Find Coefficient Gamma in Linear Combination of Vector Expression

To find the value of the scalar $\gamma$, we start by defining the given vectors: $\vec{p} = 2\hat{i} + \hat{j} + 3\hat{k}$ $\vec{q} = \hat{i} - \hat{j} + \hat{k}$ $\vec{v} = 15\hat{i} + 10\hat{j} + 6\hat{k}$

The given vector equation is: $\vec{v} = \alpha \left( 2\vec{p} + \vec{q} \right) + \beta \left( \vec{p} - 2\vec{q} \right) + \gamma \left( \vec{p} \times \vec{q} \right)$

First, let's calculate the cross product $\vec{p} \times \vec{q}$: $\vec{p} \times \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 3 \\ 1 & -1 & 1 \end{vmatrix}$ $\vec{p} \times \vec{q} = \hat{i}\big(1 - (-3)\big) - \hat{j}\big(2 - 3\big) + \hat{k}\big(-2 - 1\big) = 4\hat{i} + \hat{j} - 3\hat{k}$

Now, recall that the cross product $\vec{p} \times \vec{q}$ is orthogonal (perpendicular) to both $\vec{p}$ and $\vec{q}$. Therefore, any linear combination of $\vec{p}$ and $\vec{q}$ is also orthogonal to $\vec{p} \times \vec{q}$: $\left( 2\vec{p} + \vec{q} \right) \cdot \left( \vec{p} \times \vec{q} \right) = 0$ $\left( \vec{p} - 2\vec{q} \right) \cdot \left( \vec{p} \times \vec{q} \right) = 0$

Taking the dot product of both sides of the original vector equation with $\left( \vec{p} \times \vec{q} \right)$: $\vec{v} \cdot \left( \vec{p} \times \vec{q} \right) = \alpha \underbrace{\left[ (2\vec{p} + \vec{q}) \cdot (\vec{p} \times \vec{q}) \right]}_{0} + \beta \underbrace{\left[ (\vec{p} - 2\vec{q}) \cdot (\vec{p} \times \vec{q}) \right]}_{0} + \gamma |\vec{p} \times \vec{q}|^2$

This simplifies to: $\vec{v} \cdot \left( \vec{p} \times \vec{q} \right) = \gamma |\vec{p} \times \vec{q}|^2$

Next, we compute the required dot product and magnitude squared:

1. $\vec{v} \cdot \left( \vec{p} \times \vec{q} \right) = (15)(4) + (10)(1) + (6)(-3) = 60 + 10 - 18 = 52$
2. $|\vec{p} \times \vec{q}|^2 = 4^2 + 1^2 + (-3)^2 = 16 + 1 + 9 = 26$

Substituting these values into our simplified equation: $52 = \gamma \cdot 26$ $\gamma = \frac{52}{26} = 2$