Question

Consider the planes given by the equations x+y+4y=2, and z-(x+2y)=5.
Find a vector V parallel to the line of intersection of the planes.

Answer 1

x+y+4z=2 and z-(x+2y)=5 right?

Answer 2

The first think you want to do is get the line of intersection.

Answer 3

Actually hold on...

Answer 4

ok

Answer 5

Okay so that's my guess is to find the line of intersection.

Answer 6

I would do that by solving for z in one equation, and substituting that expression into the other equation.

Answer 7

okay so z=0.5-1x/4-1y/4

Answer 8

I would have done z = 5 + x + 2y....

Answer 9

So we get x+y+4(5+x+2y) = 2

Answer 10

okay now what?

Answer 11

Solve for y to get it into y = mx + b form.

Answer 12

our vector should just be <1, m> I'm thinking.

Answer 13

We did this the ghetto way though, I have a feeling they wanted you to do it differently.

Answer 14

We'll see. haha

Answer 15

What did you end up getting?

Answer 16

y=-2x-2 Does that look right to you?

Answer 17

Nope. http://www.wolframalpha.com/input/?i=x%2By%2B4z%3D2+%2C+z-%28x%2B2y%29%3D5

Answer 18

okay so y=-5x/9-2

Answer 19

Yep, so m = -5/9. So our vector is <1, -5/9> or <9, -5>. They didn't specify a magnitude so as long as we remain proportioned, our direction is still correct.

Answer 20

hmm , it says we got the wrong answer, when I put in <1,-5/9> and <9,-5>

Answer 21

Hmmmmm.

Answer 22

I wonder if it wants a specific magnitude...

Answer 23

Hmmm actually

Answer 24

I think we just take the cross product of the normal vector

Answer 25

when I think about it, there is no way our answer is only a two dimensional vector.

Answer 26

Yeah, I was thinking that too.

Answer 27

also our line of intersection was 2D when it should be 3D...

Answer 28

yeah

Answer 29

So do you know how to get the normal vectors? Just get them and do the cross product.

Answer 30

I can for x+y+4z=2 its <1,1,4>

Answer 31

For the other? Just use distributive property to get rid of the parenthesis.

Answer 32

z-x-2y=5?

Answer 33

Yeah, now what is the normal vector?

Answer 34

<-1,-2,1>

Answer 35

So take the cross product. That is the answer.

Answer 36

okay I got (9,-5,-1)

Answer 37

Sweet right answer!... I hate to ask because this question was such a pain, but do you think you could help me with part b) of the question

Answer 38

Ummm really quick

Answer 39

okay!

Answer 40

Find the equation of a plane through the origin which is perpendicular to the line of intersection of these two planes.
This plane is

Answer 41

This isn't any different than the problem we did in the other question. Just find the normal vectors of each plane, get the cross product. That will get you the coefficients.

Answer 42

The point we know is in the plane is (0, 0, 0), since they said it passes through the origin.

Answer 43

Gotta go... good luck.

Answer 44

Thanks man thats great!