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

Hint:

Use vector operations like dot products and cross products to simplify expressions systematically.

Answer 1

Given: \[ \vec{p} = 2\hat{i} + \hat{j} + 3\hat{k}, \quad \vec{q} = \hat{i} - \hat{j} + \hat{k}. \] Calculate: \[ 2\vec{p} + \vec{q} = 5\mathbf{i} + 3\v{j} + 7\hat{k}, \quad \hat{p} - 2\vec{q} = \hat{i} + 3\mathbf{j} - \hat{k}. \] Cross product: \[ \vec{p} \times \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
2 & 1 & 3
1 & -1 & 1 \end{vmatrix} = 4\hat{i} + \hat{j} - 3\hat{k}. \] Equating components: \[ \alpha(5\hat{i} + 3\hat{j} + 7\hat{k}) + \beta(\hat{i} + 3\hat{j} - \hat{k}) + \gamma(4\hat{i} + \mathbf{j} - 3\hat{k}) = 15\hat{i} + 10\hat{j} + 6\hat{k}. \] From the system of equations: \[ 5\alpha + \beta + 4\gamma = 15, \quad 3\alpha + 3\beta + \gamma = 10, \quad 7\alpha - \beta - 3\gamma = 6. \] Solving: \[ \alpha = \frac{7}{5}, \quad \beta = \frac{11}{5}, \quad \gamma = 2. \]