How can he prove the points $A=(a_1, a_2, a_3)$, $B=(b_1, b_2, b_3)$, $C=(c_1, c_2, c_3)$, $D=(d_1, d_2, d_3)$ belong to the same plane, as if they belong can then find plane. I know how to prove three given points belong the same plane.
[Math] How to prove four points belong to the same plane
geometry
Related Solutions
An outline of one method to find the point of intersection:
First find the equations of the line and the plane
A parameterization of the line is $$\tag{1} (x,y,z)= (o_1+d_1 t\, , o_2+d_2 t\,, o_3+d_3 t ),\quad -\infty<t<\infty. $$
To find an equation of the plane, take the cross product of the vectors $A-B$ and $B-C$. This will give you a normal vector to the plane: $(N_1, N_2, N_3)$. The equation of the plane is then, using $A$ as a point on the plane: $$\tag{2} N_1(x-a_1)+N_2(y-a_2)+N_3(z-a_3)=0. $$
Now, to find the point of intersection, substitute the information from $(1)$ $$ x=o_1+d_1 t , \quad y= o_2+d_2 t\quad z= o_3+d_3 t $$ into $(2)$ and solve for $t$. Then substitute this value of $t$ into $(1)$ to find the coordinates of the point.
I'm assuming there is a point of intersection. There may not be, or there may be infinitely many...
Yes it is a necessary and sufficient condition if you include as "point of intersection" the point at infinity in the case of parallel lines.
If the two lines $l_1: \; a_1x+b_1y+c_1=0,\quad l_2:\; a_2x+b_2y+c_2=0$ intersect in a point than $l_3=\lambda l_1 +\mu l_2$ also passes though that point. And the their $3 \times 3$ determinant is null.
Viceversa, if the determinant is null, then the rows are dependent and one can be expressed as a linear combination of the others.
-- reply to your comment --
In analytic or affine geometry, the introduction of the homogeneous coordinates render the panorama more ..homogeneous, in fact.
In particular we have the interesting duality points/lines in 2D, and points /planes in 3D lines.
in 2D, two lines always have a common point same as two points have a common line.
Translated to algebra, given
$$l_1: \; a_1x_1+b_1x_2+c_1x_0=0,\quad l_2:\; a_2x_1+b_2x_2+c_2x_0=0$$
then
$$l_1 \cap l_2$$ always has one non-trivial solution, either in terms of a common point $(x_1,x_2,x_0)$ in the above, and in terms of a line common to two points in
$$a x_{1,1}1+bx_{1,2}+cx_{1,0}=0 \; \cap \; a x_{2,1}1+bx_{2,2}+cx_{2,0}=0$$
Then you know that in a homogeneous system prescribing the common point among three lines, a necessary and sufficient condition is that the determinant of the coefficients ($3 \times 3$) be null, because otherwise (rank of the matrix $3$) you have the only solution $(0,0,0)$ which does not correspond to a defined point.
Then if the rank of the matrix is $2$, you have a homogeneous solution $\lambda (\xi, \eta, \zeta)$ which is one point,
and if the rank is $1$ you get a homogeneous solution depending on two parameters, which is a line and which is the same as any of the three in the system.
The same, reverted, for three points sharing a common line.
Best Answer
Hint: Translate by $-A$ so that we can assume $A=(0,0,0)$. Then points $A,B,C,D$ lie on a plane iff vectors $B,C,D$ do not span whole space, i.e. if $$\left| \begin{array}{ccc} b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \\ d_1 & d_2 & d_3 \end{array} \right|=0$$