Relative positions of lines and planes

Put the general point of the line, $\displaystyle {\underline{x}}={\underline{v}+t\underline{u}}$, in the plane’s equation. The result is a first-degree equation in $\displaystyle t$:

• one solution – the line cuts the plane, and the solution gives the meeting point;
• no solution – the line is parallel to the plane;
• every $\displaystyle t$ is a solution – the line lies in the plane.

In the last two cases $\displaystyle t$ drops out of the equation, since $\displaystyle {\underline{n}\cdot\underline{u}}={0}$: the direction vector is perpendicular to the normal.

Find where the line $\displaystyle {\underline{x}}={(1,2,3)+t(2,-1,1)}$ meets the plane $\displaystyle {2x+y-z}={7}$.

• The general point: $\displaystyle (1+2t,\ 2-t,\ 3+t)$.
• $\displaystyle {2(1+2t)+(2-t)-(3+t)}={7}$, that is $\displaystyle {1+2t}={7}$, and $\displaystyle {t}={3}$.
• The meeting point: $\displaystyle (7,-1,6)$. Check: $\displaystyle {14-1-6}={7}$.

With the plane $\displaystyle {x+y-z}={6}$ we get $\displaystyle {(1+2t)+(2-t)-(3+t)}={6}$, that is $\displaystyle {0}={6}$: no solution, and the line is parallel to the plane.

A line is perpendicular to a plane if and only if its direction vector is parallel to the normal. So the perpendicular from $\displaystyle P$ to the plane is $\displaystyle {\underline{x}}={P+t\underline{n}}$, and the foot of the perpendicular $\displaystyle F$ is where it meets the plane.

Example. The foot of the perpendicular from $\displaystyle P(4,-2,0)$ to the plane $\displaystyle {2x-y+2z}={1}$:

• the perpendicular: $\displaystyle (4+2t,\ -2-t,\ 2t)$;
• $\displaystyle {2(4+2t)-(-2-t)+2\cdot2t}={1}$, that is $\displaystyle {10+9t}={1}$, $\displaystyle {t}={-1}$;
• $\displaystyle {F}={(2,-1,-2)}$, and the distance $\displaystyle {PF}={|t|\cdot|\underline{n}|}={1\cdot3}={3}$.

• The normals are parallel: the planes are parallel, or they coincide when the equations are multiples of each other ($\displaystyle {x+y}={1}$ and $\displaystyle {2x+2y}={2}$).
• The normals are not parallel: the planes intersect in a line. Find two common points – for example with $\displaystyle {z}={0}$ and with $\displaystyle {z}={1}$, each time a system in two unknowns – and the line passes through them.

Example. $\displaystyle {x+y+z}={6}$ and $\displaystyle {x-y+2z}={3}$:

• $\displaystyle {z}={1}$: $\displaystyle {x+y}={5}$, $\displaystyle {x-y}={1}$, so $\displaystyle (3,2,1)$;
• $\displaystyle {z}={3}$: $\displaystyle {x+y}={3}$, $\displaystyle {x-y}={-3}$, so $\displaystyle (0,3,3)$;
• the line of intersection: $\displaystyle {\underline{x}}={(3,2,1)+t(-3,1,2)}$. Its direction vector is perpendicular to both normals.

The line $\displaystyle {\underline{x}}={(1,0,2)+t(1,2,-1)}$.

Write the meeting point in part 1, and the conclusion in part 2.

The planes $\displaystyle {x+y-z}={1}$ and $\displaystyle {x-y+z}={3}$.

Write the line of intersection, and check that it lies in the first plane: put it in the equation.

Find the foot of the perpendicular from $\displaystyle P(1,2,3)$ to the plane $\displaystyle {x+y+z}={0}$, and the distance of $\displaystyle P$ from the plane.

Explain:

1. What is the relative position of the planes $\displaystyle {2x-y+z}={3}$ and $\displaystyle {4x-2y+2z}={5}$?
2. Is the line $\displaystyle {\underline{x}}={(0,1,1)+t(-6,4,-2)}$ perpendicular to the plane $\displaystyle {3x-2y+z}={5}$?