Given that $\vec{a}$ and $\vec{b}$ are two non-zero vector. The two vectors form 4 resultant vectors such that $\vec{a} + 3\vec{b}$ and $2\vec{a} – 3\vec{b}$ are perpendicular, $\vec{a} – 4\vec{b}$ and $\vec{a} + 2\vec{b}$ are perpendicular. How can I find the angle between $\vec{a}$ and $\vec{b}$?
The answer given here is 114.09. Any help is much appreciated.
Best Answer
You are given that $\left\langle a+3b,2a-3b\right\rangle=0$ and $\left\langle a-4b,a+2b \right\rangle=0$. Hence
$$\left\langle a+3b,2a-3b\right\rangle=2\left\langle a,a\right\rangle-9\left\langle b,b\right\rangle+3\left\langle a,b\right\rangle=0$$ and $$\left\langle a-4b,a+2b \right\rangle=\left\langle a,a \right\rangle-8\left\langle b,b \right\rangle-2\left\langle a,b \right\rangle=0.$$(Here I assumed that you are working with a real inner-product space). By subtracting the second equation twice from the first, we obtain $$\left\langle b,b \right\rangle+\left\langle a,b \right\rangle=0.$$
Plugging the previous equation into the second yields $$\|a\|^2=6\|b\|^2.$$ Hence $\frac{\|b\|}{\|a\|}=\sqrt{\frac{1}{6}}.$
Now you know that $\left\langle a,b \right\rangle=\cos(\theta)\|a\|\|b\|.$ Thus, after plugging the previous results in the third equation, we get
$$\cos(\theta)=-\frac{\|b\|}{\|a\|}=-\sqrt{\frac{1}{6}}.$$