Question

If \( \vec{a} = \hat{i} + 2\hat{j} - \hat{k} \), \( \vec{b} = 2\hat{i} - \hat{j} + 3\hat{k} \), and \( \vec{c} = -\hat{i} + 3\hat{j} + 2\hat{k} \), find \( \vec{a} \cdot (\vec{b} \times \vec{c}) \).

• A. \( 15 \)
• B. \( -30 \)
• C. \( 20 \)
• D. -20

Hint:

Scalar triple product uses the determinant method for quick computation.

Answer 1

The Correct Option is B

To find \( \vec{a} \cdot (\vec{b} \times \vec{c}) \), we first calculate the cross product \( \vec{b} \times \vec{c} \).

Step 1: Calculate \( \vec{b} \times \vec{c} \)

Let \(\vec{b} = 2\hat{i} - \hat{j} + 3\hat{k}\) and \(\vec{c} = -\hat{i} + 3\hat{j} + 2\hat{k}\). The cross product \(\vec{b} \times \vec{c}\) is given by the determinant:

\(\vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 3 \\ -1 & 3 & 2 \end{vmatrix}\)

Using the formula for the determinant of a 3x3 matrix, we have:

\(\vec{b} \times \vec{c} = \hat{i}((-1)\cdot2 - 3\cdot3) - \hat{j}(2\cdot2 - 3\cdot(-1)) + \hat{k}(2\cdot3 - (-1)\cdot(-1))\)

\(\vec{b} \times \vec{c} = \hat{i}(-2 - 9) - \hat{j}(4 + 3) + \hat{k}(6 - 1)\)

\(\vec{b} \times \vec{c} = \hat{i}(-11) - \hat{j}(7) + \hat{k}(5)\)

\(\vec{b} \times \vec{c} = -11\hat{i} - 7\hat{j} + 5\hat{k}\)

Step 2: Compute \( \vec{a} \cdot (\vec{b} \times \vec{c}) \)

Given \(\vec{a} = \hat{i} + 2\hat{j} - \hat{k}\), the dot product \(\vec{a} \cdot (\vec{b} \times \vec{c})\) is calculated as:

\(\vec{a} \cdot (\vec{b} \times \vec{c}) = (1)(-11) + (2)(-7) + (-1)(5)\)

\(\vec{a} \cdot (\vec{b} \times \vec{c}) = -11 - 14 - 5\)

\(\vec{a} \cdot (\vec{b} \times \vec{c}) = -30\)

Thus, the answer is \(-30\).