Question: If the vector $x\,\hat i + y\,\hat j -2\,\hat k $ makes an angle of $\pi-\operatorname{arccos}(\frac{1}{3})$ with the positive $y$-axis, the find the value of $x$ and $y$.
What I have done so far:
I've tried using the scalar product for finding the angle;
$a\cdot b = |a||b| \operatorname{cos}(\theta).$
I've tried substituting in $y$ and $x$ but I ended up nowhere.
Best Answer
Your first attempt looks good: take $\vec{a} = (x,y,-2)$ and $\vec{b} = (0,1,0).$ You know that the angle between them will be $\theta = \pi - \arccos(\frac{1}{3})$, so $\cos (\theta) = \cos\pi\cdot\cos(\arccos(\frac{1}{3})) -\sin\pi\cdot\sin(\arccos(\frac{1}{3}))= -\frac{1}{3}.$ Since $\vec{a}\cdot\vec{b} = x\cdot0 + y\cdot 1+ -2\cdot0 = y, \, |\vec{b}| = 1$, you have that $$ y =-\frac{|\vec{a}|}{3} = -\frac{1}{3}\sqrt{x^2+y^2+4} \implies 8y^2 - x^2 = 4 .$$
Note that this is the equation of a hyperbola. So any solution $(x,y)$ of the above equation with $y < 0$ solves the problem. Try to visualize this geometrically!