# Show that there is a common line of intersection of the three given planes.

Let $x-y\sin\alpha-z\sin\beta=0,x\sin\alpha-y+z\sin\gamma=0$ and $x\sin\beta+y\sin\gamma-z=0$ be the equations of the planes such that $\alpha+\beta+\gamma=\frac{\pi}{2}$,(where $\alpha,\beta,\gamma\neq0$).Then show that there is a common line of intersection of the three given planes.

We know that three given planes,$a_1x+b_1y+c_1z+d_1=0,a_2x+b_2y+c_2z+d_2=0,a_3x+b_3y+c_3z+d_3=0$ meet in a common line if $\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \\ \end{vmatrix}=0$
if $\begin{vmatrix} 1 & -\sin\alpha & -\sin\beta \\ \sin\alpha & -1 & \sin\gamma \\ \sin\beta & \sin\gamma & -1 \\ \end{vmatrix}=0$
I simplified this determinant to get $1-\sin^2\alpha-\sin^2\beta-\sin^2\gamma-2\sin\alpha\sin\beta\sin\gamma=0$

We have to prove the trigonometric expression true in order to prove the question.But i could not prove the trigonometric expression true.

Rewrite the required result as

$$\cos^2\gamma - \sin^2 \alpha - \sin^2 \beta \overset{?}= 2\sin \alpha \sin \beta \sin \gamma$$

We'll try to show that the LHS equals the RHS. Note that $\cos \gamma = \sin (\alpha + \beta)$ and vice versa, as $\gamma = \pi/2 - (\alpha + \beta)$ $$\sin^2 (\alpha + \beta) - \sin^2 \alpha - \sin^2 \beta \\ \overset{(a)}= ~\sin^2\alpha \cos^2\beta - \sin^2\alpha + \cos^2\alpha\sin^2\beta - \sin^2\beta + 2\sin \alpha \sin \beta \cos \alpha \cos \beta \\ \overset{(b)}= ~-2\sin^2\alpha \sin^2 \beta + 2\sin \alpha \sin \beta \cos \alpha \cos \beta \\ = ~2\sin \alpha \sin \beta (\cos \alpha \cos \beta - \sin \alpha \sin \beta )\\ \overset{(c)}= 2 \sin \alpha \sin \beta \cos (\alpha + \beta)\\$$

where

$(a)$ follows from squaring $\sin (\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$ and some regrouping.

$(b)$ is $\sin^2\alpha \cos^2\beta - \sin^2\alpha = \sin^2 \alpha (\cos^2 \beta -1) = -\sin^2 \alpha \sin^2 \beta$ applied a couple of times.

$(c)$ is the standard formula for $\cos (\alpha + \beta)$