (3)