Let $\vec{p} = 2\hat{i} + \hat{j} + 3\hat{k}$ and $\vec{q} = \hat{i} - \hat{j} + \hat{k}$. If for some real numbers $\alpha, \beta,$ and $\gamma$, we have $$15\hat{i} + 10\hat{j} + 6\hat{k} = \alpha(2\vec{p} + \vec{q}) + \beta(\vec{p} - 2\vec{q}) + \gamma(\vec{p} \times \vec{q})$$ then the value of ...

The correct numerical answer is 2. Click to view the complete step-by-step calculation.

JEE Advanced 2024 Paper 2 Mathematics Question 11 Solution

Let $\vec{a} = 15\hat{i} + 10\hat{j} + 6\hat{k}$ . The given equation is $\vec{a} = (2\alpha + \beta)\vec{p} + (\alpha - 2\beta)\vec{q} + \gamma(\vec{p} \times \vec{q})$ .
Take the dot product of both sides with $(\vec{p} \times \vec{q})$ .
$\vec{a} \cdot (\vec{p} \times \vec{q}) = (2\alpha + \beta)\vec{p} \cdot (\vec{p} \times \vec{q}) + (\alpha - 2\beta)\vec{q} \cdot (\vec{p} \times \vec{q}) + \gamma |\vec{p} \times \vec{q}|^2$ .
Since $(\vec{p} \times \vec{q})$ is perpendicular to $\vec{p}$ and $\vec{q}$ , the first two terms on the RHS are zero.
$\vec{p} \times \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 3 \\ 1 & -1 & 1 \end{vmatrix} = 4\hat{i} + \hat{j} - 3\hat{k}$ .
$\vec{a} \cdot (\vec{p} \times \vec{q}) = (15)(4) + (10)(1) + (6)(-3) = 60 + 10 - 18 = 52$ .
$|\vec{p} \times \vec{q}|^2 = 4^2 + 1^2 + (-3)^2 = 16 + 1 + 9 = 26$ .
Substituting values: $52 = \gamma(26) \implies \gamma = 2$ .