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UP TGT Mathematics 2021 Official Paper

Option 3 : \(-3\vec i+\vec j+\vec k\)

**Concept:**

The** magnitude of the moment of a force **acting about a point is equal to the cross product of force about a particular point (F) and the position vector at that point(r).

i.e., the magnitude of the moment of a force (T) = r × F

**Calculation:**

Given F = \(\vec i+2\vec j+\vec k\)

which passes through a point (2, 3,1) = r(1)

and point (1, 2, -1) = r(2)

the position vector, r = r(1) - r(2)

= (2, 3,1) - (1, 2, -1)

= (1, 1, 2)

r = (1, 1, 2)

So, T = r × F

= \(\left| {\begin{array}{*{20}{c}} \hat{i}&\hat{j}&\hat{k}\\ 1&1&2\\ 1&2&1 \end{array}} \right|\)

\(=\hat{i}(1-4)-\hat{j}(1-2)+\hat{k}(2-1)\)

\(=\hat{i}(-3)+\hat{j}(1)+\hat{k}(1)\)

\(=-3\hat{i}+\hat{j}+\hat{k}\)

The momentum of the force about the point (1, 2, -1) is \(-3\hat{i}+\hat{j}+\hat{k}\).

**Hence,** the correct answer is option **3).**

CT 1: हिन्दी (आदिकाल)

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