# Calculating a vector from a known angle and a known vector

• Michael9272
In summary, the conversation is about determining Vector N' in relation to Vector A and A', and Vector N in relation to Vector A, given certain conditions and information. The participants are trying to find a solution and have different ideas and interpretations about how to approach the problem. They discuss the possibility of using cross products to solve it, but are not sure if it yields the correct answer. They also mention the importance of making sure the vector normal to the plane created by N and A remains normal after the vectors are moved. Despite not being able to fully convince themselves of the solution, they are willing to try it and verify the results through various checks.
Michael9272
I've been wrestling with this problem for several days. I really wanted to solve it myself, but at this point I just keep barking up the same wrong trees. Any help would be greatly appreciated. Here is the problem -

I am trying to determine Vector N', given the following:

Vector A = [5,3,1]
Vector N = [3,2,-21]
Vector A' = [5,3,2] <-- Vector A has moved by one unit in Z

Furthermore,

N is normal to A,
N' is normal to A',
and N' is to A' what N is to A

So far I've determined the following:

theta(A,A') = 9.172 deg = theta(N,N')
ScalarA = 5.916
ScalarN = 21.307
ScalarA' = 6.164

I really want to figure this out on my own, but I am convinced there is some simple relationship that I am missing. If someone could point me in the right direction I'd be grateful.

Would you please explain what is meant by "N' is to A' what N is to A".

Philip,

Thanks for looking at the post. I was afraid that statement would be unclear. What I meant was that as vector A moves to become vector A', vector N experiences the same motion, thus becoming N'.

If you make an "L" with your thumb and your index finger. Think of your thumb as Vector A and your index finger as Vector N. Now rotate (assume no translation) your hand to a new position, your thumb would now be A' and your index finger would be N'.

Does that make sense?

I'm envisaging your vectors A and A' as position vectors, i.e. as giving displacements of points from the origin. In that way I can make sense of your talking about the vector A 'moving' by one z unit, as I suppose you to mean that the point moves by one z unit. Vector A' is slightly longer than vector A and its direction is also different.

A and A' define a plane, P. It's quite easy to picture this plane, as it stands 'upright' on the x-y plane. I find that the vector normal to this plane P is (3a, -5a, 0), in which a is 1/sqrt(34), if the normal is to be of unit length.

Now I think that when you change A to become A' and move N to become N' in the way you describe, then N' will be at the same angle to the normal to the plane as N was. Am I right?

If so, I think I know what to do, but I'm not at all sure that I've interpreted the question properly.

One way to solve it is to define a vector V to be the cross product of N and A (N x A = V). You can say that N' x A' = V as well, so then given an N, A and A', you can solve for the N'.

In this example, N x A = [-65,108,1] = V. N' = [n1,n2,n3] and N' x A' = [ 2*n2 - 3*n3, 5*n3 - 2*n1, 3*n1 - 5*n2]; If N' x A' = V, then N' = [109, 65.3,65.2];

I'm not sure this gives you what you want, but it's a solution. I think there might be other solutions to the system of equations that would get N' close to the original direction of N.

Philip Wood said:
I'm envisaging your vectors A and A' as position vectors, i.e. as giving displacements of points from the origin. In that way I can make sense of your talking about the vector A 'moving' by one z unit, as I suppose you to mean that the point moves by one z unit. Vector A' is slightly longer than vector A and its direction is also different.

A and A' define a plane, P. It's quite easy to picture this plane, as it stands 'upright' on the x-y plane. I find that the vector normal to this plane P is (3a, -5a, 0), in which a is 1/sqrt(34), if the normal is to be of unit length.

Now I think that when you change A to become A' and move N to become N' in the way you describe, then N' will be at the same angle to the normal to the plane as N was. Am I right?

If so, I think I know what to do, but I'm not at all sure that I've interpreted the question properly.

Philip,

I think you are correct, but I haven't been able to fully convince myself of it (sorry, that third dimension is giving me fits). Intuitively it seems to make sense, though and I haven't been able to disprove it.

There are few easy checks we can do to verify the results (for one A' dot N' should equal 0).

Thanks again for the help.

Regards

timthereaper said:
One way to solve it is to define a vector V to be the cross product of N and A (N x A = V). You can say that N' x A' = V as well, so then given an N, A and A', you can solve for the N'.

In this example, N x A = [-65,108,1] = V. N' = [n1,n2,n3] and N' x A' = [ 2*n2 - 3*n3, 5*n3 - 2*n1, 3*n1 - 5*n2]; If N' x A' = V, then N' = [109, 65.3,65.2];

I'm not sure this gives you what you want, but it's a solution. I think there might be other solutions to the system of equations that would get N' close to the original direction of N.

Tim,

Thanks for the reply. I'm not sure the assumption that N' X A' = V is correct. For one, since N' is normal to A', N' dot A' = 0. Also, if V is normal to the plane created by N and A, and N and A move (becoming N' and A') you would expect V to move as well becoming V'. I can imagine a special case where V = V', but that's only for very specific circumstances (I think). Finally, If A' is similar to A, I would expect (perhaps incorrectly) N' to be somewhat similar to N.

I appreciate the help. If I've missed something please let me know.

Regards

Michael,
Using my interpretation of "N' is to A' what N is to A", as explained above, I find N' to be (5.84, 3.70, -20.2). This assumes that N' has the same magnitude (length) as N. My method uses two equations based on dot products, and I'd happily explain it, but...

What makes this problem difficult is interpreting "N' is to A' what N is to A", making me ask: who set this problem? It is, in my opinion, very poorly worded. If I'd been set it, I'd ask the teacher to re-word so it makes proper sense. If it came out of a book, I'd wonder about the competence of the author. If I'd set it to test myself, I'd try and make the wording precise, a useful exercise in itself because it will help you to clarify your ideas, but an exercise that should be abandoned if you're not getting anywhere - there are plenty of tightly-worded problems out there, I'm sure.

Philip Wood said:
Michael,
Using my interpretation of "N' is to A' what N is to A", as explained above, I find N' to be (5.84, 3.70, -20.2). This assumes that N' has the same magnitude (length) as N. My method uses two equations based on dot products, and I'd happily explain it, but...

What makes this problem difficult is interpreting "N' is to A' what N is to A", making me ask: who set this problem? It is, in my opinion, very poorly worded. If I'd been set it, I'd ask the teacher to re-word so it makes proper sense. If it came out of a book, I'd wonder about the competence of the author. If I'd set it to test myself, I'd try and make the wording precise, a useful exercise in itself because it will help you to clarify your ideas, but an exercise that should be abandoned if you're not getting anywhere - there are plenty of tightly-worded problems out there, I'm sure.

Philip,

Regarding the wording of the sentence - this problem is part of a bigger exercise posed to the class by the professor (briefly, we were asked to determine the impact on alignment if the incoming beam is stationary but the reflective surface moves. I've been able to quantify every other aspect of the problem, this is the last piece to the puzzle). The "N' is to A' what N is to A" portion came out of side discussions with the teaching assistant and probably made more sense in the context of the discussion.

In any event, your solution seems to pass the litmus tests. For one, theta(N,N') = 9.172°. Also, A' dot N' is about zero (Actually, I calculated -0.1, but this could just be a rounding error somewhere) I set up I'd be curious to hear what your approach was...

Regards

Michael. Thank you for explaining how the problem arose. Here's my method.

Notation: N' = (p, q, r).

(1) I assume that N' is to have the same magnitude (length) as N', in which case
p2 + q2 + p2 = 32 + 22 + (-21)2 = 454

(2) N' is perpendicular to A', so N'.A' = 0.
So 5p + 3q + 2r = 0

(3) A unit vector normal to the plane containing A and A' is U = (3a, -5a, 0), in which a = 1/sqrt(34).

If N' and N are both at the same angle to U, then, because N' and N have the same magnitude, N'.U = N.U. So
3ap - 5aq = 9a - 10a, that is: 3p - 5q = -1 .

I used the nice simple linear equations from (1) and (2) to express q and r in terms of p. Then I substituted these expressions for q and r into the equation from (1). This gave me a quadratic equation for p. I chose the most plausible root!

Best wishes.

IMHO, the problem that I see is that in 3-space (R3), there are an infinite amount of solutions to the equation N'.A'=0. The constraints you choose allow you to get a solution. In Philip's solution, he chose the constraints that 1) U = A x A' and 2) N'.U = N.U and 3) |N|=|N'|. Even with those 3 constraints, Philip ended up with 2 solutions. As long as you're fine with those constraints, then this is a great solution. However, you might want to check that these constraints can be used in the bigger exercise that this is a part of. I didn't think the problem all the way through when I posted my method, which doesn't even give you a good solution (I posted way too quickly).

## 1. How do you calculate a vector from a known angle and a known vector?

To calculate a vector from a known angle and a known vector, you can use trigonometric functions such as sine, cosine, and tangent. First, you will need to determine the components of the known vector, which can be done by using the angle and the magnitude of the vector. Then, you can use the trigonometric functions to calculate the components of the new vector based on the given angle.

## 2. What is the formula for calculating a vector from a known angle and a known vector?

The formula for calculating a vector from a known angle and a known vector is as follows: V = (|V| * cosθ, |V| * sinθ), where V is the new vector, |V| is the magnitude of the known vector, and θ is the given angle.

## 3. Can the direction of the new vector be determined from the known angle and vector?

Yes, the direction of the new vector can be determined from the known angle and vector. The angle itself provides information about the direction in which the new vector will be pointing, and by using the trigonometric functions, the components of the new vector can be calculated, which will also determine its direction.

## 4. How can you verify the accuracy of the calculated vector?

You can verify the accuracy of the calculated vector by using the Pythagorean theorem. The magnitude of the new vector can be calculated using the components that were determined from the given angle and vector. If the magnitude of the new vector matches the calculated magnitude, then the calculated vector is accurate.

## 5. Is there a difference between calculating a vector in 2D versus 3D?

Yes, there is a difference between calculating a vector in 2D and 3D. In 2D, the vector will only have two components, whereas in 3D, the vector will have three components. Additionally, the trigonometric functions used to calculate the components will also differ slightly in 3D. However, the overall concept and formula for calculating a vector from a known angle and vector remains the same for both 2D and 3D.

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