Finding angles in the xy plane with vectors Watch

Mad Man
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Here is the question.
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RDKGames
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(Original post by Mad Man)
Here is the question.
I assume you would know how to answer this if you were given the equation of the plane in the form ax+by+cz = d ??

If so, then you just need to realise that the equation of the xy-plane is simply z=0.
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Mad Man
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(Original post by RDKGames)
I assume you would know how to answer this if you were given the equation of the plane in the form ax+by+cz = d ??

If so, then you just need to realise that the equation of the xy-plane is simply z=0.
No can you explain?
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Mad Man
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Give me a worked answer please.
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RDKGames
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(Original post by Mad Man)
No can you explain?
Okay then, I presume you know how to find an angle between two lines which intersect in 3D, or more specifically, how to find the angle between any two vectors. Look at the diagram below.

Suppose we have some plane \Pi and a line \ell going through it. The angle between these is marked as \theta. One approch to determine it, is to firstly determine what \alpha is, which is presicely the angle between the normal to the plane, \mathbf{n}, and the line \ell. If you can do that, then it's a simple matter of realising that \alpha + \theta = 90, therefore our angle is given by \theta = 90-\alpha.

In your question, you can think of \ell as being your vector \mathbf{a}, and the plane \Pi being the xy-plane z=0. Have a go.

No fully worked solutions will be provided to you, it's not the kind of site where others do your work for you. Post your working out if you get stuck.



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Mad Man
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(Original post by RDKGames)
Okay then, I presume you know how to find an angle between two lines which intersect in 3D, or more specifically, how to find the angle between any two vectors. Look at the diagram below.

Suppose we have some plane \Pi and a line \ell going through it. The angle between these is marked as \theta. One approch to determine it, is to firstly determine what \alpha is, which is presicely the angle between the normal to the plane, \mathbf{n}, and the line \ell. If you can do that, then it's a simple matter of realising that \alpha + \theta = 90, therefore our angle is given by \theta = 90-\alpha.

In your question, you can think of \ell as being your vector \mathbf{a}, and the plane \Pi being the xy-plane z=0. Have a go.

No fully worked solutions will be provided to you, it's not the kind of site where others do your work for you. Post your working out if you get stuck.



I'm so confused, please give me a worked example. I am studying on my own so this isn't for homework.
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Mad Man
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Can you explain the answer?
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RDKGames
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(Original post by Mad Man)
I'm so confused, please give me a worked example. I am studying on my own so this isn't for homework.
Have you covered how to find angles between two vectors, or is that over your head as well?
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Mad Man
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(Original post by RDKGames)
Have you covered how to find angles between two vectors, or is that over your head as well?
I know that Cosx=x/magnitude of vector
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Mad Man
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(Original post by RDKGames)
Have you covered how to find angles between two vectors, or is that over your head as well?
But unlike the latter statement, we are using two planes (x and y) which is confusing.
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Mad Man
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RDKGames
help me please.
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Mad Man
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Help.....
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Mad Man
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(Original post by RDKGames)
Have you covered how to find angles between two vectors, or is that over your head as well?
Thanks for nothing.
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Mad Man
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(Original post by RDKGames)
Okay then, I presume you know how to find an angle between two lines which intersect in 3D, or more specifically, how to find the angle between any two vectors. Look at the diagram below.

Suppose we have some plane \Pi and a line \ell going through it. The angle between these is marked as \theta. One approch to determine it, is to firstly determine what \alpha is, which is presicely the angle between the normal to the plane, \mathbf{n}, and the line \ell. If you can do that, then it's a simple matter of realising that \alpha + \theta = 90, therefore our angle is given by \theta = 90-\alpha.

In your question, you can think of \ell as being your vector \mathbf{a}, and the plane \Pi being the xy-plane z=0. Have a go.

No fully worked solutions will be provided to you, it's not the kind of site where others do your work for you. Post your working out if you get stuck.



Oh I get it now! But why is z=0?
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RDKGames
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(Original post by Mad Man)
Can you explain the answer?
Without good visualisation skills, it's difficult to explain why this answer is logical. But here's a similar problem using the exact same approach that is easier to visualise:

Find the angle between the vector \mathbf{i} + 2 \mathbf{j} + 3\mathbf{k} and the xy-plane.


Hopefully the diagram below helps you visualise this vector, as well as the xy plane. The red plane is the xy plane, and the pink vector is our \mathbf{i} + 2\mathbf{j} + 3\mathbf{k} vector. I have labelled the three components for you on the diagram as well, the 1 across in the x direction, 2 in the y dir, and 3 up in the z dir.

In the second diagram, I have drawn on a teal length which is contained ENTIRELY in the xy plane. It's a two dimensional vector, hopefully you can see how it is the vector \mathbf{i} + 2\mathbf{j}. And because this vector is entirely in the xy plane, then our problem of finding the angle between the pink vecotr and the xy plane, reduces to finding the angle \theta between the pink vector and our teal vector.

What we have going on here are two right-angled triangles to help us do that. One that is lying flat on the xy plane, with sides 1 and 2, and another that is standing upright in the same plane as the pink and teal vector.

Using basic Pythagoras on the flat triangle, you can determine the length of the teal line, and then use basic trig on the standing triangle to determine the angle \theta, because notice that (pink length)*cos(theta) = (teal length). Hence you can work out what cos(theta) is and hence what theta is.



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RDKGames
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(Original post by Mad Man)
Oh I get it now! But why is z=0?
On the xy-plane, every single point has z coordinate as zero. So the entire plane is described by the equation z=0.
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Mad Man
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(Original post by RDKGames)
On the xy-plane, every single point has z coordinate as zero. So the entire plane is described by the equation z=0.
Thanks, how do I increase my visualisation skills?
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Sinnoh
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(Original post by Mad Man)
Thanks, how do I increase my visualisation skills?
Practice more questions with vectors and planes, including sketching them. Really they're not going to come up in A-level maths but if you're interested then have a look at the vectors topics in AS further maths.
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Mad Man
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(Original post by Sinnoh)
Practice more questions with vectors and planes, including sketching them. Really they're not going to come up in A-level maths but if you're interested then have a look at the vectors topics in AS further maths.
Can you explain the answer the textbook gave (above).
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RDKGames
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(Original post by Mad Man)
Can you explain the answer the textbook gave (above).
Look at my example I gave. If you understand the approach there, then textbook simply used the exact same approach but for a different vector.
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