This is absolutely not related to physics because it involves $3D$ geometry and vectors.
The torque of a force $\vec F = 2i + 3j$ $\hspace{0.4cm}$(in Newtons) ,
acting at a point $(1,0,4) \hspace{0.2cm}$ about the axis $\hspace{0.15cm} y=3x$
is $$\frac{6k}{\sqrt10} Nm$$
find the value of k (integer answer)
here's what i did :
So we start off by finding the closet point on the line $ y = 3x$ from $(1,0,4)$
Doing some calculation should show that the point is $(\frac{1}{10},\frac{3}{10},0)$
now what i did was, i found out the radius vector from this point to $(1,0,4)$
then take cross product with force given to get the torque.
I got this weird answer for torque : $-12i + 8j + 3.3k$
whose modulus is "not looking good" and not coherent with the answer they expect.
where'd i go wrong ? what is the formal and actual way ?
Ah, I figured it out. What you found was the torque about the point $\left(\frac{1}{10}, \frac{3}{10}, 0\right)$. The torque about the axis $y = 3 x$ in the $xy$-plane is the component of this vector in the direction of $(1, 3, 0)$, and note that the torque you found does not point in this direction. What the solution you shared seems to suggest is that it doesn't actually matter which point on the axis you choose to calculate the torque around; that torque should have the same component along the specified axis. This is because any other point on the axis will result in a radius vector that differs from yours by a scalar multiple of the direction vector of the axis, and hence the torque will differ by a vector perpendicular to the direction of the axis, which won't change the component.