Distances in space

The distance between two shapes (a point, a line, a plane) is the shortest distance between a point of one and a point of the other. The shortest segment is perpendicular to the shapes: from a point to a line or a plane – the perpendicular.

$\displaystyle A$ is a point in the plane, $\displaystyle \underline{n}$ the normal, and $\displaystyle F$ the foot of the perpendicular from $\displaystyle P$. $\displaystyle \overrightarrow{PF}$ is parallel to $\displaystyle \underline{n}$, and $\displaystyle \overrightarrow{AF}\perp\underline{n}$, so $\displaystyle {|\underline{n}\cdot\overrightarrow{AP}|}={|\underline{n}\cdot\overrightarrow{FP}|}={|\underline{n}|\cdot PF}$. Putting in the coordinates gives:

The equation is written as $\displaystyle {ax+by+cz+d}={0}$ (the free number on the left). Parallel planes $\displaystyle {ax+by+cz+d_1}={0}$ and $\displaystyle {ax+by+cz+d_2}={0}$ (with the same $\displaystyle a$, $\displaystyle b$, $\displaystyle c$): $\displaystyle {d}={\frac{|d_1-d_2|}{\sqrt{a^2+b^2+c^2}}}$. A line parallel to a plane: the distance of any of its points from the plane.

Example. $\displaystyle P(4,-2,0)$ and the plane $\displaystyle {2x-y+2z-1}={0}$: $\displaystyle {d}={\frac{|8+2+0-1|}{\sqrt{4+1+4}}}={\frac{9}{3}}={3}$ – as we found with the foot of the perpendicular.

The foot $\displaystyle X$ of the perpendicular from $\displaystyle P$ to the line $\displaystyle {\underline{x}}={\underline{v}+t\underline{u}}$ is the point where $\displaystyle \overrightarrow{PX}\perp\underline{u}$:

• write $\displaystyle {X}={\underline{v}+t\underline{u}}$ and solve $\displaystyle {(X-P)\cdot\underline{u}}={0}$ – an equation in $\displaystyle t$;
• the distance is $\displaystyle |\overrightarrow{PX}|$.

Parallel lines: the distance of a point of one from the other.

Through the line $\displaystyle \ell_1$ there is one plane parallel to $\displaystyle \ell_2$: its normal is perpendicular to both direction vectors. The distance between the lines is the distance of any point of $\displaystyle \ell_2$ from this plane.

Example. $\displaystyle {\ell_1:\ \underline{x}}={(1,0,2)+t(1,1,1)}$, $\displaystyle {\ell_2:\ \underline{x}}={(0,1,0)+s(1,0,1)}$.

• $\displaystyle {\underline{n}}={(a,b,1)}$: $\displaystyle {a+b+1}={0}$, $\displaystyle {a+1}={0}$, so $\displaystyle {\underline{n}}={(-1,0,1)}$.
• The plane through $\displaystyle (1,0,2)$: $\displaystyle {-(x-1)+(z-2)}={0}$, that is $\displaystyle {x-z+1}={0}$.
• The distance of $\displaystyle (0,1,0)$: $\displaystyle {d}={\frac{|0-0+1|}{\sqrt{2}}}={\frac{\sqrt{2}}{2}}$.

What is the distance of $\displaystyle P(1,2,3)$ from the plane $\displaystyle {2x-y+2z-1}={0}$?

What is the distance between the parallel planes $\displaystyle {2x-y+2z}={1}$ and $\displaystyle {4x-2y+4z}={20}$? (Hint: divide the second equation by 2.)

The point $\displaystyle P(3,1,4)$ and the line $\displaystyle {\underline{x}}={(1,0,0)+t(1,1,1)}$.

Find the $\displaystyle t$ of the foot of the perpendicular.

The line $\displaystyle {\underline{x}}={(1,2,3)+t(2,-1,1)}$ is parallel to the plane $\displaystyle {x+y-z}={6}$ (you saw this in the previous lesson).
What is the distance between them?

$\displaystyle {\ell_1:\ \underline{x}}={(0,0,1)+t(1,2,0)}$, $\displaystyle {\ell_2:\ \underline{x}}={(1,0,0)+s(0,1,1)}$.

Show that the lines are skew.

$\displaystyle {\underline{n}}={(a,b,1)}$ is perpendicular to both direction vectors. Find $\displaystyle a$ and $\displaystyle b$.