Vector projection of $$$\left\langle -8, 3\right\rangle$$$ onto $$$\left\langle -6, -2\right\rangle$$$

Calculate the vector projection of $$$\mathbf{\vec{v}} = \left\langle -8, 3\right\rangle$$$ onto $$$\mathbf{\vec{u}} = \left\langle -6, -2\right\rangle$$$.

The vector projection is given by $$$\text{proj}_{\mathbf{\vec{u}}}\left(\mathbf{\vec{v}}\right) = \frac{\mathbf{\vec{v}}\cdot \mathbf{\vec{u}}}{\mathbf{\left\lvert\vec{u}\right\rvert}^{2}} \mathbf{\vec{u}}.$$$

$$$\mathbf{\vec{v}}\cdot \mathbf{\vec{u}} = 42$$$ (for steps, see dot product calculator).

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

Thus, the vector projection is $$$\text{proj}_{\mathbf{\vec{u}}}\left(\mathbf{\vec{v}}\right) = \frac{42}{\left(2 \sqrt{10}\right)^{2}}\cdot \left\langle -6, -2\right\rangle = \frac{21}{20}\cdot \left\langle -6, -2\right\rangle = \left\langle - \frac{63}{10}, - \frac{21}{10}\right\rangle$$$ (for steps, see vector scalar multiplication calculator).

The vector projection is $$$\left\langle - \frac{63}{10}, - \frac{21}{10}\right\rangle = \left\langle -6.3, -2.1\right\rangle$$$A.