Vector Algebra
Cross Product and Moment
Grade None
Question:
<p>Three forces $\vec{F}_1 = \vec{i} + 2\vec{j} - 3\vec{k}$, $\vec{F}_2 = 2\vec{i} + 3\vec{j} + 4\vec{k}$ and $\vec{F}_3 = \vec{i} - \vec{j} + \vec{k}$ acting on a particle at the point $(0, 1, 2)$. The magnitude of the moment of the forces about the point $(1, -2, 0)$ is</p>
<p>(a) $2\sqrt{35}$</p>
<p>(b) $6\sqrt{10}$</p>
<p>(c) $4\sqrt{7}$</p>
<p>(d) None of these</p>
Step-by-Step Solution
Key Concept: The moment of forces about a point is found using the cross product of position vector from that point to the line of action and the total force vector.
Step 1: Calculate total force $\vec{F} = \vec{F}_1 + \vec{F}_2 + \vec{F}_3 = (\hat{i} + 2\hat{j} - 3\hat{k}) + (2\hat{i} + 3\hat{j} + 4\hat{k}) + (\hat{i} - \hat{j} + \hat{k}) = 4\hat{i} + 4\hat{j} + 2\hat{k}$ Step 2: Find position vector $\vec{PA} = (0-1)\hat{i} + (1-(-2))\hat{j} + (2-0)\hat{k} = -\hat{i} + 3\hat{j} + 2\hat{k}$ Step 3: Calculate moment $\vec{M} = \vec{PA} \times \vec{F} = (-\hat{i} + 3\hat{j} + 2\hat{k}) \times (4\hat{i} + 4\hat{j} + 2\hat{k})$ Step 4: Using determinant form: $\vec{M} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 3 & 2 \\ 4 & 4 & 2 \end{vmatrix} = \hat{i}(6-8) - \hat{j}(-2-8) + \hat{k}(-4-12) = -2\hat{i} + 10\hat{j} - 16\hat{k}$ Step 5: Magnitude $|\vec{M}| = \sqrt{4 + 100 + 256} = \sqrt{360} = 6\sqrt{10}$ ∴ Answer is (b) $6\sqrt{10}$
Correct Answer: B