# Angle between $\left\langle 6, 2\right\rangle$ and $\left\langle -4, -6\right\rangle$

The calculator will find the angle between the vectors $\left\langle 6, 2\right\rangle$ and $\left\langle -4, -6\right\rangle$, with steps shown.
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Calculate the angle between the vectors $\mathbf{\vec{u}} = \left\langle 6, 2\right\rangle$ and $\mathbf{\vec{v}} = \left\langle -4, -6\right\rangle$.

### Solution

First, calculate the dot product: $\mathbf{\vec{u}}\cdot \mathbf{\vec{v}} = -36$ (for steps, see dot product calculator).

Next, find the lengths of the vectors:

$\mathbf{\left\lvert\vec{u}\right\rvert} = 2 \sqrt{10}$ (for steps, see vector length calculator).

$\mathbf{\left\lvert\vec{v}\right\rvert} = 2 \sqrt{13}$ (for steps, see vector length calculator).

Finally, the angle is given by $\cos{\left(\phi \right)} = \frac{\mathbf{\vec{u}}\cdot \mathbf{\vec{v}}}{\mathbf{\left\lvert\vec{u}\right\rvert} \mathbf{\left\lvert\vec{v}\right\rvert}} = \frac{-36}{\left(2 \sqrt{10}\right)\cdot \left(2 \sqrt{13}\right)} = - \frac{9 \sqrt{130}}{130}$ (in case of complex numbers, we need to take the real part of the dot product).

$\phi = \operatorname{acos}{\left(- \frac{9 \sqrt{130}}{130} \right)} = \left(\frac{180 \operatorname{acos}{\left(- \frac{9 \sqrt{130}}{130} \right)}}{\pi}\right)^0$

Angle in radians: $\phi = \operatorname{acos}{\left(- \frac{9 \sqrt{130}}{130} \right)}\approx 2.480549484739106$A.
Angle in degrees: $\phi = \left(\frac{180 \operatorname{acos}{\left(- \frac{9 \sqrt{130}}{130} \right)}}{\pi}\right)^0\approx 142.125016348901798^0.$A