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

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

Aicharya · Accepted Answer

$15 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}=\alpha(2 \overrightarrow{\mathrm{p}}+\overrightarrow{\mathrm{q}})+\beta(\overrightarrow{\mathrm{p}}-2 \overrightarrow{\mathrm{q}})+\gamma(\overrightarrow{\mathrm{p}} 	imes \overrightarrow{\mathrm{q}})$ taking dot with $(\overrightarrow{\mathrm{p}} 	imes \overrightarrow{\mathrm{q}})$ $\left|\begin{array}{ccc}15 & 10 & 6 \ 2 & 1 & 3 \ 1 & -1 & 1\end{array}ight|=0+0+\gamma\left(\mathrm{p}^2 \mathrm{q}^2-(\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{q}})^2ight) \quad\left[\because(\overrightarrow{\mathrm{p}} 	imes \overrightarrow{\mathrm{q}})^2+(\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{q}})^2=\mathrm{p}^2 \mathrm{q}^2ight]$ $\begin{aligned} & \Rightarrow 52=26 \gamma \ & 	herefore \gamma=2\end{aligned}$

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…

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