Finding angles in the xy plane with vectors

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 $z=0$.

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.

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?

Have you covered how to find angles between two vectors, or is that over your head as well?

Have you covered how to find angles between two vectors, or is that over your head as well?

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.

Can you explain the answer?

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 $z=0$.

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.

Can you explain the answer the textbook gave (above).