# Ask Uncle Colin: A Vector Problem

Dear Uncle Colin,

I’m told that the resultant of the vectors $\bb{a} = 3\bi - 2\bj$ and $\bb{b} = p\bi - 2p\bj$ is parallel to $\bb{c} = 2\bi - 3\bj$ - and I need to find $p$ and the resultant. I don’t really know where to begin!

- A Rotten Resultant, Or Worse

Hi, ARROW, and thanks for your message! Let’s break the question down.

We’re looking at the *resultant* of two vectors, which means “what you get when you add them together”. In this case, that’s $\bb{a} + \bb{b} = (3+p)\bi + (-2-2p)\bj$. (Because I’ve spend the last few decades losing minus signs, I try to remember to put them in place explicitly after the + in between components here.)

Now, this resultant is *parallel* to $\bb{c}$, which means it’s a positive multiple of it. For some value of $k>0$, $(3+p)\bi + (-2-2p)\bj = 2k\bi - 3k\bj$.

That can only be true if the $\bi$ components are equal and the $\bj$ components are equal, so $3+p=2k$ and $2+2p = 3k$. We’ve got a simultaneous equation!

Since we’re looking for $p$, let’s eliminate $k$. Multiplying the first by three and the second by two gives:

- $9 + 3p = 6k$
- $4 + 4p = 6k$

So, $9+3p = 4+4p$ and $p=5$.

The resultant is then $8\bi - 12 \bj$, which is indeed $4\bb{c}$.

Hope that helps!

- Uncle Colin