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I have 3 vectors $$A\langle0,-1,-1\rangle$$ $$B\langle-1,0,-1\rangle$$ $$C \langle-1,-1,-2\rangle$$ Where C is $A+B$

Simply put I wanted to find the angle between $A$ and $B$

Using dot product I found that $A\cdot B = 1$ and then dividing by the magnitudes $||A||=\sqrt{2}$, $||B|| = \sqrt{2}$, and $||C||=\sqrt{6}$

Giving me an angle of $60^\circ$

However when I find the angle between $A$ and $C$ I find an angle of $30^\circ$ and this is the same answer I get for the angle between $B$ and $C$.

$A\cdot C=3$ then $cos\theta=\frac{3}{\sqrt{2}\sqrt{6}}$ where $\theta=30^\circ$ This is the same for $B\cdot C$

Adding all of these angles together I only get a grand total of $120^\circ$, how is this possible as I have to have $180^\circ$

Can anyone help me figure out what I am missing or where I went wrong?

  • 2
    The formula $\cos\theta=(A\cdot B)/(||A||\,||B||)$ works, when $A$ and $B$ are the two sides of the triangle **that point away from the vertex of the angle**. If $C=A+B$ is the third side of the triangle then only $-A$ and $B$ can start from the same vertex. Draw a picture (of a plane triangle) to see this! – 2017-02-06

4 Answers 4

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The problem is that there are two possible angles for any dot product, $\theta$ and $\pi - \theta$ (Draw the intersecting lines and you will see the adjacent supplementary angles.)

Since 60 degrees is too small, try 180 degrees - 60 degrees = 120 degrees. That will work.

You could also try 180 degrees - 30 degrees = 150 degrees but that would be too large.

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    I am a bit confused.Cos$(\pi-\theta)$ not equal to Cos$\theta$.What you have said is correct for cross product. – 2017-02-07
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To expand on Jyrki Lahtonen’s comment, you’re measuring the wrong angles. Your triangle has vertices at points $A$, $B$ and $C$, so you should be measuring angles that have these three points as vertices, but all of the angles that you’re computing have the origin as their vertex instead. For instance, to find $\angle{ABC}$ of your triangle, you should be measuring the angle between the vectors $A-B$ and $C-B$—the two displacement vectors from $B$ to the other two vertices—instead of the displacement vectors to points $A$ and $C$ from the origin.

(By the way, this has nothing to do with the points being three-dimensional. You’d have the same problem in two dimensions.)

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A,B and C are position vectors of points of a triangle.They do not make up the side of the triangles.B-A, C-B, A-C are the vectors which make up the sides of the triangle.

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Your vectors $A$ , $B$ and $C$ form a triangle if and only if $A + B + C = 0$

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    Only if A + B = C which is true.| Or -C, etc as directionality does not matter in the triangle. – 2017-02-06
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    -1, as this answer is incorrect. The mentioned vectors do in fact form a triangle. – 2017-02-06