3D Coordinate Geometry - Intersection of Planes

There are three possible relationships between two planes in a three-dimensional space; they can be parallel, identical, or they can be intersecting. Comparing the normal vectors of the planes gives us much information on the relationship between the two planes. If the normal vectors are parallel, the two planes are either identical or parallel. If the normal vectors are not parallel, then the two planes meet and make a line of intersection, which is the set of points that are on both planes. The figure below depicts two intersecting planes.

ParallelAngleBisector

The way to obtain the equation of the line of intersection between two planes is to find the set of points that satisfies the equations of both planes. Since the equation of a plane consists of three variables and we are given two equations (since we have two planes), solving the simultaneous equations will give a relation between the three variables, which is equivalent to the equation of the intersection line. The example below demonstrates how this process is done.

Find the equation of the intersection line of the following two planes:

$\begin{aligned} \alpha : x+y+z&=1 \\ \beta : 2x+3y+4z&=5. \end{aligned}$

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Eliminating $z$ gives

$2x=-y-1, \qquad (1)$

and eliminating $y$ gives

$2x=2z-4. \qquad (2)$

Hence, from (1) and (2), the equation of the intersection line between the two planes $\alpha$ and $\beta$ is

$2x=-y-1=2z-4 \implies x=\frac{y+1}{-2} = z-2.\ _\square$

Do the following two planes $\alpha$ and $\beta$ meet?

$\begin{aligned} \alpha : 2x + y - z &= 6 \\ \beta : -4x - 2y +2z &= -5 \end{aligned}$

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The normal vectors of the planes are $\vec{n_{\alpha}}= (2, 1, -1)$ and $\vec{n_{\beta}}=(-4, -2, 2),$ respectively.

Since $-2\vec{n_{\alpha}}=\vec{n_{\beta}},$ the normal vectors of the two planes are parallel, which implies that the two planes $\alpha$ and $\beta$ are either parallel or identical.

The point $(3,0,0)$ is on plane $\alpha$ but not $\beta,$ which implies that the two planes are not identical. Therefore the two planes are parallel and do not meet each other. $_\square$

What is the condition in which the following two planes $\alpha$ and $\beta$ meet each other?

$\begin{aligned} \alpha : 3x + ay -2z &= 5 \\ \beta : 6x + by -4z &= 3 \end{aligned}$

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The normal vectors of the two planes $\alpha$ and $\beta$ are $\vec{n_{\alpha}}= (3,a,-2)$ and $\vec{n_{\beta}}=(6,b,-4) ,$ respectively.

Notice that when $b=2a ,$ the two normal vectors are parallel. In this case, since $2\times5\neq3,$ the two planes are not identical but parallel.

Since two planes in a three-dimensional space always meet if they are not parallel, the condition for $\alpha$ and $\beta$ to meet is $b\neq2a.$ $_ \square$

What is equation of the line of intersection between the following two planes $\alpha$ and $\beta?$

$\begin{aligned} \alpha : x-y+4z&=2 \\ \beta : x+2y-2z&=4 \end{aligned}$

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Eliminating $x$ by subtracting the two equations gives

$6z=3y-2. \qquad (1)$

Eliminating $y$ by multiplying the first equation by 2 and adding the second equation gives

$6z=-3x+8. \qquad (2)$

Hence, from (1) and (2) the equation of the line of intersection is

$-3x+8=3y-2=6z. \ _ \square$

What is the volume surrounded by the $xy$-plane, $yz$-plane, $xz$-plane, and the plane $x+y+z=4?$

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ParallelAngleBisector

The four planes make a tetrahedron, as shown in the figure above.

The $x$-, $y$-, and $z$-intercepts of the plane $x+y+z=4$ are $A=(4,0,0) , B=(0,4,0),$ and $C=(0,0,4) ,$ respectively.

Hence, the volume $V$ of the tetrahedron is

$\begin{aligned} V &= (\text{area of base}) \times (\text{height}) \times \frac{1}{3} \\ &= \left(4\cdot4\cdot\frac{1}{2}\right) \times 4\times \frac{1}{3} \\ &= \frac{32}{3}. \ _ \square \end{aligned}$