A force \vec{F}=2\hat{i}-\lambda\hat{j}+5\hat{k} is applied at the point A(1,2,5). If its moment about the point B(-1,-2,3) is 16\hat{i}-6\hat{j}+2\lambda\hat{k} then what is the value of \lambda?

  1. A. -2 ✓
  2. B. 0
  3. C. 1
  4. D. 2

Correct Answer: A. -2

Explanation

The moment vector is \vec{M} = \vec{r} \times \vec{F}, where \vec{r} = A - B = \langle 2, 4, 2 \rangle. The cross product is (2\hat{i} + 4\hat{j} + 2\hat{k}) \times (2\hat{i} - \lambda\hat{j} + 5\hat{k}) = \hat{i}(20 + 2\lambda) - \hat{j}(10 - 4) + \hat{k}(-2\lambda - 8) = (20 + 2\lambda)\hat{i} - 6\hat{j} - (2\lambda + 8)\hat{k}. We are given \vec{M} = 16\hat{i} - 6\hat{j} + 2\lambda\hat{k}. Equating the \hat{i} components: 20 + 2\lambda = 16 \implies 2\lambda = -4 \implies \lambda = -2. (This is consistent with the \hat{k} component: -(-4 + 8) = -4, which matches 2(-2)).

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