[Math] Plane through the intersection of two given planes.

3dvectors

If I have two planes $$\mathbf r\cdot\mathbf {\hat n_1}=p_1\\ \mathbf r\cdot\mathbf {\hat n_2}=p_2$$

If they intersect somewhere then the intersection will form a line, if $\mathbf r'$ is the position vector of a point on the intersection line then I have $$\mathbf r'\cdot\mathbf {\hat n_1}=p_1\\ \mathbf r'\cdot\mathbf {\hat n_2}=p_2$$

If I combine these two equations I would get $$\mathbf r'\cdot(\mathbf {\hat n_1}+\mathbf{\hat n_2})=p_1+p_2$$
This looks like a vector equation of a plane.

But if we see in internet or in books we would find there is scalar $\lambda$ multiplied.

$$\mathbf r'\cdot(\mathbf {\hat n_1}+\lambda\mathbf{\hat n_2})=p_1+\lambda p_2$$

So my question is why there is $\lambda$ there? Please also give me the geometrical image of the plane which is passing through the intersection of those two planes. I'm not able to imagine how would that plane look like.

Best Answer

We have to make some considerations:

  1. The intersection of two planes could be empty if the planes are parallel, but you already state that you assume an intersection, so this is not a problem.
  2. The intersection of two planes which are the same is just the plane itself. We will deal with this case later.
  3. Suppose now you have two distinct, non-parallel planes. You write the equations of each plane as $\vec{r}\cdot\hat{n}_1=p_1$ and $\vec{r}\cdot\hat{n}_2=p_2$. Now, if I multiply each of these equations by a constant, the equations remain true. For instance, $\vec{r}\cdot\hat{n}_1=p_1$ implies $\vec{r}\cdot(A\hat{n}_1)=Ap_1$, and similarly I can get $\vec{r}\cdot(B\hat{n}_2)=Bp_2$.
  4. These two planes are distinct and non-parallel, so they intersect in a line. As you say, points on this line have to satisfy both plane equations simultaneously, so I can describe the line by the system of equations $$ \begin{cases} \vec{r}\cdot(A\hat{n}_1)=Ap_1 \\\vec{r}\cdot(B\hat{n}_2)=Bp_2 \end{cases}. $$
  5. As you also pointed out, the combination of these equations $$\vec{r}\cdot(A\hat{n}_1+B\hat{n}_2)=Ap_1+Bp_2$$ looks like the equation of a plane for given $A$ and $B$, because it is so. This is not the equation of a line, many more points satisfy it. Let us divide through by $A$ (assuming it is not zero), and denote $B/A$ by $\lambda$. We obtain: $$\vec{r}\cdot(\hat{n}_1+\lambda\hat{n}_2)=p_1+\lambda p_2.$$ Once again, for a fixed $\lambda$, this is the equation of a plane.
  6. However, we obtain a line if we require that this equation is true for any value of $\lambda$. Remember that the values of $A$ and $B$ were arbitrary, and therefore so is $\lambda$. The intersecting line should satisfy this new equation for all $\lambda$. For instance, if I choose $\lambda=0$ and $\lambda=1$, I recover the system: $$ \begin{cases} \vec{r}\cdot\hat{n}_1=p_1 \\\vec{r}\cdot(\hat{n}_1+\hat{n}_2)=p_1+p_2 \end{cases}, $$ which is equivalent to the system we had in point 4.
  7. This is all still true if the two planes are the same, but in this case the system of equations is redundant because $\hat{n}_1=\hat{n}_2$ and $p_1=p_2$. This gives $$ \begin{cases} \vec{r}\cdot(A\hat{n}_1)=Ap_1 \\\vec{r}\cdot(B\hat{n}_1)=Bp_1 \end{cases} $$ and $$\vec{r}\cdot[(\lambda+1)\hat{n}_1]=(\lambda+1) p_1,$$ which only define the same plane again, not a line.
  8. As for the picture, I found this image online: enter image description here The red and green planes intersect in a line, and the blue one is another plane which passes through the line.

I hope this helps!

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