vectors orthogonal etc.

**antman** · March 2nd 2009, 06:14 PM

Which of the vectors u1=(1,2), u2=(0,1), u3=(-2,-4), u4=(-2,1), u5=(2,4), u6=(-6,3) are
a. orthogonal?
b. in the same direction?
c. in opposite directions?

my answers
a. u1 & u4, u1 & u6, u3 & u4, u3 & u6, u4 & u5, u5 & u6 because their dot products are 0
b. u1, u2 & u5 AND u4 & u6
c. u3 & u5

does anyone know if I did this right? did I miss anything?

**Soroban** · March 2nd 2009, 06:40 PM

Hello, antman!

Given the vectors: . $\begin{array}{ccc}u_1&=&\langle1,2\rangle \\ u_2&=&\langle0,1\rangle \\u_3&=&\langle\text{-}2,\text{-}4\rangle \\u_4&=& \langle\text{-}2,1\rangle \\u_5&=&\langle2,4\rangle \\ u_6&=&\langle\text{-}6,3\rangle \end{array}$

which are:

a) orthogonal?
b) in the same direction?
c) in opposite directions?

My answers

$a)\;\;u_1, u_4,\;\;u_1, u_6,\;\;u_3,u_4,\;\;u_3,u_6,\;\;u_4,u_5,\;\;u_5,u_ 6$ . . . . Right!

$b)\;\;u_1, {\color{red}\rlap{//}}u_2,u_5,\;\;u_4,u_6$

$c)\;\;u_3, u_5,\;\;{\color{blue}u_1,u_3}$

Pretty good work . . .

**antman** · March 2nd 2009, 06:56 PM

I understand where I went wrong and what I missed from your help, but is there a specific equation to use as a proof? For example, the vectors in the same direction seem to be multiples of one another.

**Reckoner** · March 2nd 2009, 07:35 PM

Originally Posted by antman

For example, the vectors in the same direction seem to be multiples of one another.

If $u_1$ and $u_2$ are multiples of one another (and thus parallel), then you should be able to find a scalar $c$ such that $u_1=cu_2.$ Can you find such a $c?$

If you setup a system of equations to find this value, you will see that the system is inconsistent. Therefore, $\langle1,2\rangle$ and $\langle0,1\rangle$ are not parallel. And if they are not parallel, they cannot be in the same direction.

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