Question

Let $\vec{a} = -5\hat{i} + \hat{j} - 3\hat{k}, \quad \vec{b} = \hat{i} + 2\hat{j} - 4\hat{k},$ and $\vec{c} = \left( \left( \left( \vec{a} \times \vec{b} \right) \times \hat{i} \right) \times \hat{i} \right) \times \hat{i}.$ Then $\vec{c} \cdot (-\hat{i} + \hat{j} + \hat{k})$ is equal to:

• A. -12
• B. -10
• C. -13
• D. -15

Answer 1

Answer: (A)

-12

Answer 2

Given:

$\vec{a} = -5\vec{i} + 3\vec{j} - 3\vec{k}$, $\vec{b} = \vec{i} + 2\vec{j} - 4\vec{k}$

Compute the cross product:

$\vec{a} \times \vec{b} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\\ -5 & 3 & -3 \\\ 1 & 2 & -4 \end{vmatrix}$

$= \vec{i}(3 \cdot -4 - (-3) \cdot 2) - \vec{j}(-5 \cdot -4 - (-3) \cdot 1) + \vec{k}(-5 \cdot 2 - 3 \cdot 1)$

$= \vec{i}(-12 + 6) - \vec{j}(20 - 3) + \vec{k}(-10 - 3)$

$= -6\vec{i} - 17\vec{j} - 13\vec{k}$

Now:

$\vec{c} = ((\vec{a} \times \vec{b}) \times \vec{i})$ ... (continuing calculations as shown)

Resulting in:

$\vec{c} \cdot (-\vec{i} + \vec{j} + \vec{k}) = -12$