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:

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

Answer 2

Given vectors \( \vec{a} = -5\hat{i} + \hat{j} - 3\hat{k} \), \( \vec{b} = \hat{i} + 2\hat{j} - 4\hat{k} \), and \( \vec{c} = \left( \left( \left( \vec{a} \times \vec{b} \right) \times \hat{i} \right) \times \hat{i} \right) \times \hat{i} \), we need to compute \( \vec{c} \cdot (-\hat{i} + \hat{j} + \hat{k}) \).

Concept Used:

The vector triple product identity is \( \vec{u} \times (\vec{v} \times \vec{w}) = (\vec{u} \cdot \vec{w}) \vec{v} - (\vec{u} \cdot \vec{v}) \vec{w} \). Also, cross product is anticommutative: \( \vec{u} \times \vec{v} = - \vec{v} \times \vec{u} \). For cross products with \( \hat{i} \), we can compute step by step using determinant form.

Step-by-Step Solution:

Step 1: Compute \( \vec{a} \times \vec{b} \).

\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -5 & 1 & -3 \\ 1 & 2 & -4 \end{vmatrix} = \hat{i}(1 \cdot (-4) - (-3) \cdot 2) - \hat{j}((-5) \cdot (-4) - (-3) \cdot 1) + \hat{k}((-5) \cdot 2 - 1 \cdot 1) \] \[ = \hat{i}(-4 + 6) - \hat{j}(20 + 3) + \hat{k}(-10 - 1) = 2\hat{i} - 23\hat{j} - 11\hat{k} \]

Let \( \vec{p} = \vec{a} \times \vec{b} = 2\hat{i} - 23\hat{j} - 11\hat{k} \).

Step 2: Compute \( \vec{p} \times \hat{i} \).

\[ \vec{p} \times \hat{i} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -23 & -11 \\ 1 & 0 & 0 \end{vmatrix} = \hat{i}( (-23)\cdot 0 - (-11)\cdot 0 ) - \hat{j}( 2\cdot 0 - (-11)\cdot 1 ) + \hat{k}( 2\cdot 0 - (-23)\cdot 1 ) \] \[ = -\hat{j}(0 + 11) + \hat{k}(0 + 23) = -11\hat{j} + 23\hat{k} \]

Let \( \vec{q} = \vec{p} \times \hat{i} = -11\hat{j} + 23\hat{k} \).

Step 3: Compute \( \vec{q} \times \hat{i} \).

\[ \vec{q} \times \hat{i} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & -11 & 23 \\ 1 & 0 & 0 \end{vmatrix} = \hat{i}( (-11)\cdot 0 - 23\cdot 0 ) - \hat{j}( 0\cdot 0 - 23\cdot 1 ) + \hat{k}( 0\cdot 0 - (-11)\cdot 1 ) \] \[ = -\hat{j}(0 - 23) + \hat{k}(0 + 11) = 23\hat{j} + 11\hat{k} \]

Let \( \vec{r} = \vec{q} \times \hat{i} = 23\hat{j} + 11\hat{k} \).

Step 4: Compute \( \vec{c} = \vec{r} \times \hat{i} \).

\[ \vec{r} \times \hat{i} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 23 & 11 \\ 1 & 0 & 0 \end{vmatrix} = \hat{i}( 23\cdot 0 - 11\cdot 0 ) - \hat{j}( 0\cdot 0 - 11\cdot 1 ) + \hat{k}( 0\cdot 0 - 23\cdot 1 ) \] \[ = -\hat{j}(0 - 11) + \hat{k}(0 - 23) = 11\hat{j} - 23\hat{k} \]

Thus, \( \vec{c} = 11\hat{j} - 23\hat{k} \).

Step 5: Compute \( \vec{c} \cdot (-\hat{i} + \hat{j} + \hat{k}) \).

\[ \vec{c} \cdot (-\hat{i} + \hat{j} + \hat{k}) = (0\hat{i} + 11\hat{j} - 23\hat{k}) \cdot (-\hat{i} + \hat{j} + \hat{k}) \] \[ = 0\cdot(-1) + 11\cdot 1 + (-23)\cdot 1 = 0 + 11 - 23 = -12 \]

Therefore, \( \vec{c} \cdot (-\hat{i} + \hat{j} + \hat{k}) = \mathbf{-12} \).