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Linear Algebra 6 Online
OpenStudy (konradzuse):

Do there exist scalars k and l such that the vectors u = (2,k,6) , v = (l,5,3) , and w = (1,2,3) are mutually orthogonal with respect to the Euclidean inner product?

OpenStudy (konradzuse):

The answer says no, but I originally thought that since k and l were u2 and v1 that itonly matters the 3rd spot, but I realized that you do (u1v1w1) +etc....

OpenStudy (konradzuse):

Isn't it possible that we could find something = 0? Since that is what orthagonal means.

OpenStudy (konradzuse):

Oh weait can scalars be negative?

OpenStudy (anonymous):

if all three vectors were orthogonal, then:\[u\cdot w = 0\Longrightarrow (2)(1)+(k)(2)+(6)(3)=0\]\[2+2k+18=0\Longrightarrow 2k=-20\Longrightarrow k=-10\]Similarly,since v and w are orthogonal, we get that l must be -19.

OpenStudy (anonymous):

yes scalars can be negative.

OpenStudy (konradzuse):

So we are actually solving for them...? I thought it was any number..

OpenStudy (konradzuse):

or those would be the numbers to = 0?

OpenStudy (anonymous):

Since we are asking "does there exist", the question is asking "is there any one such number k and l." Its not the same as "for all/any k and l."

OpenStudy (konradzuse):

It doens't say 1 though? Or does that "mutually orthagonal" mean something?

OpenStudy (konradzuse):

Do there exist scalars k and l such that the vectors

OpenStudy (anonymous):

If u and w are orthogonal, then k would have to be -10. If k is any other number, they wont be orthogonal since the inner/dot product wouldnt come out to zero.

OpenStudy (anonymous):

mutually orthogonal means that all three vectors are perpendicular to each other.

OpenStudy (anonymous):

The thing is, there is no way these three vectors can be mutually orthogonal. Since k would have to be -10, and l would have to be -19. Then u and v wouldnt be orhtogonal. There is no way to get all three to be perpendicular at the same time.

OpenStudy (konradzuse):

I see, makes sense, thanks!

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