Identify the point of application of force

The force is applied at the intersection of:

• plane $x=2$
• x-axis

On the x-axis, we have $y=0$ and $z=0$.

Since also $x=2$, the point of application is: $A=(2,0,0)$

Given force vector

$\vec F = 4\hat i + 3\hat j + 4\hat k$

Point about which torque is to be calculated

$P=(2,3,4)$

Position vector from point $P$ to point of application $A$

$\vec r = \overrightarrow{PA} = (2-2)\hat i + (0-3)\hat j + (0-4)\hat k$ $\vec r = 0\hat i -3\hat j -4\hat k$

Torque formula

Torque about point $P$ is: $\vec \tau = \vec r \times \vec F$

So,

Compute the cross product

$\vec \tau = \hat i[(-3)(4)-(-4)(3)] - \hat j[(0)(4)-(-4)(4)] + \hat k[(0)(3)-(-3)(4)]$

$\vec \tau = \hat i(-12+12) - \hat j(0+16) + \hat k(0+12)$

$\vec \tau = 0\hat i -16\hat j +12\hat k$

Magnitude of torque

$|\vec \tau| = \sqrt{0^2+(-16)^2+12^2}$ $|\vec \tau| = \sqrt{256+144}$ $|\vec \tau| = \sqrt{400} = 20$

Nearest integer

$20$

Comparison with stored answer

Derived answer = $20$

Stored correct answer = $20$

Hence, the answer matches.