http://img35.imageshack.us/img35/993...arproblem3.png

Not sure how to do this one, can I get some tips? Thanks.

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- Sep 18th 2009, 02:54 PMLovesMeTrikebeginner problem linear algebra
http://img35.imageshack.us/img35/993...arproblem3.png

Not sure how to do this one, can I get some tips? Thanks. - Sep 18th 2009, 04:19 PMgalactus
Try using $\displaystyle cos{\theta}=\frac{u\cdot v}{||u|| \;\ ||v||}$

If you get 1, then $\displaystyle cos^{-1}(1)=0$. That means there is no angle between them and they are parallel.

$\displaystyle u\cdot v$ is the dot product of u and v.

i.e. $\displaystyle u\cdot v=([-6,3,3],[-4,2,2])=36$

$\displaystyle ||u|| \;\ ||v||=\sqrt{(-6)^{2}+3^{2}+3^{2}}\sqrt{(-4)^{2}+2^{2}+2^{2}}=36$

$\displaystyle cos{\theta}=\frac{36}{36}=1$

$\displaystyle cos^{-1}(1)=0$

Parallel. - Sep 19th 2009, 09:49 AMLovesMeTrike
I think it is much simpler than that, we haven't covered any of that material you mentioned. We've only covered topics to do with vectors. I think it has something to do with scalar multiples maybe?

- Sep 19th 2009, 11:27 PMaman_cc
Here is the criteria: u1 ans u2 are parallel if their co-ordinates are in the same proportion. so u1 and u3 are parallel: -6/-4 = 3/2 = 3/2.

so u1 = k x u3 hence parallel. To check if the direction is same check the sign of k