# Mat angle question

I tried making cos x =c and sinx=s but it,s getting me nowhere and so confused how to finish this.
(edited 8 months ago)
Well, you've let cos be c and sin be s, that's a good start. It's a simultaneous equation after all, so... solve it and see what happens?

EDIT: It really depends on how much you know about systems of equations, which renders this problem trivial. Big Spoilers:

Spoiler

Of course, if you don't know this, ignore it for now.
(edited 8 months ago)
Try the right question this time ...
as well as solving them, you could "sketch"/reason about each of the lines (as a function of theta) so ~y = m(theta)x+c(theta). What do you notice?
(edited 8 months ago)
Original post by tonyiptony
Well, you've let cos be c and sin be s, that's a good start. It's a simultaneous equation after all, so... solve it and see what happens?

EDIT: It really depends on how much you know about systems of equations, which renders this problem trivial. Big Spoilers:

Spoiler

Of course, if you don't know this, ignore it for now.

Ok I solved it algebraically and I got 2s+y=c
so 2 sin theta+y= cos theta and so whatever the value of cos theta and sine theta as you,re adding a certain value of y it,ll work for all values of theta. Also where did you get the rotation anticlowise from as you don,t know what direction it,s going.
Original post by realed
Ok I solved it algebraically and I got 2s+y=c
so 2 sin theta+y= cos theta and so whatever the value of cos theta and sine theta as you,re adding a certain value of y it,ll work for all values of theta. Also where did you get the rotation anticlowise from as you don,t know what direction it,s going.

Hmm... hard to parse what you're saying, but that's basically it. When you solve it outright, no value of theta makes x or y undefined.

As for the rotation thing, it's almost kind of a definition. So obviously you wouldn't know if you've never met them before (perfectly normal at this stage). But it would help if you understand what matrices are as transformations (see 3blue1brown's Essence of Linear Algebra).
(edited 8 months ago)
Original post by realed
Ok I solved it algebraically and I got 2s+y=c
so 2 sin theta+y= cos theta and so whatever the value of cos theta and sine theta as you,re adding a certain value of y it,ll work for all values of theta. Also where did you get the rotation anticlowise from as you don,t know what direction it,s going.

If you get the gradients of each line they're -tan(theta) and cot(theta) so theyre perpenicular (product -1) so what can you say about the solution/intersection point existing?
(edited 8 months ago)
Original post by mqb2766
If you get the gradients of each line they're -tan(theta) and cot(theta) so theyre perpenicular (product -1) so what can you say about the solution/intersection point existing?

That it intersects at 1 point
Original post by tonyiptony
Hmm... hard to parse what you're saying, but that's basically it. When you solve it outright, no value of theta makes x or y undefined.

As for the rotation thing, it's almost kind of a definition. So obviously you wouldn't know if you've never met them before (perfectly normal at this stage). But it would help if you understand what matrices are as transformations (see 3blue1brown's Essence of Linear Algebra).

Wait a sec is the rotation anti-clockwise come from the fact that the rotation matrix is costheta -sin theta
Sin theta cos theta
which looks like the thing in the question and you times it by your original point to get an invariant point of (2,1)
Also you get the determinent thing is 1 by doing ad-bc and because the determinant is non-zero it means that it has an inverse so it can work for any value of theta.
(edited 8 months ago)
Original post by realed
That it intersects at 1 point

So that pretty much answers the question?
Original post by mqb2766
So that pretty much answers the question?

I guess but where do you get the gradient is -tan theta and cos theta
Original post by realed
I guess but where do you get the gradient is -tan theta and cos theta

for a line
ax + by = c
then
y = -a/b x + c/b
Each of the equations is just an equation of the line. So a,b=... here and m=-a/b and intercept c/b. For mat youd be expected to know/transform between the "two" equations of a line.

You need to make sure b!=0 when you rearrange (vertical line), but thats easy to reason about here as the simultaneous equations become trivial.

So really you dont need to solve the equations, just think about what they represent.
(edited 8 months ago)
Original post by realed
Wait a sec is the rotation anti-clockwise come from the fact that the rotation matrix is costheta -sin theta
Sin theta cos theta
which looks like the thing in the question and you times it by your original point to get an invariant point of (2,1)
Also you get the determinent thing is 1 by doing ad-bc and because the determinant is non-zero it means that it has an inverse so it can work for any value of theta.

Hmm... I'm not understanding what your line of logic is. Your second sentence doesn't quite follow your first. But explaining it almost means teaching you the topic on systems of equations (perhaps again). But briefly speaking:

Hopefully, you know a system of equations can be rewritten in the form of Ax=b, A being the matrix of coefficients, x being the solution vector, and b is... well.. the vector whatever's on RHS (I call it the constant vector). b is not the invariant vector (invariant means something else). Might be a good exercise to write down what this Ax=b thing is for this particular question. Remember, this is something you write down - you don't need to calculate anything.
So the solution we seek for is x=A^{-1}b. But that is only valid if A^{-1} exists in the first place. Then the argument on determinant kicks in.

This idea of determinant to check for unique solutions works for any systems.
I talked briefly about rotation here mostly as a sidenote. It gives a geometric meaning as to what the system is trying to solve.