# Find vectors that are perpendicular & parallel

1. Sep 14, 2013

### Painguy

1. The problem statement, all variables and given/known data

1)Give a unit vector
a) in the same direction as v=2i +3j
b)Perpendicular to v

2)
a) Find a vector perpendicular to the plane z=2+3x-y
b)Find a vector parallel to the plane

2. Relevant equations

3. The attempt at a solution

1) a)
Method 1:
u=v/||v|| so i get u=2/√13i + 3/√13j

Method 2:
||u||*||v||cos(θ)=2
arccos(2/√13)=56.31*=θ
||u||=1 so u=cos(56.31)i + sin(56.31)j=2/√13i + 3/√13j

Method 3:
u.v=||v||
(xi + yj).(2i +3j)=√13
2x + 3y=√13

Here i get stuck. How do i do this and is there any other way?

b)
Method 1:
u.v=||u||*||v||cos(θ)=0
1*√13cos(θ)=0

I get stuck here aswell

Method 2: I realized that if i take the angle between v & the x axis I should get the angle between u and the x axis so I can take sin and cos of those angles to get the unit vector perpendicular
u=-3/√13i+2/√13j

The reason I'm doing all these ways is because I'm curious of the most efficient and proper way to solve the problem. My class is currently covering dot products and I was expecting it all to be as simple as u.v=(xi + yj).(2i +3j)=√13 or (xi + yj).(2i +3j)=0 then I would solve algebraically, but I cant seem to figure out how to do it that way.

2)
a) I know that the normal vector is orthogonal to the pane so n=-3i+j+k

is there another way to do this?

b)
n.v=||n||*||v||cos(θ)=0
(-3i+j+k).(xi+yj +zk)=0
(-3x +y +z)=0

I get the equation of a similar plane but not the one they originally gave me.

2. Sep 15, 2013

### vela

Staff Emeritus
If you rewrite your original equation slightly, you have
$$\vec{u}\cdot \frac{\vec{v}}{\|\vec{v}\|} = 1.$$ Any vector that has a component in the direction of $\vec{v}$ with length 1 will satisfy that equation. You'd need another condition to choose only the vector in the same direction as $\vec{v}$.

From here, it follows that $\theta = \pm 90^\circ$. Not very useful.

Do you know about the cross product?

If you think about it, you should see there are an infinite number of vectors that are parallel to the plane, so it's no surprise that the equation you ended up with has an infinite number of solutions.

Do you see what the difference is between the plane described by -3x+y+z=0 and the original plane is?

3. Sep 15, 2013

### voko

You got stuck here because for any vector v there are infinitely many vectors whose dot product with v is v. Take any vector u. You can represent it as a sum of t and n, which are orthogonal to each other, and t is parallel with v and n is perpendicular to v. Clearly u.v = t.v. So if t is a unit vector, then you get the desired result no matter what its n component is.

To use this method, you need an additional condition on u: namely, that it is a unit vector!

The situation is similar here. Now the parallel component must be zero, and the perpendicular component is arbitrary, unless you constrain it with an additional requirement.

If you find two vectors in that plane - can you obtain one perpendicular to them both?

4. Sep 15, 2013

### Painguy

Well the 1st part of the 1st question is asking that vector u be a unit vector which is why i had u.v=||v||.
I can only think of 1 unit vectors that is in the same direction as v
1) u=2/√13i + 3/√13j

I'm aware that there are infinitely many vectors that are parallel with v since I can just scale v, but in this case I'm trying to figure out an algebraic way of finding a unit vector u in the same direction as v. I suppose thats exactly what I did with method 1 & 2? I guess my mistake in method 3 was that i had u=(xi + yj) thereby implying that it is not a unit vector My mistake :tongue:

I am aware of the cross product, but the rest of the class has not yet been introduced to it.

For the second question regarding parallel vectors to the plane, can I choose any 2 points I want on the plane? For example If i choose points x=3 & y=2 i could plug it into the equation z=2+3x-y & get the 1st point P(3,2,9). Then I could set x=0, y=0 and get Q(0,0,2).
PQ=-3i -2j -7j
Just to check that the above vector is in fact parallel to the plane i take the dot product of PQ with the normal.

PQ.n=(-3)(-3)+(-2)(1)+(-7)(1)=0

Does that seem right?

5. Sep 15, 2013

### vela

Staff Emeritus
Oh, I see what you meant. Yes, if you threw in the condition $x^2+y^2=1$, you'd be able to solve for x and y.

If you have two vectors parallel to the plane, their cross product would be perpendicular to the plane. You could also use the dot product to solve for a vector perpendicular to both vectors.

Yes. Did you notice that (-3, -2, -7) is a solution to the equation -3x+y+z=0 you found earlier?