Vector IN the yz- plane?

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Homework Statement


Two vectors are given as A=3x+y-2z and B=x-z. A third vector C lying in the yz-plane is perpendicular to A and its magnitude is 7 units. Find the angle between C and B.


Homework Equations


cross and dot product


The Attempt at a Solution


I can't imagine of way, a vector lying on the yz plane can be perpendicular to a 3d vector which has 3 nonzero unit vector components. An explanation on this is much appreciated.
 

Answers and Replies

  • #2
HallsofIvy
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A vector in the "yz-plane" has the form <0, y, z>. Further, two vectors are perpendicular if and only if there dot product is 0. <a, b, c> will be perpendicular to <0, x, y> if and only if by+ cz= 0. For example, <a, z, -y> is perpendicular to <0, y, z>.
 
  • #3
Ray Vickson
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Homework Statement


Two vectors are given as A=3x+y-2z and B=x-z. A third vector C lying in the yz-plane is perpendicular to A and its magnitude is 7 units. Find the angle between C and B.


Homework Equations


cross and dot product


The Attempt at a Solution


I can't imagine of way, a vector lying on the yz plane can be perpendicular to a 3d vector which has 3 nonzero unit vector components. An explanation on this is much appreciated.
A = 3x+y-2z and B=x-z are numbers, not vectors! So, my first question is: what do you mean? Have you copied the questions exactly as they were given to you?
 
  • #4
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I'm not capable of writing exactly as they are given, they are vectors.
 
  • #5
Ray Vickson
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I'm not capable of writing exactly as they are given, they are vectors.
Then I am not capable of helping you.
 
  • #6
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A = 3x+y-2z and B=x-z are numbers, not vectors! So, my first question is: what do you mean? Have you copied the questions exactly as they were given to you?
I'm not capable of writing exactly as they are given, they are vectors.
I'm guessing they're something like this:
a = 3i + j - 2k
b = i - k

They could also be written like this:
a = <3, 1, -2>
b = <1, 0, -1>
 
  • #7
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yeah, i guess i couldn't think of writing it in that way. they were just meant to be x, y and z's with hats. i'll probably be more thoughtful next time.
HallsofIvy, I have found a result using dot product but what really bothered me with this question is not the solution but some things that i couldn't grasp in it. i will look into it tomorrow. thanks for your answer.
Ray Vickson, i hope you will be capable to help me with questions that are well written in the future.
 
  • #8
Ray Vickson
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yeah, i guess i couldn't think of writing it in that way. they were just meant to be x, y and z's with hats. i'll probably be more thoughtful next time.
HallsofIvy, I have found a result using dot product but what really bothered me with this question is not the solution but some things that i couldn't grasp in it. i will look into it tomorrow. thanks for your answer.
Ray Vickson, i hope you will be capable to help me with questions that are well written in the future.
Yes, no problem.

Even if you can't figure out how to write something symbolically, you can explain in words what you mean; for example, you could say that x stands for the unit vector along the x-axis, etc (but I would discourage that choice of notation, for reasons I won't go into here). Or, you could write e_x for the unit vector along the x-axis (making sure to explain your notation). Or, you could say that x really means <1,0,0>, where those are the three components of the vector. Or, you could forget about using x,y and z, and just write what they mean: <1,0,0>, <0,1,0> and <0,0,1>. There are lots of ways to do it; just not the way you tried!
 
  • #9
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I see now that i've had some misconceptions about the problem. I've tried to solve the problem but i am not sure about the result.

Vector A = 3i + j - 2k and vector C is perpendicular to it and it is also lying on the yz plane.

A.C = 0 = 3.Ci + 1.Cj -2.Ck Ci=0 since it is on they yz plane
Cj - 2.Ck= 0 Cj=2Ck
Vector C has the magnitude of 7:
C= 2Ck.j + Ck.k

7² = 49 = 4Ck² + Ck²
Ck = 7/**√5

C= 14/√5.j + 7/√5.k I need the angle between B and C.

B= i - k it is given. It has the magnitude of √2

B.C = 1.0 + 0. 14/√5 -1.7/√5 = √2.7.cosθ

-1/√10 = cosθ
arccos(-1/√10) = 108.4

Sorry for the messy math, do you think this is the answer?
 
  • #10
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I see now that i've had some misconceptions about the problem. I've tried to solve the problem but i am not sure about the result.

Vector A = 3i + j - 2k and vector C is perpendicular to it and it is also lying on the yz plane.
Vector A is very clear, but it would be clearer if you wrote C in a similar form before you start working with it.

The components you give below for vector C are difficult to understand, as Ci could be interpreted as C * i and similarly for Cj and Ck. To make things clearer, let's define C this way:
C = c2j + c3k

I'm using the given information that this vector lies in the y-z plane.
A.C = 0 = 3.Ci + 1.Cj -2.Ck Ci=0 since it is on they yz plane
Cj - 2.Ck= 0 Cj=2Ck
Vector C has the magnitude of 7:
C= 2Ck.j + Ck.k

7² = 49 = 4Ck² + Ck²
The notation above is awful. On the right side, it looks like C * k2.
Ck = 7/**√5
??? What does this (above) mean?
C= 14/√5.j + 7/√5.k I need the angle between B and C.

B= i - k it is given. It has the magnitude of √2

B.C = 1.0 + 0. 14/√5 -1.7/√5 = √2.7.cosθ
This is correct, but it's difficult to follow your reasoning, since you're leaving so much out, and your notation is so unclear.

1.0 and -1.7 appear to be the numbers 1.0, and -1.7, rather than the products 1 * 0 and -1 * 7. It's common practice to use * to indicate multiplication.
-1/√10 = cosθ
arccos(-1/√10) = 108.4

Sorry for the messy math, do you think this is the answer?
What units for 108.4? Radians, degrees?
 
  • #11
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I've done some remediations on it; it should be clearer now.
Vector A= 3i + j - 2k
Vector C = C2j + C3k
I'm using the given information that this vector lies in the y-z plane.

A.C = 0 = 3*0 + 1*C2 - 2*C3
C2 - 2.C3= 0
C2=2C3
C= 2C3j + C3.k
Using the given magnitude of the vector C, which is 7.

7² = 49 = 4C32 +C32

C= 14/√5*j + 7/√5*k
I need the angle between B and C.

B= i - k it is given. It has the magnitude of √2

B.C = 1*0 + 0*14/√5 -1*7/√5 = √2*7*cosθ

-1/√10 = cosθ
arccos(-1/√10) = 108.4°
 
  • #12
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Looks good. Thanks for making the extra effort to make it more readable!
 
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