Question

Find the two unit vectors in R^2 that are perpendicular to the line y=-2

Answer 1

Consider any two points on \(y\). Let them be \(P_1\) and \(P_2\). \(P_2-P_1\) is the directional vector of \(y\).

Answer 2

So for example... \((0,-2)\) and \((1,-2)\).

Answer 3

\[
(1,-2)-(0,-2) = \langle 1,0\rangle
\]

Answer 4

We can find perpendicular vectors based on the fact that the dot product will equal 0.

Answer 5

so is <1,0> the vector that represents y=-2? or is that one of my perpendicular vectors

Answer 6

It represents the direction of \(y=-2\).

Answer 7

Consider the vector: \(\langle a,b\rangle\)

Answer 8

\[
0 = \langle a,b\rangle \cdot \langle 1,0\rangle = a+0=a
\]

Answer 9

Okay that makes sense. Sorry if I sound completely stupid with this. I'm fairly new to vectors. My professor basically showed us how to add and subtract vectors and dabbled into multiplying them then assigned us this stuff without explaining any of it. So I'm completely lost.

Answer 10

So \(a=0\). Since it is a unit vector then: \[
1^2 = \langle a,b\rangle \cdot \langle a,b\rangle = a^2+b^2 = 0^2+b^2 = b^2\implies b^2=1^2=1
\]

Answer 11

Now, do you know the solutions to: \[
b^2 = 1
\]?

Answer 12

No, you completely lost me

Answer 13

Do you know how to compute: \[
\langle a,b\rangle \cdot \langle a,b\rangle
\]?

Answer 14

a^2+b^2 ?

Answer 15

Yes. Now we already established that \(a=0\), in order to make the vector perpendicular.

Answer 16

okay

Answer 17

Now, do you know what a unit vector is?

Answer 18

a vector with the length of 1

Answer 19

?

Answer 20

do you know how to calculate the length of a vector?

Answer 21

ermmm. I want to say sqrt(a^2+b^2) but I don't think that's right

Answer 22

That is correct.

Answer 23

The best way to remember it is \(\sqrt{v\cdot v}\).

Answer 24

So in our case \[
\sqrt{a^2+b^2}
\]

Answer 25

Now we are saying: \[
1 =\sqrt{a^2+b^2}
\]

Answer 26

We square both sides: \[
1^2 = a^2+b^2
\]

Answer 27

We already determined that \(a=0\) so \(a^2=0\).

Answer 28

\[
1 = b^2
\]

Answer 29

Does this make sense?

Answer 30

so one of the perpendicular vectors is <0,1>?

Answer 31

Yes, that is true...
But you haven't solved for \(b\)! What did you get?

Answer 32

It is quadratic. It could have 2 solutions.

Answer 33

I am so confused...

Answer 34

Okay... let me put it this way...\[
1=x^2
\]So\[
0 = x^2-1
\]

Answer 35

okay, that makes more sense.

Answer 36

So as you can see, there are two values for \(b\)! That is how come we get two possible unit vectors.