Let p = 2i + j + 3k and q = i - j + k. If for some real numbers α, β and γ, we have

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

Answer 1

Correct answer: 2

\(2 \vec{p}+\vec{q}=5 \hat{i}+\hat{j}+7 \hat{k}\)

\(\vec{p}-2 \vec{q}=0 \hat{i}+3 \hat{j}+\hat{k} \)

\(\vec{p} \times \vec{q}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 3 \\ 1 & -1 & 1 \end{array}\right|=\hat{i}(4)-\hat{j}(-1)+\hat{k}(-3)\)

\(=4 \hat{i}+\hat{j}-3 \hat{k}\)

\(15 \hat{i}+10 \hat{j}+6 \hat{k}=\alpha(5 \hat{i}+\hat{j}+7 \hat{k})+\beta(3 \hat{j}+\hat{k})+\gamma(4 \hat{i}+\hat{j}-3 \hat{k})\)

\( \therefore 15=5 \alpha+4 \gamma \)

\( 10=\alpha+3 \beta+\gamma\)

\(6=7 \alpha+\beta-3 \gamma \)

\(\therefore \alpha=\frac{7}{5}, \beta=\frac{11}{5}, \gamma=2\)

\(\therefore \gamma=2\)