Equation of a Plane Passing Through the Line of Intersection of Two Given Planes

Theorem 6.22. The vector equation of a plane passing through the line of intersection of the planes $\vec r\cdot\vec n_1=d_1$ and $\vec r\cdot\vec n_2=d_2$ is

Proof. Regroup the left side as $\vec r\cdot(\vec n_1+\lambda\vec n_2)-(d_1+\lambda d_2)=0$, i.e. $\vec r\cdot\vec n=d$ with $\vec n=\vec n_1+\lambda\vec n_2$ and $d=d_1+\lambda d_2$ — this IS a plane's standard equation, for every value of $\lambda$. If $\vec r_1$ is the position vector of any point of the LINE of intersection, it satisfies both $\vec r_1\cdot\vec n_1=d_1$ and $\vec r_1\cdot\vec n_2=d_2$ simultaneously, so it satisfies the combined equation too — hence every plane in this one-parameter family passes through the whole line, for whatever $\lambda$ is chosen.

Cartesian form. For planes $a_1x+b_1y+c_1z=d_1$ and $a_2x+b_2y+c_2z=d_2$, the family is

$(a_1x+b_1y+c_1z-d_1)+\lambda(a_2x+b_2y+c_2z-d_2)=0.$ …