Question

For given vectors \( \vec{a} = -\hat{i} + \hat{j} + 2\hat{k} \) and \( \vec{b} = 2\hat{i} - \hat{j} + \hat{k} \) where \( \vec{c} = \vec{a} \times \vec{b} \) and \( \vec{d} = \vec{c} \times \vec{b} \). Then the value of \( (\vec{a}-\vec{b}) \cdot \vec{d} \) is:

• A. \( -35 \)
• B. \( 53 \)
• C. \( -52 \)
• D. \( 25 \)

Hint:

When multiple vector operations are involved, compute step-by-step and keep vectors in component form to avoid sign errors.

Answer 1

The Correct Option is A

Concept:

Cross product of vectors is found using the determinant method.
Dot product is computed as the sum of products of corresponding components.
Step 1: Write vectors in component form. \[ \vec{a} = (-1,\,1,\,2), \quad \vec{b} = (2,\,-1,\,1) \]
Step 2: Compute \( \vec{c} = \vec{a} \times \vec{b} \). \[ \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
-1 & 1 & 2
2 & -1 & 1 \end{vmatrix} \] \[ \vec{c} = \hat{i}(1\cdot1 - 2\cdot(-1)) - \hat{j}((-1)\cdot1 - 2\cdot2) + \hat{k}((-1)\cdot(-1) - 1\cdot2) \] \[ \vec{c} = 3\hat{i} + 5\hat{j} - \hat{k} \]
Step 3: Compute \( \vec{d} = \vec{c} \times \vec{b} \). \[ \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
3 & 5 & -1
2 & -1 & 1 \end{vmatrix} \] \[ \vec{d} = \hat{i}(5\cdot1 - (-1)\cdot(-1)) - \hat{j}(3\cdot1 - (-1)\cdot2) + \hat{k}(3\cdot(-1) - 5\cdot2) \] \[ \vec{d} = 4\hat{i} - 5\hat{j} - 13\hat{k} \]
Step 4: Compute \( \vec{a}-\vec{b} \). \[ \vec{a}-\vec{b} = (-1-2,\,1-(-1),\,2-1) = (-3,\,2,\,1) \]
Step 5: Compute the dot product. \[ (\vec{a}-\vec{b}) \cdot \vec{d} = (-3)(4) + (2)(-5) + (1)(-13) \] \[ = -12 - 10 - 13 = -35 \]