Let $a_1x+b_1y+c_1z+d_1=0$ and $a_2x+b_2y+c_2z+d_2=0$ be the equations that describe two planes. In my lecture notes, it's written that $a_1x+b_1y+c_1z+d_1=0$ and $a_2x+b_2y+c_2z+d_2=0$ have a straight line as their intersection if
$$y=-\frac{c_1a_2-c_2a_1}{b_2c_1-c_2b_1}x-\frac{c_1d_2-c_2d_1}{b_2c_1-c_2b_1}$$ and
$$z=-\frac{b_1a_2-b_2a_1}{b_2c_1-c_2b_1}x-\frac{b_1d_2-b_2d_1}{b_2c_1-c_2b_1}$$
I don't see how we get this condition.
Any help will be appreciated.
Thank you
On
Given the wording of the problem (excluding the rank zero case) one way of formulating the condition for the planes to have an intersection line is equivalent to for the matrix $$\begin{pmatrix} a_1&b_1& c_1\\ a_2&b_2& c_2\end{pmatrix}\quad(\bigstar),$$ to not have rank one. This is then equivalent to for the matrix to have rank two, or the existence of a non-zero two by two minor.
Consider the system $\begin{align}a_1x+b_1y+c_1z+d_1&=0\\ a_2x+b_2y+c_2z+d_2&=0\end{align}$ or $\begin{pmatrix} a_1&b_1& c_1&d_1\\ a_2&b_2& c_2&d_2\end{pmatrix}\begin{pmatrix}x\\y\\z\\1\end{pmatrix}=0.$
Let's investigate when this intersection is a line in affine 3-space.
If we first assume the case where the first minor is non-zero (the one you get by deleting column 1 of ($\bigstar$)), as the OPs lecture notes rightly say this assumption does give a line, because then we can solve by multiplying the first row by $c_2$ (not very interesting if it is zero) and subtracting the second row multiplied by $c_1$ (again, not very interesting if it is zero) i.e. $\begin{pmatrix} a_1c_2-a_2c_1&b_1c_2-b_2c_1& 0 &d_1c_2-d_2c_1\\ a_2&b_2&c_2&d_2\end{pmatrix}$ and similarly the second row multiplied by $b_1$ subtract the first row multiplied by $b_2$ to get $\begin{pmatrix} a_1c_2-a_2c_1&b_1c_2-b_2c_1& 0 &d_1c_2-d_2c_1\\ a_2b_1-a_1b_2&0&c_2b_1-c_1b_2&d_2b_1-d_1b_2\end{pmatrix},$ now take the non-zero minor outside: $$(b_1c_2-b_2c_1)\cdot\begin{pmatrix} \frac{a_1c_2-a_2c_1}{b_1c_2-b_2c_1}&1& 0 &\frac{d_1c_2-d_2c_1}{b_1c_2-b_2c_1}\\ \frac{a_2b_1-a_1b_2}{b_1c_2-b_2c_1}&0&1&\frac{d_2b_1-d_1b_2}{b_1c_2-b_2c_1}\end{pmatrix},$$ dropping it and multiplying the matrix by $\begin{pmatrix}x\\y\\z\\1\end{pmatrix},$ to get $$\begin{align}\frac{a_1c_2-a_2c_1}{b_1c_2-b_2c_1} x+y+\frac{d_1c_2-d_2c_1}{b_1c_2-b_2c_1}=0\\\frac{a_2b_1-a_1b_2}{b_1c_2-b_2c_1}x+z+\frac{d_2b_1-d_1b_2}{b_1c_2-b_2c_1}=0\end{align}.$$
What we now have is a parametrization of the solution line: $$\begin{align}x&=t\\y&=-\frac{c_1a_2-c_2a_1}{b_2c_1-c_2b_1}t-\frac{c_1d_2-c_2d_1}{b_2c_1-c_2b_1}\\z&=-\frac{b_2a_1-b_1a_2}{b_2c_1-c_2b_1}t-\frac{b_2d_1-b_1d_2}{b_2c_1-c_2b_1},&\end{align}.$$ This is a line since it's of dimension 1, it's linearly depending on one parameter, we can even find two points defining it by setting $t=0$ to get $(x,y,z)=(0,-\frac{c_1d_2-c_2d_1}{b_2c_1-c_2b_1},-\frac{b_2d_1-b_1d_2}{b_2c_1-c_2b_1})$ and $t=1$ to get $(1,\frac{c_2 (a_1 + d_1) - c_1 (a_2 + d_2)}{b_2c_1-c_2b_1},\frac{b_1 (a_2 + d_2) - b_2 (a_1 + d_1)}{b_2c_1-c_2b_1}).$
But, there are planes for which we can't take $x$ as the free variable, say we couldn't divide because the chosen first minor did vanish but still $(\bigstar)$ could have rank two: $\begin{align} (a_1c_2-a_2c_1)x&+&(b_1c_2-b_2c_1)y&& &&+&(d_1c_2-d_2c_1)=0\\ (a_2b_1-a_1b_2)x&+& &&(c_2b_1-c_1b_2)z&&+&(d_2b_1-d_1b_2)=0\end{align}.$ This first strategy then likely fails (could still make sense in projective 3-space, homogenizing the equations with a new variable $w$). But then we could still get a line as the intersection, because then maybe $y$ works as the free variable $\begin{pmatrix} 1&\star& 0&\star\\ 0&\star& 1&\star\end{pmatrix}$ and we could divide by the second minor, or $z;$ $\begin{pmatrix} 1&0&\star&\star\\ 0&1&\star&\star\end{pmatrix},$ and divide by the third minor.
The strategy could still fail, but then the matrix $(\bigstar)$ wouldn't have a non-zero two by two minor and have rank less than 2. Making the planes parallel, since the normal vectors of the planes would be scalar multiples of each other, i.e. defining the same or opposite direction in space. And, yes the planes could then even coincide, for you could have supplied essentially the same equation twice.
In order to find the intersection between the planes $a_1x+b_1y+c_1z+d_1=0$ and $a_2x+b_2y+c_2z+d_2=0$, we have to solve the system of equations: \begin{cases} &a_1x+b_1y+c_1z+d_1=0;\\ &a_2x+b_2y+c_2z+d_2=0. \end{cases} We can rewrite it as \begin{cases} &b_1y+c_1z=-a_1x-d_1;\\ &b_2y+c_2z=-a_2x-d_2. \end{cases} In the matrix notation, we have $$\begin{pmatrix} b_1 & c_1\\ b_2 & c_2 \end{pmatrix}\begin{pmatrix} y\\ z \end{pmatrix}=\begin{pmatrix} -a_1x-d_1\\ -a_2x-d_2 \end{pmatrix}\Longrightarrow\begin{pmatrix} y\\ z \end{pmatrix}=\begin{pmatrix} b_1 & c_1\\ b_2 & c_2 \end{pmatrix}^{-1}\begin{pmatrix} -a_1x-d_1\\ -a_2x-d_2 \end{pmatrix}.$$ Suppose that $b_2c_1-c_2b_1\neq0$. Then $$\begin{pmatrix} b_1 & c_1\\ b_2 & c_2 \end{pmatrix}^{-1}=-\frac{1}{b_2c_1-c_2b_1}\begin{pmatrix} c_2 & -c_1\\ -b_2 & b_1 \end{pmatrix}.$$ Substituting this into the previous equation, we get $$\begin{pmatrix} y\\ z \end{pmatrix}=-\frac{1}{b_2c_1-c_2b_1}\begin{pmatrix} c_2 & -c_1\\ -b_2 & b_1 \end{pmatrix}\begin{pmatrix} -a_1x-d_1\\ -a_2x-d_2 \end{pmatrix}.$$ It follows that $$\begin{pmatrix} y\\ z \end{pmatrix}=-\frac{1}{b_2c_1-c_2b_1}\begin{pmatrix} (c_1a_2-c_2a_1)x+(c_1d_2-c_2d_1)\\ (b_2a_1-b_1a_2)x+(b_2d_1-b_1d_2) \end{pmatrix}.$$ Therefore, \begin{align} y&=-\frac{c_1a_2-c_2a_1}{b_2c_1-c_2b_1}x-\frac{c_1d_2-c_2d_1}{b_2c_1-c_2b_1};\\ z&=-\frac{b_2a_1-b_1a_2}{b_2c_1-c_2b_1}x-\frac{b_2d_1-b_1d_2}{b_2c_1-c_2b_1}, \end{align} which are the equations that describe a straight line.