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 $<m,n,o>$, $b_1 = <x_1,y_1,z_1>$ and $b_2 = <x_2,y_2,z_2>$
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 $b_2$

Thanks a lot in advance.


1 Answer 1


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

$$\cos\theta =\frac{(a_1\times b_1)\cdot (a_2\times b_2)} { | a_1\times b_1| | a_2\times b_2|}$$


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