316. Find the projection of the vector $\bar{a}=3 \bar{i}+4 \bar{j}-\bar{k}$ onto the direction of the vector $\bar{b}=\bar{i}+\bar{j}$.
Solution. It is known that

![](https://cdn.mathpix.com/cropped/2024_05_22_1b9a1c1720bae4af5bc5g-073.jpg?height=87&width=340&top_left_y=837&top_left_x=419)

The cosine of the angle between vectors $\bar{b}$ and $\bar{a}$ is found using formula (16):

$$
\cos (\widehat{b}, \bar{a})=\frac{3 \cdot 1+4 \cdot 1+(-1) \cdot 0}{\sqrt{3^{2}+4^{2}+(-1)^{2}} \sqrt{1^{2}+1^{2}}}=\frac{7}{\sqrt{52}}
$$

Therefore,

$$
\pi p_{\bar{b}} \bar{a}=\sqrt{26} \cdot \frac{7}{\sqrt{52}}=\frac{7 \sqrt{2}}{2}
$$

This problem can be solved more concisely if we note that

$$
\operatorname{\pi p}_{\bar{b}} \bar{a}=a \cos \widehat{(\bar{b}, \bar{a})}=\frac{\bar{a} \bar{b}}{b}
$$

Using this formula, we get

$$
\operatorname{пr}_{\bar{b}} \bar{a}=\frac{(3 \bar{i}+4 \bar{j}-\bar{k})(\bar{i}+\bar{j})}{\sqrt{2}}=\frac{3+4}{\sqrt{2}}=\frac{7 \sqrt{2}}{2} .
$$