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 $60^\circ$.

  • $\begingroup$ Last 2 lines unclear. Please frame the question properly. $\endgroup$ Nov 9, 2015 at 14:42
  • $\begingroup$ @Aniket Please see a revision of my phrasing. Let me know if this makes sense? $\endgroup$ Nov 9, 2015 at 14:47
  • $\begingroup$ Yes your question makes sense now. $\endgroup$ Nov 9, 2015 at 14:48
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    $\begingroup$ 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. $\endgroup$ Nov 9, 2015 at 14:56
  • $\begingroup$ @uniquesolution Thanks! Sorry just noticed this now... Will add these. $\endgroup$ Nov 9, 2015 at 14:57

1 Answer 1


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.

  • $\begingroup$ Thanks. This makes sense. Now in this question they have not been specific about whether it is obtuse or acute would it be safe to merely provide the acute or perhaps both? $\endgroup$ Nov 9, 2015 at 15:03
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    $\begingroup$ For being most accurate, it is better to provide both answers mentioning the cases. $\endgroup$ Nov 9, 2015 at 15:06

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