charlygyloavao9

2022-10-08

Find the direction cosines of the a vector in the plane of $\stackrel{\to }{b}=<2,1,3>$ and $\stackrel{\to }{c}=<-1,2,1>$ and perpendicular to $\stackrel{\to }{c}$
Let the vector $\stackrel{\to }{a}=⟨x,y,z⟩$
Since it is perpendicular to $\stackrel{\to }{c}$
$-x+2y+z=0$
And
$|\begin{array}{ccc}x& y& z\\ 2& 1& 3\\ -1& 2& 1\end{array}|$
$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}-5x-7y+5z=0$
since a, b, c are coplanar
So y=0
Then x=y
So the answer should be $⟨\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}}⟩$ , but that is not correct.
What is wrong with my solution?

Piper Pruitt

Let $\stackrel{\to }{r}=p\stackrel{\to }{b}+q\stackrel{\to }{c}$, take dot product of this with $\stackrel{\to }{c}$, and use $\stackrel{\to }{r}.\stackrel{\to }{c}=0$, then
$\stackrel{\to }{r}.\stackrel{\to }{c}\stackrel{\to }{=}p\stackrel{\to }{b}.\stackrel{\to }{c}+q\stackrel{\to }{c}.\stackrel{\to }{c}\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}p=-q\frac{{c}^{2}}{\stackrel{\to }{b}.\stackrel{\to }{c}}=-q\frac{6}{3}=-2q.$
So the required vector $\stackrel{\to }{r}$ is:
$\stackrel{\to }{r}=q\left[-2\stackrel{\to }{b}+\stackrel{\to }{c}\right]=q\left(-5\stackrel{\to }{i}-5\stackrel{\to }{k}\right).$
q is arbitrary real number. Next $\stackrel{^}{r}=\frac{\stackrel{\to }{i}+\stackrel{\to }{k}}{\sqrt{2}}$
It also means that

namely, the vector triple product.

Aidyn Crosby

Note that the projection of $\stackrel{\to }{b}$ onto the direction perpendicular to $\stackrel{\to }{c}$ is
$\stackrel{\to }{b}-\left(\stackrel{\to }{b}\cdot \stackrel{\to }{c}\right)\frac{\stackrel{\to }{c}}{|c{|}^{2}}=<2,1,3>-\frac{1}{2}<-1,2,1>=<\frac{5}{2},0,\frac{5}{2}>$
Normalize to get $\stackrel{\to }{a}=<\frac{1}{\sqrt{2}},0,\frac{1}{\sqrt{2}}>$

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