Line enclosing an angle with another line

Question

A question states,

Lines $L_1:x+\sqrt 3y=2$, and $L_2:ax+by=1$ meet at $P$ and enclose an angle of $45^\mathrm{o}$ between them. A line $L_3:y=\sqrt 3x$, also passes through $P$. Find $a^2+b^2$.

The solutions I've seen say the answer is 2, taking the line $\frac{1-\sqrt3}{2}x+\frac{1+\sqrt3}{2}y=1 $ to be the required line, and that makes sense. However, the question seems wrong to me, as there's the other line $(2+\sqrt 3) x-y=1$ which satisfies the conditions and yet doesn't give the same value of $a^2+b^2$.

It seems like a trivial mistake, but I've seen it done this way in multiple solutions, so is it indeed a mistake or am I missing something?

Answer 1

I see my mistake now; I initially had the figure down right, with the line passing through $(\frac 1 2, \frac{\sqrt3}{2} ) $ and making a $-75°$ with the x-axis being the one I had been having trouble with. I just ended up using $\tan75°=(2+\sqrt3)$ instead of its negative in the point-slope form, and that threw me off.

Answer 2

Let $\frac{a}{b}=m$.

Thus, $$\left|\frac{m-\frac{1}{\sqrt3}}{1+\frac{m}{\sqrt3}}\right|=1$$ and $$\frac{a}{2}+\frac{b\sqrt3}{2}=1,$$ which gives

$$m-\frac{1}{\sqrt3}=-1-\frac{m}{\sqrt3}$$ which is $$(\sqrt3+1)m=1-\sqrt3,$$ which is $$m=-\frac{\sqrt3-1}{\sqrt3+1},$$ which is $$m=-\frac{(\sqrt3-1)^2}{2},$$ which is $$m=\sqrt3-2$$ or $$m-\frac{1}{\sqrt3}=1+\frac{m}{\sqrt3}$$ which is $$(\sqrt3-1)m=\sqrt3+1,$$ which is $$m=\frac{\sqrt3+1}{\sqrt3-1},$$ which is $$m=\frac{(\sqrt3+1)^2}{2},$$ which is $$m=2+\sqrt3.$$

In the first case we obtain:

$$\frac{a}{b}=\sqrt3-2$$ and $$a+b\sqrt3=2,$$ which gives $$(\sqrt3-2)b+b\sqrt3=2,$$ which is $$(\sqrt3-1)b=1,$$ which is $$b=\frac{\sqrt3+1}{2}$$ and $$a=\frac{(\sqrt3-2)(\sqrt3+1)}{2}=\frac{1-\sqrt3}{2}$$

The second case gives $$\frac{a}{b}=2+\sqrt3$$ and $$a+b\sqrt3=2,$$ which gives $$(2+\sqrt3)b+b\sqrt3=2,$$ which is $$(\sqrt3+1)b=1,$$ which is $$b=\frac{\sqrt3-1}{2}$$ and $$a=\frac{(2+\sqrt3)(\sqrt3-1)}{2}=\frac{1+\sqrt3}{2}$$ and both cases give $$a^2+b^2=2.$$

Answer 3

From $L_1$ and $L_3$ it is easy to determine that the intersection point is $P = (\dfrac{1}{2}, \dfrac{\sqrt{3}}{2} )$.

The normal to $L_1$ is $(1, \sqrt{3})$ so it makes an angle of $60^\circ$ with the $x$-axis. Hence line $L_2$'s normal must make an angle of either $105^\circ$ or $15^\circ$. Working with unit normals, we get

$n_1 = (\cos 105^\circ, \sin 105^\circ) = (\dfrac{1}{\sqrt{2}} (\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}), \dfrac{1}{\sqrt{2}}(\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}))$

and

$n_2 = (\cos 105^\circ, \sin 105^\circ) = (\dfrac{1}{\sqrt{2}} (\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}), \dfrac{1}{\sqrt{2}}(-\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}))$

In the first case, the equation of $L_2$ is $t n_1 ( (x, y) ) = t n_1 \cdot P = 1$

And we want to solve for $t$ that will make $t n_1 \cdot P = 1$

By direct evaluation,

$\begin{equation} \begin{split} n_1 \cdot P &= (\dfrac{1}{2\sqrt{2}} (\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}) + \dfrac{\sqrt{3}}{2\sqrt{2}}(\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}) \\ &= \dfrac{1}{\sqrt{2}} ( \dfrac{1}{4} +\dfrac{3}{4} -\dfrac{\sqrt{3}}{4} +\dfrac{\sqrt{3}}{4} ) = \dfrac{1}{\sqrt{2}} \end{split} \end{equation} $

And this means that $t = \sqrt{2} $ which in turn implies that $a^2 + b^2 = t^2 = 2$

Note that $(a,b) = t n_1 = \sqrt{2} n_1 = \dfrac{1}{2}(1-\sqrt{3}, 1 + \sqrt{3})$

For the second case, we follow the same steps:

$\begin{equation} \begin{split} n_2 \cdot P &= \dfrac{1}{\sqrt{2}} ( \dfrac{1}{2}(\dfrac{1}{2}+\dfrac{\sqrt{3}}{2})+\dfrac{\sqrt{3}}{ 2}(-\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}) \\ &= \dfrac{1}{\sqrt{2}} (\dfrac{1}{4} + \dfrac{3}{4} + \dfrac{3}{4} -\dfrac{\sqrt{3}}{4} ) = \dfrac{1}{\sqrt{2}} \end{split} \end{equation}$

Thus the value of $t$ is again $\sqrt{2}$ , making $a^2 + b^2 = t^2 = 2 $

and $(a,b) = t n_2= \sqrt{2} n_2 = \dfrac{1}{2}(1+\sqrt{3}, -1+ \sqrt{3})$