Question:medium

If \( \bar{a} = 2\hat{i} + 3\hat{j} + 4\hat{k}, \bar{b} = \hat{i} - 2\hat{j} - 2\hat{k}, \bar{c} = -\hat{i} + 4\hat{j} + 3\hat{k} \) and if \( \bar{d} \) is vector perpendicular to both \( \bar{b} \) and \( \bar{c} \), \( \bar{a} \cdot \bar{d} = 18 \), then \( |\bar{a} \times \bar{d}|^2 = \)}

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Lagrange's Identity: $|\bar{a} \times \bar{b}|^2 + (\bar{a} \cdot \bar{b})^2 = |\bar{a}|^2 |\bar{b}|^2$.
Updated On: May 16, 2026
  • 640
  • 680
  • 720
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Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Understanding the Question:
We need to find vector \( \bar{d} \) first using the perpendicularity and dot product conditions, then calculate the magnitude squared of the cross product.
Step 2: Key Formula or Approach:
1. \( \bar{d} = \lambda (\bar{b} \times \bar{c}) \).
2. \( \bar{a} \cdot \bar{d} = 18 \implies \lambda \bar{a} \cdot (\bar{b} \times \bar{c}) = 18 \).
3. \( |\bar{a} \times \bar{d}|^2 = |\bar{a}|^2 |\bar{d}|^2 - (\bar{a} \cdot \bar{d})^2 \).
Step 3: Detailed Explanation:
Calculate \( \bar{b} \times \bar{c} \):
\[ \bar{b} \times \bar{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
1 & -2 & -2
-1 & 4 & 3 \end{vmatrix} = \hat{i}(-6 + 8) - \hat{j}(3 - 2) + \hat{k}(4 - 2) = 2\hat{i} - \hat{j} + 2\hat{k} \] Let \( \bar{d} = \lambda (2, -1, 2) \).
Given \( \bar{a} \cdot \bar{d} = 18 \implies (2, 3, 4) \cdot \lambda(2, -1, 2) = 18 \implies \lambda(4 - 3 + 8) = 18 \implies 9\lambda = 18 \implies \lambda = 2 \).
So, \( \bar{d} = (4, -2, 4) \).
Now, calculate magnitudes:
\( |\bar{a}|^2 = 2^2 + 3^2 + 4^2 = 4 + 9 + 16 = 29 \).
\( |\bar{d}|^2 = 4^2 + (-2)^2 + 4^2 = 16 + 4 + 16 = 36 \).
Calculate \( |\bar{a} \times \bar{d}|^2 \):
\[ |\bar{a} \times \bar{d}|^2 = (29)(36) - (18)^2 = 1044 - 324 = 720 \] Step 4: Final Answer:
The value is 720.
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