Question

Let \(\vec{a} = \hat{i} + 2\hat{j} + \hat{k}\), \(\vec{b} = 3(\hat{i} - \hat{j} + \hat{k})\). Let \(\vec{c}\) be the vector such that \(\vec{a} \times \vec{c} = \vec{b}\) and \(\vec{a} \cdot \vec{c} = 3\). Then \(\vec{a} \cdot ((\vec{c} \times \vec{b}) - \vec{b} \cdot \vec{c})\) is equal to:

• A. 32
• B. 24
• C. 20
• D. 36

Answer 1

The Correct Option is B

Given vectors:
\(\vec{a} = i + 2j + k, \quad \vec{b} = 3(i - j + k)\)

Let \( \vec{c} \) be a vector such that \( \vec{a} \times \vec{c} = \vec{b} \) and \( \vec{a} \cdot \vec{c} = 3 \). We need to evaluate:

\(\vec{a} \cdot \left[ (\vec{c} \times \vec{b}) - \vec{b} - \vec{c} \right]\)

Step 1. Expression Simplification: Consider:

  \(\vec{a} \cdot \left[ (\vec{c} \times \vec{b}) - \vec{b} - \vec{c} \right] = \vec{a} \cdot (\vec{c} \times \vec{b}) - \vec{a} \cdot \vec{b} - \vec{a} \cdot \vec{c} \quad \text{...(i)}\)

Step 2. Given Conditions: It is given that:

  \(\vec{a} \times \vec{c} = \vec{b}\)

  Therefore:

 \(\vec{a} \cdot (\vec{c} \times \vec{b}) = \vec{b} \cdot \vec{b} = |\vec{b}|^2\)

  Calculating the magnitude:

  \(\vec{b} = 3(i - j + k)\)

 \(|\vec{b}|^2 = 3^2[(1)^2 + (-1)^2 + (1)^2] = 27\)

  Thus:

 \(\vec{a} \cdot (\vec{c} \times \vec{b}) = 27 \quad \text{...(ii)}\)

Step 3. Calculating \( \vec{a} \cdot \vec{b} \):

 \(\vec{a} \cdot \vec{b} = (1)(3) + (2)(-3) + (1)(3) = 3 - 6 + 3 = 0 \quad \text{...(iii)}\)

Step 4. Given \( \vec{a} \cdot \vec{c} \):

\(\vec{a} \cdot \vec{c} = 3 \quad \text{...(iv)}\)

Step 5. Final Calculation: Substituting the values from (ii), (iii), and (iv) into (i):
  
  \(\vec{a} \cdot \left[ (\vec{c} \times \vec{b}) - \vec{b} - \vec{c} \right] = 27 - 0 - 3 = 24\)