# How do I use the given vector equation to resolve vector $p$ into a parallel and perpendicular component?

I am working on the following problem: Here's what I've done so far:  I know that dotting the first component with q should equal one to show that it is parallel and dotting the second component with q should equal to 0 to show that it is equal to zero to show that it is perpendicular. I haven't been getting those two results. Please help!

The parallel component is obtained by the scalar product of $q$ with the dot product of $p$ and $q$, after normalization of $q$.

Hence $$\left(p\cdot\frac{q}{\|q\|}\right)\frac{q}{\|q\|}=\frac{p\cdot q}{\|q\|^2}q$$ i.e.

$$\frac{(3,-2,-1)\cdot(2,-2,3)}{2^2+(-2)^2+3^2}(2,-2,3)=\frac{7}{17}(2,-2,3).$$

The perpendicular component is the difference

$$(3,-2,-1)-\frac{7}{17}(2,-2,3).$$

If you multiply that by $q$, you get $7-7$.

• With that being said, are my initial calculations for splitting $p$ correct? May 30, 2018 at 10:10
• @AugieJavax98: there is enough information in my answer for you to check. At first sight, $14\ne 13\cdot2$, but I noticed a sign error in my answer.
– user65203
May 30, 2018 at 10:12
• Ok thanks. By the way, you cubed 3. Isn't it supposed to be squared? May 30, 2018 at 10:13
• @AugieJavax98: I am fixing the typos.
– user65203
May 30, 2018 at 10:14

I know that dotting the first component with q should equal one to show that it is parallel

Not in general: $$p_\parallel \cdot q = \lVert p_\parallel \rVert \lVert q \rVert \cos \angle(p_\parallel, q) = \lVert p_\parallel \rVert$$

and dotting the second component with q should equal to 0 to show that it is equal to zero to show that it is perpendicular.

$$(p \times q)_1 = -6/\sqrt{17} - (-1 \cdot -2/\sqrt{17}) = -8 / \sqrt{17}$$