Find vector parallel to yz plane and perpendicular to other given vector

[Raziel2701]

1. The problem statement, all variables and given/known data
Find a vector u that is parallel to the yz-plane and perpendicular to the vector v=<5,0,4>.



2. The attempt at a solution

I first tried to find a unit vector parallel to the yz-plane, I then crossed this vector with vector v, but then I remembered that the resulting vector would be perpendicular to both, so that wouldn't work.

Let me be more precise:

1. I found a unit vector to (0,1,1)

I got (0,1/sqrt(2), 1/sqrt(2))

2. I was going to cross this vector with v (5,0,4) but then I realized that the resulting vector would be perpendicular both the yz plane and vector v so I didn't follow through.


So then :

1. I crossed u(a,b,c) with the yz plane (0,1,1) and got:
(b-c)i -(a)j +(a)k and set it equal to zero because I want them to be parallel.

Clarifying that I'm using a,b, and c as my unknowns.

2.I tried dotting this vector I found with vector v(5,0,4) and I set them equal to zero but now I have an equation with three unknowns but I don't know how to solve them.

Here's what I got:

5b-5c +4a=0

[tiny-tim]

Hi Raziel2701! :wink:
Find a vector u that is parallel to the yz-plane and perpendicular to the vector v=<5,0,4>.

oooh, you're making this so complicated :cry: …

forget cross products, forget unit vectors …

what's a typical vector parallel to the yz-plane? …

now when is it perpendicular to <5,0,4> ? :smile:

[Raziel2701]

I don't think I'm following. Apparently any vector of the form <0,y,z,> as long as y and z are equal should be parallel to the yz plane correct?

So if that is right then that I get, but as to when it is perpendicular to vector <5,0,4>, I honestly don't know. I'm rather stubborn about using the dot product to find what those vector components would be, but I can't get the right answer yet.

Any further hints would be great.

[tiny-tim]

I don't think I'm following. Apparently any vector of the form <0,y,z,> as long as y and z are equal should be parallel to the yz plane correct?

No, you are following, you just think you aren't. :biggrin:

Yes, any (0,b,c) will do it. :smile:
So if that is right then that I get, but as to when it is perpendicular to vector <5,0,4>, I honestly don't know. I'm rather stubborn about using the dot product to find what those vector components would be, but I can't get the right answer yet.


Why this aversion to the dot product … it's much simpler than the cross product anyway! :smile:

So, what is (0,b,c).(5,0,4) ? And what does it need to be? :wink:

[Raziel2701]

Ok, so I got a valid answer, but there's still something that seems wrong.

So I got <0,2,0> for my answer, the middle component can be any number apparently, and this vector is indeed perpendicular with the vector <5,0,4> but unless my math is wrong, the dot product of my answer and <0,1,1> does not give me an angle of zero to indicate that the two vectors are parallel but rather, I get an angle of 45 degrees.

So what's up with that?

And thank you for your help, I like the puzzling, yet helpful advice rather than straight out giving me an answer and an explanation.

[tiny-tim]

Yes, <0,b,0> for any value of b will do.

It lies in the yz-plane, and it's perpendicular to <5,0,4> because the dot product is 5*0 + 0*b + 4*0.

(0,1,1) is a vector you invented in your first post … i did say to forget it! :wink:

ok, now just go back to your fist post and check why you didn't need a unit vector, and why the cross product wouldn't work …

you can use cross product to make a new vector perpendicular to a given vector, but not to check that two given vectors are perpendicular … for that, you need the dot product. :smile: