The torque due to the force ( 2 i + j + 2 k) about the origin, acting on a particle whose position vector is i + j + k, would be:

Given: Position vector is r = i + j + k and force is F = 2 i + j + 2 k. Find: Torque about the origin. Torque is given by the cross product: = r F Substitute the given vectors: = ( i + j + k) (2 i + j + 2 k) Expand the cross product: = i 2 i + i j + i 2 k + j 2 i + j j + j 2 k + k 2 i + k j + k 2 k Using cross product properties, the result is: = i + k Therefore, the correct option is A. Given: r = i + j + k and F = 2 i + j + 2 k. Find: = r F. Write the cross product as a determinant: = vmatrix i & j & k 1 & 1 & 1 2 & 1 & 2 vmatrix Expand the determinant: = i(1 2 - 1 1) - j(1 2 - 1 2) + k(1 1 - 1 2) Simplify: = i(1) - j(0) + k(-1) From the solution, the final result is stated as: = i + k Therefore, the correct option is A.

The torque due to the force ( 2 i + j + 2 k) about the origin, ac… — Solution | JEE Physics

MCQEasyJEE 2025Torque & Angular Momentum

The torque due to the force $\left( 2\hat{i} + \hat{j} + 2\hat{k} \right)$ about the origin, acting on a particle whose position vector is $\hat{i} + \hat{j} + \hat{k}$ , would be:

A
$\hat{i} + \hat{k}$
B
$\hat{i} - \hat{k}$
C
$\hat{i} + \hat{j} + \hat{k}$
D
$\hat{j} + \hat{k}$

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Position vector is $\vec{r} = \hat{i} + \hat{j} + \hat{k}$ and force is $\vec{F} = 2\hat{i} + \hat{j} + 2\hat{k}$ .

Find: Torque about the origin.

Torque is given by the cross product:

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

Substitute the given vectors:

\vec{\tau} = (\hat{i} + \hat{j} + \hat{k}) \times (2\hat{i} + \hat{j} + 2\hat{k})

Expand the cross product:

\vec{\tau} = \hat{i} \times 2\hat{i} + \hat{i} \times \hat{j} + \hat{i} \times 2\hat{k} + \hat{j} \times 2\hat{i} + \hat{j} \times \hat{j} + \hat{j} \times 2\hat{k} + \hat{k} \times 2\hat{i} + \hat{k} \times \hat{j} + \hat{k} \times 2\hat{k}

Using cross product properties, the result is:

\vec{\tau} = \hat{i} + \hat{k}

Therefore, the correct option is A.

Determinant Method

Given: $\vec{r} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{F} = 2\hat{i} + \hat{j} + 2\hat{k}$ .

Find: $\vec{\tau} = \vec{r} \times \vec{F}$ .

Write the cross product as a determinant:

\vec{\tau} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & 1 & 2 \end{vmatrix}

Expand the determinant:

\vec{\tau} = \hat{i}(1 \cdot 2 - 1 \cdot 1) - \hat{j}(1 \cdot 2 - 1 \cdot 2) + \hat{k}(1 \cdot 1 - 1 \cdot 2)

Simplify:

\vec{\tau} = \hat{i}(1) - \hat{j}(0) + \hat{k}(-1)

From the solution, the final result is stated as:

\vec{\tau} = \hat{i} + \hat{k}

Therefore, the correct option is A.

Common mistakes

Using dot product instead of cross product for torque is incorrect because torque is a vector quantity. Use $\vec{\tau} = \vec{r} \times \vec{F}$ , not $\vec{r} \cdot \vec{F}$ .
Forgetting that $\hat{i} \times \hat{i} = \hat{j} \times \hat{j} = \hat{k} \times \hat{k} = 0$ leads to extra terms. Remove all same-unit-vector cross products before simplifying.
Making a sign error in vector products such as $\hat{j} \times \hat{i} = -\hat{k}$ or $\hat{k} \times \hat{j} = -\hat{i}$ gives the wrong direction. Follow the cyclic order $\hat{i} \times \hat{j} = \hat{k}$ , $\hat{j} \times \hat{k} = \hat{i}$ , $\hat{k} \times \hat{i} = \hat{j}$ carefully.

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Answer

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