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Question: If force $\overrightarrow{F} = 3\hat{i} + 4\hat{j} - 2\hat{k}$ acts on a particle having position ve...

If force F→=3i^+4j^−2k^\overrightarrow{F} = 3\hat{i} + 4\hat{j} - 2\hat{k} acts on a particle having position vector 2i^+j^+2k^2\hat{i} + \hat{j} + 2\hat{k} then, the torque about the origin will be:

A

3i^+4j^−2k^3\hat{i} + 4\hat{j} - 2\hat{k}

B

−10i^+10j^+5k^-10\hat{i} + 10\hat{j} + 5\hat{k}

C

10i^+5j^−10k^10\hat{i} + 5\hat{j} - 10\hat{k}

D

10i^+j^−5k^10\hat{i} + \hat{j} - 5\hat{k}

Answer

−10i^+10j^+5k^-10\hat{i} + 10\hat{j} + 5\hat{k}

Explanation

Solution

To find the torque about the origin, we use the formula for torque τ→=r→×F→\overrightarrow{\tau} = \overrightarrow{r} \times \overrightarrow{F}, where r→\overrightarrow{r} is the position vector and F→\overrightarrow{F} is the force vector.

Given:

Position vector r→=2i^+j^+2k^\overrightarrow{r} = 2\hat{i} + \hat{j} + 2\hat{k}

Force vector F→=3i^+4j^−2k^\overrightarrow{F} = 3\hat{i} + 4\hat{j} - 2\hat{k}

We calculate the cross product r→×F→\overrightarrow{r} \times \overrightarrow{F} using the determinant method:

τ→=∣i^j^k^21234−2∣\overrightarrow{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 2 \\ 3 & 4 & -2 \end{vmatrix}

Expand the determinant:

τ→=i^((1)(−2)−(2)(4))−j^((2)(−2)−(2)(3))+k^((2)(4)−(1)(3))\overrightarrow{\tau} = \hat{i} \left( (1)(-2) - (2)(4) \right) - \hat{j} \left( (2)(-2) - (2)(3) \right) + \hat{k} \left( (2)(4) - (1)(3) \right) τ→=i^(−2−8)−j^(−4−6)+k^(8−3)\overrightarrow{\tau} = \hat{i} \left( -2 - 8 \right) - \hat{j} \left( -4 - 6 \right) + \hat{k} \left( 8 - 3 \right) τ→=i^(−10)−j^(−10)+k^(5)\overrightarrow{\tau} = \hat{i} \left( -10 \right) - \hat{j} \left( -10 \right) + \hat{k} \left( 5 \right) τ→=−10i^+10j^+5k^\overrightarrow{\tau} = -10\hat{i} + 10\hat{j} + 5\hat{k}