1. ## vectors orthogonal etc.

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?

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?

2. 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?

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

3. 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.

4. 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.