beginner problem linear algebra

**LovesMeTrike** · September 18th 2009, 02:54 PM

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

**galactus** · September 18th 2009, 04:19 PM

Try using $cos{\theta}=\frac{u\cdot v}{||u|| \;\ ||v||}$

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

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

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

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

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

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

Parallel.

**LovesMeTrike** · September 19th 2009, 09:49 AM

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?

**aman_cc** · September 19th 2009, 11:27 PM

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

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