Finding the acute angle between two lines (or between two vectors)

What is an acute angle?

An acute angle is an angle that’s less than $90^\circ$, like this:

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If we want to find the acute angle between two lines, we can convert the lines to standard vector form and then use the formula

$\cos{\theta}=\frac{a\cdot b}{|a||b|}$

where $a$ and $b$ are the given vectors, $a\cdot{b}$ is the dot product of the vectors, $|a|$ is the magnitude of the vector $a$ (its length) and $|b|$ is the magnitude of the vector $b$ (its length). We can find the magnitude of both vectors using the distance formula

$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

for a two-dimensional vector where the point $(x_1,y_1)$ is the origin $(0,0)$.

If the formula above gives a result that’s greater than $90^\circ$, then we’ve found the obtuse angle between the lines. To find the acute angle, we just subtract the obtuse angle from $180^\circ$, and we’ll get the acute angle.

How do we find the acute angle between two lines, when the lines are defined by vectors?

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Converting lines into vectors, then finding the angle between those two vectors

Example

Find the acute angle between the lines.

$x+3y=2$

$3x-6y=5$

First we’ll convert the lines to standard vector form.

$x+3y=2$

$a=\langle1,3\rangle$

and

$3x-6y=5$

$b=\langle3,-6\rangle$

Before we can use our formula, we need to find the dot product of $a$ and $b$.

$a\cdot{b}=(1)(3)+(3)(-6)$

$a\cdot{b}=3-18$

$a\cdot{b}=-15$

Now we need to find the length of each vector using the distance formula.

$|a|=\sqrt{(1-0)^2+(3-0)^2}$

$|a|=\sqrt{1+9}$

$|a|=\sqrt{10}$

and

$|b|=\sqrt{(3-0)^2+(-6-0)^2}$

$|b|=\sqrt{9+36}$

$|b|=\sqrt{45}$

If we want to find the acute angle between two lines, we can convert the lines to standard vector form.

Plugging $a\cdot{b}=-15$, $|a|=\sqrt{10}$, and $|b|=\sqrt{45}$ into the formula, we get

$\cos{\theta}=\frac{-15}{\sqrt{10}\sqrt{45}}$

$\cos{\theta}=\frac{-15}{\sqrt{450}}$

$\cos{\theta}=\frac{-15}{\sqrt{450}}$

$\cos{\theta}=\frac{-15}{\sqrt{225\cdot2}}$

$\cos{\theta}=\frac{-15}{15\sqrt{2}}$

$\cos{\theta}=\frac{-1}{\sqrt{2}}$

Rationalize the denominator.

$\cos{\theta}=\frac{-1}{\sqrt{2}}\left(\frac{\sqrt{2}}{\sqrt{2}}\right)$

$\cos{\theta}=\frac{-\sqrt{2}}{2}$

Looking at the top half of the unit circle, we can see that

$\theta=\frac{3\pi}{4}=135^\circ$

Since the answer is greater than $90^\circ$, we’ve found the obtuse angle between the lines. To find the acute angle, we’ll just subtract this value from $180^\circ$.

$\theta=180^\circ-135^\circ$

$\theta=45^\circ$

The acute angle between the lines is $45^\circ$.

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