Scalar triple product can be computed efficiently using the determinant method.
To determine \( \vec{p} \cdot (\vec{q} \times \vec{r}) \), first compute the cross product \( \vec{q} \times \vec{r} \), then compute the dot product with \( \vec{p} \). The given vectors are: \( \vec{p} = 3\hat{i} - \hat{j} + 2\hat{k} \), \( \vec{q} = \hat{i} + 4\hat{j} - \hat{k} \), \( \vec{r} = 2\hat{i} - 3\hat{j} + 5\hat{k} \).
Step 1: Calculate the cross product \( \vec{q} \times \vec{r} \).
| \(\hat{i}\) | \(\hat{j}\) | \(\hat{k}\) | |
|---|---|---|---|
| \(\vec{q}\) | 1 | 4 | -1 |
| \(\vec{r}\) | 2 | -3 | 5 |
The cross product \( \vec{q} \times \vec{r} \) is computed as:
\( \vec{q} \times \vec{r} = \left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 4 & -1 \\ 2 & -3 & 5 \end{array}\right| \)
= \( \hat{i}(4 \cdot 5 - (-1) \cdot (-3)) - \hat{j}(1 \cdot 5 - (-1) \cdot 2) + \hat{k}(1 \cdot (-3) - 4 \cdot 2) \)
= \( \hat{i}(20 - 3) - \hat{j}(5 + 2) + \hat{k}(-3 - 8) \)
= \( \hat{i} \cdot 17 - \hat{j} \cdot 7 - \hat{k} \cdot 11 \)
= \( 17\hat{i} - 7\hat{j} - 11\hat{k} \)
Step 2: Calculate the dot product \( \vec{p} \cdot (\vec{q} \times \vec{r}) \).
\( \vec{p} \cdot (\vec{q} \times \vec{r}) = (3\hat{i} - \hat{j} + 2\hat{k}) \cdot (17\hat{i} - 7\hat{j} - 11\hat{k}) \)
= \( 3 \cdot 17 + (-1) \cdot (-7) + 2 \cdot (-11) \)
= \( 51 + 7 - 22 \)
= \( 36 \)
The result is \( 36 \).