# Calculate the angle between 2 vectors without a calculator

Given vectors $\underline u = (1, 1, 0)$ and $\underline v = (0, 1, 1)$

I understand how to derive the answer before converting this to degrees:

$cos\theta =\left(\frac{u⋅v}{||u||⋅||v||}\right)$

My workings on this question:

$||u|| = \sqrt{a^2 +b^2 + c^2} = \sqrt{(1)^2 + (1)^2 + (0)^2} = \sqrt{2}$ $||v|| = \sqrt{a^2 +b^2 + c^2} = \sqrt{(0)^2 + (1)^2 + (1)^2} = \sqrt{2}$

$u⋅v = (1)(0) + (1)(1) + (0)(1) = 1$

$cos\theta =\left(\frac{1}{\sqrt{2}\sqrt{2}}\right) = \frac{1}{2}$
$\theta = cos^{-1}\left(\frac{1}{2}\right) = 60^\circ$

When I arrive at $\frac{1}{2}$ which is the second last line of my workings is it correct that this is $60^\circ$?

I am using the 1st quadrant from the unit circle and $\frac{1}{2}$ for $cos\theta$ to derive $60^\circ$.

• Last 2 lines unclear. Please frame the question properly. Nov 9, 2015 at 14:42
• @Aniket Please see a revision of my phrasing. Let me know if this makes sense? Nov 9, 2015 at 14:47
• Yes your question makes sense now. Nov 9, 2015 at 14:48
• You can still improve the formulation of the question by introducing $u,v$ right at the outset, as in "Find the angle between the vectors $u=(1,1,0)$ and $v=(0,1,1)$. Otherwise the reader needs to decipher these vectors from your solution. Nov 9, 2015 at 14:56
• @uniquesolution Thanks! Sorry just noticed this now... Will add these. Nov 9, 2015 at 14:57

In general, $$\theta = \cos^{-1} \left(\frac{1}{2}\right) = 2n\pi + \frac{\pi}{3}$$ where $n \in \mathbb{N}$.
However for your specific calculation in this vector problem, $60^{\circ}$ is right. However the answer might also be $(360^{\circ}-60^{\circ})=300^{\circ}$ depending which angle between the two vectors is required, acute or obtuse.
Moreover I advise you to write your answer in the following way i.e. after $$\cos \theta = \frac{1}{2}$$ follow it up with $$\theta = 60^{\circ}$$ because as I stated that the thing you have written implies some general things which is not the case here.