Components of Velocity in the Direction of a Vector $i-3j+2k$

Question

I'm trying to solve this problem:

A particle moves along the curve $$x=2t^2$$ $$y=t^2-4t$$ $$z=3t-5$$ where $$t$$ is the time. Find the components of its velocity at $t=1$ in the direction $$i-3j+2k$$ I started by solving for velocity: $$V=\frac{dx}{dt}i+\frac{dy}{dt}j+\frac{dz}{dt}k$$ This gave me: $$V=4ti+(2t-4)j+3k$$ At $t=1$, we have: $$V=4i-2j+3k$$ Now the issue is how to find the components of this velocity in the direction of the vector: $$i-3j+2k$$

Answer 1

Just project onto that vector: $\frac{(4 i - 2 j + 3 k)\cdot (i - 3 j + 2 k)}{|i - 3 j + 2 k|}$

This is a scalar quantity, the length (norm) of the projection of the velocity onto $(i - 3 j + 2 k)$. If you want a vectorial quantity just multiply by the normalized vector $(i - 3 j + 2 k)/|i - 3 j + 2 k|$ again : $\frac{((4 i - 2 j + 3 k)\cdot (i - 3 j + 2 k))}{|i - 3 j + 2 k|} \frac{(i - 3 j + 2 k)}{|i - 3 j + 2 k|}$