The moment of the force, F = 4 i + 5 j - 6 k at (2, 0, -3) , about the point (2, -2, -2) , is given by: - Physics MCQ NEET (UG) 2018 2026 | ExamTest.live

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The moment of the force, F = 4 \hat{i} + 5 \hat{j} - 6 \hat{k} at (2, 0, - 3), about the point (2, - 2, - 2), is given by:

Solution & Step-by-step Explanation

1. Position vector of point of application: r_{P} = 2 \hat{i} + 0 \hat{j} - 3 \hat{k} .2. Position vector of point about which torque is calculated: r_{O} = 2 \hat{i} - 2 \hat{j} - 2 \hat{k} .3. r = r_{P} - r_{O} = (2 - 2) \hat{i} + (0 - (- 2)) \hat{j} + (- 3 - (- 2)) \hat{k} = 2 \hat{j} - \hat{k} .4. Torque τ = r \times F = (2 \hat{j} - \hat{k}) \times (4 \hat{i} + 5 \hat{j} - 6 \hat{k}) . τ = \hat{i} 04 \hat{j} 25 \hat{k} - 1 - 6 τ = \hat{i} (- 12 - (- 5)) - \hat{j} (0 - (- 4)) + \hat{k} (0 - 8) τ = - 7 \hat{i} - 4 \hat{j} - 8 \hat{k}

Try it yourself before checking the explanation above.

The moment of the force, F = 4 \hat{i} + 5 \hat{j} - 6 \hat{k} at (2, 0, - 3), about the point (2, - 2, - 2), is given by:

- 7 \hat{i} - 8 \hat{j} - 4 \hat{k}

- 4 \hat{i} - \hat{j} - 8 \hat{k}

- 8 \hat{i} - 4 \hat{j} - 7 \hat{k}

- 7 \hat{i} - 4 \hat{j} - 8 \hat{k}

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