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!
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2 Answers
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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 $7-7$.
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$\begingroup$ With that being said, are my initial calculations for splitting $p$ correct? $\endgroup$ May 30, 2018 at 10:10
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1$\begingroup$ @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. $\endgroup$– user65203May 30, 2018 at 10:12
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$\begingroup$ Ok thanks. By the way, you cubed 3. Isn't it supposed to be squared? $\endgroup$ May 30, 2018 at 10:13
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$\begingroup$ @AugieJavax98: I am fixing the typos. $\endgroup$– user65203May 30, 2018 at 10:14
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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.
Check your calculation:
$$ (p \times q)_1 = -6/\sqrt{17} - (-1 \cdot -2/\sqrt{17}) = -8 / \sqrt{17} $$
etc.
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$\begingroup$ Will go through again and check. Thanks $\endgroup$ May 30, 2018 at 10:11