Cross product gives required vector:-
| i"""""j"""""k |
|0""""-2""""1|
|1""""0"""""3|
- 6i + j + 2k is perpendicular to vectors u and v.
2007-10-01 07:07:13
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answer #1
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answered by Como 7
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Okay, I think with i,j,k being vector, then i0-2j+k=0 and
i+0j+3k=0.
So, lets multiply first equation by -3, giving you 6j-3k=0.
And add i+3k=0, giving you 6j+i=0, or i=-6j.
And you know that k=2j.
So, just plug in any number in i,j, or k.
Well, just let i=1.SO, 1=-6j, so, j=-1/6.
And k=2*(-1/6)=-1/3.
So, that gives you (1,-1/6,-1/3), or to get all whole numbers,
multiply by 6, giving you (6,-1,-2).
2007-09-27 21:26:25
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answer #2
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answered by yljacktt 5
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