JEE Advanced 2024 Paper 2 · Q11 · Scalar & Vector Products

Question

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
$$15\hat{i} + 10\hat{j} + 6\hat{k} = \alpha\bigl(2\vec{p} + \vec{q}\bigr) + \beta\bigl(\vec{p} - 2\vec{q}\bigr) + \gamma\bigl(\vec{p} 	imes \vec{q}\bigr),$$ then the value of $\gamma$ is ______.

SolveKar AI · Accepted Answer

Compute $2\vec{p} + \vec{q} = 5\hat{i} + \hat{j} + 7\hat{k}$ and $\vec{p} - 2\vec{q} = 0\hat{i} + 3\hat{j} + \hat{k}$. Cross product $\vec{p} 	imes \vec{q} = \det\!\begin{pmatrix}\hat{i}&\hat{j}&\hat{k}\2&1&3\1&-1&1\end{pmatrix} = 4\hat{i} + \hat{j} - 3\hat{k}$. Equating $\hat{i}, \hat{j}, \hat{k}$ components:
$5\alpha + 4\gamma = 15$,
$\alpha + 3\beta + \gamma = 10$,
$7\alpha + \beta - 3\gamma = 6$.
From $3\cdot$(third) $-$ (second): $20\alpha - 10\gamma = 8$, i.e. $10\alpha - 5\gamma = 4$. Doubling the first: $10\alpha + 8\gamma = 30$; subtract: $13\gamma = 26 \Rightarrow \gamma = 2$. (Then $\alpha = 7/5$, $\beta = 11/5$.) Hence $\gamma = 2$.

JEE Advanced 2024 Paper 2 · Q11 · Scalar & Vector Products

Solution