Question

Find two vectors U and V.
Given:
U+V=<4,0>
U is parallel to <0,-1>
V is perpendicular to <0,-1>

U=?
V=?

Answer 1

I have a question...

Answer 2

Do you understand the dot product?

Answer 3

A little, we just went over that yesterday in class but this is supposedly homework for the section before the dot product so I don't believe its suppose to be used here

Answer 4

Okay, well, notice that U and V are perpendicular.

Answer 5

Since U is parallel to \(\langle 0,-1\rangle \), it is in the same direction. Thus anything perpendicular to \(\langle 0,-1\rangle \) is perpendicular to U.

Answer 6

This means that \[
u\cdot v = 0
\]

Answer 7

This means that: \[
u_1v_1 + u_2v_2 = 0
\]

Answer 8

We also know that \[
u_1+v_1 = 4
\]and \[
u_2+v_2 = 0
\]

Answer 9

Oh wait, one more thing! \[
cu_1 = 0 \\
cu_2 = -1
\]If they are parallel, then the components are proportional!

Answer 10

Okay, so how exactly do I put that all together to get U and V?

Answer 11

Well \[
\mathbf u = \langle u_1,u_2\rangle
\]

Answer 12

\[
cu_1 = 0 \implies u_1 = 0
\]

Answer 13

\[
u_1 +v_1 = 4\implies 0+v_1 = 4 \implies v_1=4
\]

Answer 14

\[
\mathbf u\cdot \mathbf v = 0 \implies u_1v_1 + u_2v_2 = 0 \implies (0)(4) + u_2v_2=0\implies u_2v_2 = 0
\]

Answer 15

\[
u_2+v_2 = 0\implies u_2 = -v_2
\]

Answer 16

\[
u_2v_2=0\implies u_2(-u_2) = 0\implies -u_2^2=0\implies u_2 = 0
\]

Answer 17

Wait. something isn't right here.

Answer 18

I seem to be getting \(\mathbf u = \mathbf 0 = \langle 0,0\rangle\) and \(\mathbf v = \langle 4,0\rangle\).
I suppose that makes sense. It's just a bit weird.

Answer 19

I'm not sure you can say that \(\mathbf 0\) is parallel to any vector.

Answer 20

I typed it into my webwork and it said its correct. must just be one of those weird problems. my professor makes up a lot of his own problems so it might have just been a weird one that he didn't think through

Answer 21

The thing is, for all vectors \(\mathbf 0 \cdot \mathbf v = 0\).

Answer 22

So technically \(\mathbf 0\) is perpendicular to all vectors.

Answer 23

Also, \(c \cdot \mathbf 0 = \mathbf 0\), so \(\mathbf 0\) can only be said to be perpendicular to itself.

Answer 24

I suppose if you let \(c =0\), then for all vectors \(c\mathbf v = \mathbf 0\).