Finding the perpendicular components of a vector

Question

Vector $\vec{V} = 3\vec{i}+4\vec{j}$ and also force vector $\vec{F} = 9\vec{i}+12\vec{j}$

"Find the component of $\vec{F}$ parallel to $\vec{V}$"

Now i know that by the dot product $F \cdot D = ||\vec{F}|| \space ||\vec{V}|| \times cos(\theta) = F_{1}V_{1} + F_{2}V_{2} + F_{3}V_{3} $

Accordingly when $\lambda\vec{V} = \vec{F} $ then the two are parallel

But it wants the "component of $\vec{F}$ that parallel to $\vec{V}$" what does it meant to be a component parallel how do we calculate this?

Now i've had hints that it has something to do with trig ratios i.e. "SOH CAH TOA" and it seems that parallel component is a vector of the style $\vec{X} = y\vec{i}+z\vec{j}$

Where y and z calculated.

Answer 1

The parallel component of $\textbf{F}$ parallel to $\textbf{V}$ just means a vector with magnitude $|\textbf{F}| \cos \theta$ where $\theta$ is the angle between $\textbf{F}$ and $\textbf{V}$, and pointing in the direction of $\textbf{V}$. This is also sometimes called "the projection of $\textbf{F}$ onto $\textbf{V}$", or $\textrm{proj}_\textbf{V} \textbf{F}$

Since $\textbf{F}$ is already parallel to $\textbf{V}$, the parallel component is just $\textbf{F}$ itself. For a general solution

$$\textrm{proj}_\textbf{V} \textbf{F} = |\textbf{F}| \cos \theta \frac{\textbf{V}}{|\textbf{V}|} = |\textbf{F}| \frac{\textbf{F}\cdot \textbf{V}}{|\textbf{F}||\textbf{V}|} \frac{\textbf{V}}{|\textbf{V}|} = \frac{\textbf{F}\cdot \textbf{V}}{|\textbf{V}|^2} \textbf{V} $$

The perpendicular vector is just the difference of $\textbf{F}$ and the projection of $\textbf{F}$

$$ \textbf{F}_{\bot \textbf{V}} = \textbf{F} - \textrm{proj}_\textbf{V} \textbf{F} $$

Answer 2

The steps are:

1. You find the magnitude of the projection using the dot product. In your case is 15

2. You determine the unit-vector of the projected vector. In your case (3/5,4/5)

3. You determine the projected vector by magnitude at 1 and unit-vector at 2. In your case, the result is (9, 12)