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A vector question

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chid:
Hi everyone.

Vector a=xi+yj+zk is perpendicular to vectors -i-j+5k and i-j+k, and has a magnitude of √14 units. Find a.

I know that using the dot product and magnitude x,y,z can be determined after solving simultaneous equations. I was just curious: can the cross product (not part of Spec course) be used in this case? (And if so, is it actually easier?)

Thanks.  :)

enwiabe:
yes it can!

The cross product of -i-j+5k and i-j+k is a vector perpendicular to both vectors

                       |i    j  k|
Cross product =  |-1-1 5|  = i(-1 + 5) - j(-1 - 5) + k(1 + 1) = 4i + 6j + 2k = b (the new vector)
                       |1 -1 1|
Now you want a magnitude of sqrt(14) units. The magnitude of that vector is sqrt(4^2 + 6^2 + 2^2) = sqrt(16 + 36 + 4) = sqrt(56) = 2sqrt(14), so that magnitude is TWO multiplied by sqrt(14). Therefore you want to halve b to get the required vector = 2i + 3j + k. And we know that we got the right answer because it fell out so nicely. Rule #1 of VCE maths. :P
                       

dcc:
The cross product will not always give ALL of the answers to this particular type of question, but you can generally solve these questions by solving the simultaneous equations:







Then multiplying the resultant unit vector(s) by to get the required vector(s).

enwiabe:
Uh, actually, it will give you a scalar multiple of the answer to this question. Always. :P

Neobeo:
Don't forget that there are two answers.

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