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???.
