vec a=vec m+3vec n , vec b=7vec m-5vec n , vec c=vec m-4vec…
$\vec a=\vec m+3\vec n$, $\vec b=7\vec m-5\vec n$, $\vec c=\vec m-4\vec n$, $\vec d=7\vec m-2\vec n$. Find the angle between $\vec m$ and $\vec n$ if $\vec a\perp\vec b$ and $\vec c\perp\vec d$.
Explanation of Method
The angle between m and n from two perpendicularity conditions.
Step 1: Two equations
\[ 7\vec m^{\,2}+16\,\vec m\cdot\vec n-15\vec n^{\,2}=0\quad(\mathrm I),\qquad 7\vec m^{\,2}-30\,\vec m\cdot\vec n+8\vec n^{\,2}=0\quad(\mathrm{II}). \]
Step 2: Solving
\[ (\mathrm I)-(\mathrm{II}):\ 46\,\vec m\cdot\vec n=23\,|\vec n|^2\Rightarrow \vec m\cdot\vec n=\tfrac12|\vec n|^2;\qquad (\mathrm{II}):\ 7|\vec m|^2=7|\vec n|^2. \]
\[ \cos\varphi=\frac{\tfrac12|\vec n|^2}{|\vec n|^2}=\frac12. \]
\[\varphi=60^\circ\qquad(|\vec m|=|\vec n|)\]
Result
\[\boxed{\,\varphi=60^\circ\qquad(|\vec m|=|\vec n|)\,}\]
Remark
Note. This is Problem 78 with $\vec p,\vec q$ renamed to $\vec m,\vec n$.