Question

If force \(\vec F=3\hat i+4\hat j−2\hat k\) acts on a particle having position vector \(2\hat i+\hat j+2\hat k\) then, the torque about the origin will be

• A. \(3\hat i+4\hat j−2\hat k\)
• B. \(-10\hat i+10\hat j+5\hat k\)
• C. \(10\hat i+5\hat j-10\hat k\)
• D. \(10\hat i+\hat j-5\hat k\)

Answer 1

The Correct Option is B

To find the torque (\vec{\tau}) about the origin when a force acts on a particle, we use the vector cross product formula:

\(\vec{\tau} = \vec{r} \times \vec{F}\)

where:

• \(\vec{r}\) is the position vector, given as \(2\hat{i} + \hat{j} + 2\hat{k}\)
• \(\vec{F}\) is the force vector, given as \(3\hat{i} + 4\hat{j} - 2\hat{k}\)

The cross product of two vectors \(\vec{A} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}\) and \(\vec{B} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}\) is:

\(\vec{A} \times \vec{B} = (a_2b_3 - a_3b_2)\hat{i} + (a_3b_1 - a_1b_3)\hat{j} + (a_1b_2 - a_2b_1)\hat{k}\)

Substituting the values from the question:

• a_1 = 2, a_2 = 1, a_3 = 2 (components of \(\vec{r}\))
• b_1 = 3, b_2 = 4, b_3 = -2 (components of \(\vec{F}\))

Calculating each component of the torque:

• (\text{i-component}) = (1 \cdot -2) - (2 \cdot 4) = -2 - 8 = -10
• (\text{j-component}) = (2 \cdot 3) - (2 \cdot -2) = 6 + 4 = 10
• (\text{k-component}) = (2 \cdot 4) - (1 \cdot 3) = 8 - 3 = 5

Therefore, the torque vector is:

\(\vec{\tau} = -10\hat{i} + 10\hat{j} + 5\hat{k}\)

The correct answer is:
\(-10\hat i+10\hat j+5\hat k\)

This solution shows the logical and mathematical reasoning required to derive the torque using the cross product of vectors. Each step is crucial for understanding how the forces and positions contribute to torque around a point.