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?
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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....
Isn't it possible that we could find something = 0? Since that is what orthagonal means.
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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.
yes scalars can be negative.
So we are actually solving for them...? I thought it was any number..
or those would be the numbers to = 0?
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."
It doens't say 1 though? Or does that "mutually orthagonal" mean something?
Do there exist scalars k and l such that the vectors
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.
mutually orthogonal means that all three vectors are perpendicular to each other.
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.