JEE Advance - Mathematics (2024 - Paper 2 Online - No. 11)
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 $\gamma$ is ________.
Explanation
$$15 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}=\alpha(2 \overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}})+\beta(\overrightarrow{\mathrm{p}}-2 \overrightarrow{\mathrm{q}})+\gamma(\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}})$$
taking dot with $$(\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}})$$
$$ \begin{aligned} & \left|\begin{array}{ccc} 15 & 10 & 6 \\ 2 & 1 & 3 \\ 1 & -1 & 1 \end{array}\right|=0+0+{\gamma}\left(\mathrm{p}^2 \mathrm{q}^2-(\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{q}})^2\right) \quad\left[\because(\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}})^2+(\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{q}})^2=\mathrm{p}^2 \mathrm{q}^2\right] \\ & \Rightarrow 52=26 \boldsymbol{\gamma} \\ & \therefore \gamma=2 \end{aligned}$$
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