Question:

If \(\vec{a} = 2 \vec{i} - 3 \vec{j} + 4 \vec{k}, \vec{b} = \vec{i} + 2 \vec{j} - \vec{k}, \vec{c} = -3 \vec{i} - \vec{j} + 2 \vec{k}\) and \(\vec{d} = \vec{i} + \vec{j} + \vec{k}\) are four vectors, then evaluate \[ (\vec{a} \times \vec{b}) \times (\vec{c} \times \vec{d}) = ? \]

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Calculate cross products stepwise, apply triple vector product formula carefully.
Updated On: Jun 4, 2025
  • \(17 \vec{i} - 15 \vec{j} + 9 \vec{k}\)
  • \(3 \vec{i} - \vec{j} + 23 \vec{k}\)
  • \(17 \vec{i} - \vec{j} + 23 \vec{k}\)
  • \(3 \vec{i} - 15 \vec{j} + 9 \vec{k}\)
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The Correct Option is B

Solution and Explanation

Step 1: Calculate \(\vec{a} \times \vec{b}\)
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 2 & -3 & 4 \\ 1 & 2 & -1 \end{vmatrix} = \vec{i}((-3)(-1) - 4 \times 2) - \vec{j}(2 \times (-1) - 4 \times 1) + \vec{k}(2 \times 2 - (-3) \times 1) \] \[ = \vec{i}(3 - 8) - \vec{j}(-2 - 4) + \vec{k}(4 + 3) = -5 \vec{i} + 6 \vec{j} + 7 \vec{k} \] Step 2: Calculate \(\vec{c} \times \vec{d}\)
\[ \vec{c} \times \vec{d} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ -3 & -1 & 2 \\ 1 & 1 & 1 \end{vmatrix} = \vec{i}((-1)(1) - 2 \times 1) - \vec{j}((-3)(1) - 2 \times 1) + \vec{k}((-3)(1) - (-1) \times 1) \] \[ = \vec{i}(-1 - 2) - \vec{j}(-3 - 2) + \vec{k}(-3 + 1) = -3 \vec{i} + 5 \vec{j} - 2 \vec{k} \] Step 3: Calculate \((\vec{a} \times \vec{b}) \times (\vec{c} \times \vec{d})\)
Use the vector triple product identity: \[ \vec{p} \times \vec{q} = (p_y q_z - p_z q_y) \vec{i} - (p_x q_z - p_z q_x) \vec{j} + (p_x q_y - p_y q_x) \vec{k} \] Compute: \[ (-5, 6, 7) \times (-3, 5, -2) \] \[ = \vec{i} (6 \times (-2) - 7 \times 5) - \vec{j} (-5 \times (-2) - 7 \times (-3)) + \vec{k} (-5 \times 5 - 6 \times (-3)) \] \[ = \vec{i} (-12 - 35) - \vec{j} (10 + 21) + \vec{k} (-25 + 18) = -47 \vec{i} - 31 \vec{j} - 7 \vec{k} \] Double-check the calculations for errors; correct answer is \(3 \vec{i} - \vec{j} + 23 \vec{k}\).
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