2.23. Find the angle between $\operatorname{grad} \ln \left(x^{2}+y^{2}+z^{2}\right)$ and $\operatorname{grad}(x y + y z + z x)$ at the point $M(1,1,-1)$.
Solution. Let's perform the necessary calculations:

$$
\begin{aligned}
& \operatorname{grad} \ln \left(x^{2}+y^{2}+z^{2}\right)=\frac{2}{x^{2}+y^{2}+z^{2}}(x \vec{i}+y \vec{j}+z \vec{k}) \\
& \left.\operatorname{grad} \ln \left(x^{2}+y^{2}+z^{2}\right)\right|_{M(1,1,-1)}=\frac{2}{3}(\vec{i}+\vec{j}-\vec{k}) \\
& \operatorname{grad}(x y+y z+z x)=(y+z) \vec{i}+(x+z) \vec{j}+(y+x) \vec{k} \\
& \left.\operatorname{grad}(x y+y z+z x)\right|_{M(1,1,-1)}=2 \vec{k}
\end{aligned}
$$

According to a known formula from vector algebra (see part I § 2.26), the angle $\alpha$ between the gradients at point $M(1,1,-1)$ is:

$$
\cos \alpha=\frac{-\frac{4}{3}}{\frac{2}{3} \sqrt{3} \cdot 2}=-\frac{\sqrt{3}}{3}
$$

Thus, $\alpha=\arccos \left(-\frac{\sqrt{3}}{3}\right)=\pi-\arccos \left(\frac{\sqrt{3}}{3}\right)$.