Vector projection of $$$\left\langle 2, -3, 1\right\rangle$$$ onto $$$\left\langle 4, -1, -3\right\rangle$$$

Calculate the vector projection of $$$\mathbf{\vec{v}} = \left\langle 2, -3, 1\right\rangle$$$ onto $$$\mathbf{\vec{u}} = \left\langle 4, -1, -3\right\rangle$$$.

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

$$$\mathbf{\vec{v}}\cdot \mathbf{\vec{u}} = 8$$$ (for steps, see dot product calculator).

$$$\mathbf{\left\lvert\vec{u}\right\rvert} = \sqrt{26}$$$ (for steps, see vector magnitude calculator).

Thus, the vector projection is $$$\text{proj}_{\mathbf{\vec{u}}}\left(\mathbf{\vec{v}}\right) = \frac{8}{\left(\sqrt{26}\right)^{2}}\cdot \left\langle 4, -1, -3\right\rangle = \frac{4}{13}\cdot \left\langle 4, -1, -3\right\rangle = \left\langle \frac{16}{13}, - \frac{4}{13}, - \frac{12}{13}\right\rangle$$$ (for steps, see vector scalar multiplication calculator).

The vector projection is $$$\left\langle \frac{16}{13}, - \frac{4}{13}, - \frac{12}{13}\right\rangle\approx \left\langle 1.230769230769231, -0.307692307692308, -0.923076923076923\right\rangle.$$$A