Example $\boldsymbol{d}$ Given that $a$ and $b$ are non-zero vectors, and $a+3b$ is perpendicular to $7a-5b$, $a-4b$ is perpendicular to $7a-2b$, find the angle between $a$ and $b$.
Analysis According to the necessary and sufficient conditions for perpendicularity and the angle formula to solve.
Solution Since $(a+3 b) \perp(7 a-5 b),(a-4 b) \perp(7 a-2 b)$.
We can $\left\{\begin{array}{l}(\boldsymbol{a}+3 \boldsymbol{b}) \cdot(7 \boldsymbol{a}-5 \boldsymbol{b})=0 \\ (\boldsymbol{a}-4 \boldsymbol{b}) \cdot(7 \boldsymbol{a}-2 \boldsymbol{b})=0\end{array}\right.$
That is $\left\{\begin{array}{l}7 \boldsymbol{a}^{2}+16 \boldsymbol{a} \cdot \boldsymbol{b}-15 \boldsymbol{b}^{2}=0 \\ 7 \boldsymbol{a}^{2}-30 \boldsymbol{a} \cdot \boldsymbol{b}+8 \boldsymbol{b}^{2}=0\end{array}\right.$
Solving, we get $\left\{\begin{array}{l}b^{2}=2 a \cdot b \\ a^{2}=2 a \cdot b\end{array}\right.$
That is $|\boldsymbol{a}|=|\boldsymbol{b}|=\sqrt{2 \boldsymbol{a} \cdot \boldsymbol{b}}$
Since $\cos \theta=\frac{\boldsymbol{a} \cdot \boldsymbol{b}}{|\boldsymbol{a}||\boldsymbol{b}|}=\frac{\boldsymbol{a} \cdot \boldsymbol{b}}{(\sqrt{2 \boldsymbol{a} \cdot \boldsymbol{b}})^{2}}=\frac{1}{2}$
And since $\theta \in[0, \pi]$
Therefore, the angle $\theta$ between $\boldsymbol{a}$ and $\boldsymbol{b}$ is $\frac{\pi}{3}$.