How do I split a vector into components parallel and perpendicular to a known line?

Question

Given $\mathbf{F}=\langle 7.20,−12.0,28.2\rangle\text{ Newtons}$, find the component of $\mathbf{F}$ that acts perpendicular to member DA such that the vector addition of the perpendicular and parallel components of $\mathbf{F}$ ($\mathbf{F}=\mathbf{F}_\perp + \mathbf{F}_\parallel$) with respect to DA equals $\mathbf{F}$. Express your answer in component form.

$\mathbf{A} = (−5.60,3.68,5.76)$ and $\mathbf{D} = (0,2.72,3.00)$

By finding the position vector of DA, which was $\langle -5.6,0.96,2.76\rangle$, then getting the unit vector of DA, which was $\langle -0.886,0.152,0.437\rangle$ and dotting it with the force vector I got the magnitude of the force along DA, which was $4.11\text{ Newtons}$.

Now I am not sure how to continue this problem. I know that the magnitude of $\mathbf{F}$ minus the magnitude of the force along DA equals the magnitude of $\mathbf{F}_\perp$, but how do I get it in components?

Answer 1

First find the components of $\mathbf{F}_\parallel$. You have the magnitude of the parallel component, $F_\parallel$. You also know the direction of the parallel component, $\hat{\mathbf{F}}_\parallel$. Using these two equations, you can get the components of $\mathbf{F}_\parallel$: $\mathbf{F}_\parallel = F_\parallel \hat{\mathbf{F}}_\parallel$. Now you know the components of $\mathbf{F}_\parallel$.

To get the components of $\mathbf{F}_\perp$, use $\mathbf{F}$ = $\mathbf{F}_\parallel$ + $\mathbf{F}_\perp$. Rearranging gives $\mathbf{F}_\perp$ = $\mathbf{F}-\mathbf{F}_\parallel$. Expessing this equation in component form gives you the components of $\mathbf{F}_\perp$.

By the way you are wrong about "The magnitude of $\mathbf{F}$ minus the magnitude of the force along DA equals the magnitude of $\mathbf{F}_\perp$". You meant to say the squares of the magnitude.