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Find vectors that are perpendicular & parallel

  • Thread starter Painguy
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  • #1
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Homework Statement



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

Homework Equations



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.

Thanks for your help in advance.
 

Answers and Replies

  • #2
vela
Staff Emeritus
Science Advisor
Homework Helper
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Homework Statement



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

Homework Equations



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?
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}##.

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

I get stuck here as well.
From here, it follows that ##\theta = \pm 90^\circ##. Not very useful.

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?
Do you know about the cross product?

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.
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
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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?
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!

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

I get stuck here aswell
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.

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?
If you find two vectors in that plane - can you obtain one perpendicular to them both?
 
  • #4
120
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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||.
1)Give a unit vector
a) in the same direction as v=2i +3j
b)Perpendicular to 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 might be misunderstanding your answers :confused:

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
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,580
1,196
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:
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.

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

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

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