# Angle between $\left\langle 1, 0, -1\right\rangle$ and $\left\langle 1, 1, -2\right\rangle$

The calculator will find the angle between the vectors $\left\langle 1, 0, -1\right\rangle$ and $\left\langle 1, 1, -2\right\rangle$, with steps shown.
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Calculate the angle between the vectors $\mathbf{\vec{u}} = \left\langle 1, 0, -1\right\rangle$ and $\mathbf{\vec{v}} = \left\langle 1, 1, -2\right\rangle$.

### Solution

First, calculate the dot product: $\mathbf{\vec{u}}\cdot \mathbf{\vec{v}} = 3$ (for steps, see dot product calculator).

Next, find the lengths of the vectors:

$\mathbf{\left\lvert\vec{u}\right\rvert} = \sqrt{2}$ (for steps, see vector length calculator).

$\mathbf{\left\lvert\vec{v}\right\rvert} = \sqrt{6}$ (for steps, see vector length calculator).

Finally, the angle is given by $\cos{\left(\phi \right)} = \frac{\mathbf{\vec{u}}\cdot \mathbf{\vec{v}}}{\mathbf{\left\lvert\vec{u}\right\rvert} \mathbf{\left\lvert\vec{v}\right\rvert}} = \frac{3}{\left(\sqrt{2}\right)\cdot \left(\sqrt{6}\right)} = \frac{\sqrt{3}}{2}$ (in case of complex numbers, we need to take the real part of the dot product).

$\phi = \operatorname{acos}{\left(\frac{\sqrt{3}}{2} \right)} = \frac{\pi}{6} = 30^0$

$\phi = \frac{\pi}{6}\approx 0.523598775598299$A
$\phi = 30^0$A