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

The Correct Option is A

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:

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