Evaluating a general solution to $5\cos \theta -12\sin \theta = 13$ using vectors

Question

Given that $$5\cos \theta -12\sin \theta = 13$$ I'm trying to evaluate a general solution for this equation. It appears I'll be using vector product.

My equation is equivalent to

$$\langle (5,12), (\cos\theta, \sin\theta)\rangle = 13$$

which yields (by Cauch Schwarz Inequality) $$|\langle (5,12), (\cos\theta, \sin\theta)\rangle| \le \|(5,12)\|\|(\cos\theta, \sin\theta)\| = 13$$

This is where I'm stuck.

Regards

Answer 1

Recall that given two vectors in $\mathbb{R^2}$ or $\mathbb{R^3}$ $u$ and $v$ by dot product we have

$$u\cdot v=|u||v|\cos \theta$$

and since $-1\le \cos \theta \le 1$ we have

$$-|u||v|\le u\cdot v\le |u||v|\iff |u\cdot v|\le |u||v|$$

Since $|\cos \theta|=1$ when $\theta=0, \pi$ the equality holds if and only if $u$ and $v$ are multiple vectors.

The result can be generalized for any dimension and it is known as Cauchy-Schwarz inequality.

Answer 2

Hint. \begin{align} 5\cos \theta -12\sin \theta = 13 &\implies -12\sin \theta = 13-5\cos \theta \\ &\implies (-12\sin \theta )^2= (13-5\cos \theta)^2 \\ &\implies 144(1-\cos^2 \theta)= 169-130\cos \theta +\cos^2\theta \\ &\implies 144-144\cos^2 \theta= 169-130\cos \theta +\cos^2\theta \\ &\implies 145\cos^2 \theta-130\cos\theta+25=0 \end{align}