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