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

Calculate the angle between the vectors $$$\mathbf{\vec{u}} = \left\langle 6, 2\right\rangle$$$ and $$$\mathbf{\vec{v}} = \left\langle -4, -6\right\rangle$$$.

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).

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

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

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