# Angle between two planes with a common side using vectors

Plane $$\pi_1$$ is formed by vectors $$a_1$$ and $$b_1$$ and plane $$\pi_2$$ is formed by vectors $$a_1$$ and $$b_2$$.
With these vectors given, establish the angle between planes $$\pi_1$$ and $$\pi_2$$ using dot and cross product operations.

I assumed $$a_1$$ to be $$$$, $$b_1 = $$ and $$b_2 = $$
I found normals to both planes using cross product operations, but I get stuck while determining the angle using the formula $$cos\theta = \frac{a.b}{\sqrt a^2 \sqrt b^2}$$ where both a and b are the vectors orthogonal to planes $$\pi_1$$ and $$\pi_2$$ respectively.

The solution I get seems to be extremely long and convoluted, and It doesn't seem to work out. I was wondering if there is another shorter way to solve this. The answer must be in terms of $$a_1$$, $$b_1$$ and $$b_2$$

The normal vector of $$\pi_1$$ is $$a_1\times b_1$$ and the normal vector of $$\pi_2$$ is $$a_2\times b_2$$. Then, the angle $$\theta$$ between $$\pi_1$$ and $$\pi_2$$ is
$$\cos\theta =\frac{(a_1\times b_1)\cdot (a_2\times b_2)} { | a_1\times b_1| | a_2\times b_2|}$$