5.8. Find the modulus of the projection of vector $\bar{a}(7; -4)$ onto the axis parallel to vector $\bar{b}(-8; 6)$.

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5.8. Найти модуль проекции вектора $\bar{a}(7 ;-4)$ на ось, параллельную вектору $\bar{b}(-8 ; 6)$.
5.8. We have $O K=$ proj $_{\bar{b}} \bar{a}=|\bar{a}| \cos \varphi$ (Fig. 5.5). From here, using

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-601.jpg?height=244&width=381&top_left_y=1151&top_left_x=48)

Fig. 5.5 the equality $\quad \bar{a} \bar{b}=|\bar{a}||\vec{b}| \cos \varphi, \quad$ we find proj $_{\bar{b}} \bar{a}=\frac{\bar{a} \bar{b}}{|\bar{b}|}$. Since $\bar{b}$ is the direction vector of the axis onto which the vector is projected,

$$
\operatorname{proj}_{\bar{b}} \bar{a}=\frac{7(-8)+(-4) 6}{\sqrt{64+36}}=\frac{-56-24}{10}=-8 \text {, i.e. }\left|\operatorname{proj}_{\bar{b}} \bar{a}\right|=8
$$

Answer: 8.