We continue our study of three-dimensional space by considering lines and planes in $\mathbb{R}^3.$
Part I
Lines in Spaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaace!
Lines in Spaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaace!
Lines in Space
Just as in two dimensions, it takes two points to uniquely determine a line. In figure below we choose two arbitrary points.
Lines in Space
If we let $t$ vary as a parameter, the equation $\overrightarrow{PQ}=t{\bf v}$ describes a line passing through the points $P=(x_0,y_0,z_0)$ and $Q=(x,y,z).$
Lines in Space
If we let $t$ vary as a parameter, the equation $\overrightarrow{PQ}=t{\bf v}$ describes a line passing through the points $P=(x_0,y_0,z_0)$ and $Q=(x,y,z).$
Letting where ${\bf r} = \langle x, y, z \rangle$ and ${\bf r}_0 = \langle x_0, y_0, z_0 \rangle,$ we may write this equation in several forms. $$ \begin{array}{ll} \langle x-x_0,y-y_0,z-z_0\rangle=t\langle a,b,c\rangle&\\ \langle x,y,z\rangle-\langle x_0,y_0,z_0\rangle=t\langle a,b,c\rangle&\\ {\bf r}-{\bf r}_0=t{\bf v}& \mbox{}\\ {\bf r}=t{\bf v}+{\bf r}_0&\\ \end{array} $$ Rewriting the first in the above list as $\langle x-x_0,y-y_0,z-z_0\rangle=\langle ta,tb,tc\rangle$ and breaking the equation up into components we have $x-x_0=at,\,\,y-y_0=bt,\,\,z-z_0=ct,$ or $$ \begin{array}{ll} x=at+x_0\\y=bt+y_0\\z=ct+z_0&\\ \end{array} $$ which is the parametric representation of a line in space.
Lines in Space
Yet another form we will encounter is obtained by eliminating the parameter $t.$ That is, solve each equation in $$ \begin{array}{ll} x=at+x_0\\y=bt+y_0\\z=ct+z_0&\\ \end{array} $$ for $t$ to get $\displaystyle \frac{x-x_0}{a}=t,$ $\displaystyle \frac{y-y_0}{b}=t,$ and $\displaystyle \frac{z-z_0}{c}=t.$ Then $$ \frac{x-x_0}{a}=\frac{y-y_0}{b}=\displaystyle \frac{z-z_0}{c} $$ This last equation is called the symmetric equation of a line.
Example
Find parametric and symmetric equations of the line passing through points $(1, -3, 2)$ and $(5, -2, 8).$
Solution
We first find the vector form of the equation.
Let $P=(1, -3, 2)$ and $Q=(5, -2, 8).$ Then $$\overrightarrow{PQ}={\bf v}=\langle 5-1,-2-(-3),8-2 \rangle=\langle 4,1,6 \rangle$$ The vector form of the equation is then $$ {\bf r}(t)={\bf v}t+{\bf r}_0=\langle 4,1,6 \rangle t + \langle 1,-3,2 \rangle $$ In component form we have $$ \begin{array}{rl} x(t)&=4t+1\\ y(t)&=t-3\\ z(t)&=6t+2 \end{array} $$ Eliminating the parameter, the symmetric equation is then $$ \frac{x-1}{4}=y+3=\frac{z-2}{6} $$
Special Note
The vector ${\bf r}$ will now hold a special place in our hearts. We will use this as the vector notation for a curve in space. That is, in addition to the parametric notation for a curve in space, $$ x=x(t)\\ y=y(t)\\ z=z(t), $$ we may write the above more compactly as $$ {\bf r}(t)=\langle x(t),y(t),z(t)\rangle $$
Line Segments
When modelling certain situations, such as the straight-line motion of a particle in space, we may want to model this motion from an initial known point $P=(x_0,y_0,z_0)$ to a terminal known point $Q=(x_1,y_1,z_1).$
Letting ${\bf p}=\langle x_0,y_0,z_0\rangle$ and ${\bf q}=\langle x_1,y_1,z_1\rangle,$ the vector ${\bf q}-{\bf p}$ is parallel to the line passing through $P$ and $Q.$
Just as before, the motion along the line determined by these points can be described as ${\bf r}={\bf p}+t\overrightarrow{PQ},$ or, $$ \begin{array}{ll} &{\bf r}={\bf p}+t({\bf q}-{\bf p})\\ \implies &{\bf r}={\bf p}+t{\bf q}-t{\bf p}\\ \implies &{\bf r}=(1-t){\bf p}+t{\bf q}\\ \end{array} $$ This last equation is handy for modelling straight line motion from $P$ and $Q$ where $0\leq t \leq 1.$
We see that ${\bf r}(0)=(1-0){\bf p}+0\cdot{\bf q}={\bf p}$ and ${\bf r}(1)=(1-1){\bf p}+1\cdot{\bf q}={\bf q}.$
Parametric Form of a Line Segment
The vector equation for a line segment ${\bf r}=(1-t){\bf p}+t{\bf q}={\bf p}+t({\bf q}-{\bf p})$ can be written in component (parametric) form as $$ x = x_0 + t(x_1 - x_0)\\ y = y_0 + t(y_1 - y_0)\\z = z_0 + t(z_1 - z_0) $$ where $0 \leq t \leq 1.$
Example
Find parametric equations of the line segment between points $(-1, 3, 6)$ and $(-8, 2, 4).$
Solution
Letting $P=(-1, 3, 6)$ and $Q=(-8, 2, 4),$ $$ \begin{array}{rll} x &= -1 + t(-8 - (-1))&=-7t-1\\ y &= 3 + t(2-3)&=-t+3\\ z &= 6 + t(4 -6)&=-2t+6 \end{array} $$ where $0\leq t \leq 1.$
Distance Between a Point and a Line
Let $L$ be a line in space and let $M$ be any point not on the line.
Our goal is to find the distance $D$ between the point $M$ and the line $L.$
Distance Between a Point and a Line
Suppose $L$ has direction vector ${\bf v}.$ Choose a point $P$ on $L.$ The area of the parallelogram formed by $\overrightarrow{PM}$ and ${\bf v}$ can be expressed in two ways:
1) as the magnitude of the cross product $\Vert\overrightarrow{PM}\times{\bf v}\Vert$
2) as the product of base and height, that is, the length of ${\bf v}$ and the distance $D.$
Equating these two quantities, we have $D\cdot \Vert{\bf v}\Vert=\Vert\overrightarrow{PM}\times{\bf v}\Vert.$
Distance Between a Point and a Line
Let $L$ be a line in space passing through point $P$ with direction vector ${\bf v}.$ If $M$ is any point not on $L,$ then the distance $D$ from $M$ to $L$ is $$ D=\frac{\Vert\overrightarrow{PM}\times{\bf v}\Vert}{\Vert{\bf v}\Vert} $$
Example
Find the distance between the point $M = (1, 1, 3)$ and line $\displaystyle \frac{x-3}{4}=\frac{y+1}{2}=z-3.$
Solution
To compute $$ D=\frac{\Vert\overrightarrow{PM}\times{\bf v}\Vert}{\Vert{\bf v}\Vert} $$ we need to find a ${\bf v}$ parallel to the line $L$ and a point $P$ on the line.
The symmetric equation for $L$ is $\displaystyle \frac{x-3}{4}=\frac{y+1}{2}=z-3,$ which immediately gives us a candidate ${\bf v}=\langle 4,2,1\rangle.$
To find a point $P$ on the line, we just need a point $P=(x_0,y_0,z_0)$ which satisfies the equation.
Let's find a point which satisfies $\displaystyle \frac{x-3}{4}=\frac{y+1}{2}=z-3\color{magenta}{=0}.$ (Note that we could have just as easily chosen $\color{magenta}{1},$ $\color{magenta}{-2},$ or any other number.)
Then $$ \frac{x-3}{4}=0\\\frac{y+1}{2}=0\\z-3=0 $$ which gives $x=3,$ $y=-1,$ and $z=3.$ Thus $P=(3,-1,3).$ Since $M=(1,1,3),$
Thus, $\overrightarrow{PM}=\langle 1-3,1-(-1),0 \rangle=\langle -2,2,0 \rangle$
Next, we compute $\overrightarrow{PM}\times{\bf v}.$ $$ \begin{array}{lll} \displaystyle\overrightarrow{PM}\times{\bf v} &\displaystyle=\left|\begin{array}{ccc}{\bf i}&{\bf j}&{\bf k}\\ -2&2&0 \\ 4&2&1 \\\end{array}\right| &\mbox{}\\ \displaystyle &\displaystyle= (2-0){\bf i}-(-2-0){\bf j}+(-4-8){\bf k} &\mbox{}\\ \displaystyle &\displaystyle= 2{\bf i}+2{\bf j}-12{\bf k} &\mbox{}\\ \displaystyle &\displaystyle= \langle 2,2,-12 \rangle &\mbox{}\\ \end{array} $$ Now, $\Vert {\bf v}\Vert=\sqrt{4^2+2^2+1^2}=\sqrt{21}$ and $\Vert\overrightarrow{PM}\times{\bf v}\Vert=\sqrt{2^2+2^2+(-12)^2}=\sqrt{152}.$
Thus, $$ \begin{array}{lll} \displaystyle D&\displaystyle=\frac{\Vert\overrightarrow{PM}\times{\bf v}\Vert}{\Vert{\bf v}\Vert} &\mbox{}\\ \displaystyle &\displaystyle= \frac{\sqrt{152}}{\sqrt{21}}&\mbox{}\\ \displaystyle &\displaystyle= \frac{2\sqrt{38}}{\sqrt{21}}&\mbox{}\\ \displaystyle &\displaystyle\approx 2.690370837 &\mbox{}\\ \end{array} $$
When Lines in Space Intersect... or not
When it comes to comparing two lines in space, there are four possibilities.
Example
Determine if the lines given by $$x = 2s - 1, y = s - 1, z = s - 4$$ and $$x = t - 3, y = 3t + 8, z = 5 - 2t$$ equal, parallel but not equal, skew, or intersecting.
Solution
A direction vector for the first line is ${\bf v}=\langle 2,1,1 \rangle$ and for the second line is ${\bf w}=\langle 1,3,-2 \rangle.$
Since these vectors are not scalar multiples of one another, they are not parallel.
Thus, these two lines must either intersect at a single point, or they are skew lines.
Now, if these lines intersect at a point, it must be for some $s$ and some $t$ that $$ \begin{cases} 2s-1=t-3\\ s-1=3t+8\\ s-4=5-2t\\ \end{cases} $$ We rewrite this system as $$ \begin{cases} t=2s+2\\ s=3t+9\\ s=9-2t\\ \end{cases} $$ Substituting the first equation, $t=2s+2,$ into the second equation gives $$ s=3t+9=3(2s+2)+9=6s+6+9=6s+15 $$ which gives $-5s=15$ so that $s=-3.$
But when we substitute the first equation, $t=2s+2,$ into the third equation, we have $$ s=9-2t=9-2(2s+2)=9-4s-4=5-4s $$ which gives $5s=5$ so that $s=1$ which contradicts the conclusion above that $s=-3.$
Thus, there are not values of $s$ and $t$ which simultaneously satisfy all three equations.
It follows that these two lines are skew lines.
Part II
The Plane! The Plane!
The Plane! The Plane!
Planes
While a line is determined by two points, a plane is determined by three points.
However, instead of points determining planes, we will use our knowledge of vectors to generate planes.
Planes
Consider a vector ${\bf n}=\langle a,b,c \rangle.$ The plane is all vectors $\overrightarrow{PQ}$ orthogonal to ${\bf n}.$ That is, $$ {\bf n}\cdot \overrightarrow{PQ}=0. $$
We note that a vector orthogonal to a plane is called a normal vector.
The equation is called the vector equation of the plane.
Planes
Letting $P=(x_0,y_0,z_0)$ be the initial point of the normal vector ${\bf n}=\langle a,b,c \rangle$ and $Q=(x,y,z)$ be the terminal point of $\overrightarrow{PQ},$ we may rewrite the vector equation ${\bf n}\cdot \overrightarrow{PQ}=0$ as $$ \color{#ea5f25}{\langle a,b,c \rangle} \cdot \color{#134379}{\langle x-x_0,y-y_0,z-z_0 \rangle}=0 $$
The above may be written as a scalar equation for a plane. $$ a(x-x_0)+b(y-y_0)+c(z-z_0)=0 $$
Planes
The equation for a plane $$ a(x-x_0)+b(y-y_0)+c(z-z_0)=0 $$ may also be written as $$ ax+by+cz+d=0 $$ where $d=-ax_0 - by_0 - cz_0.$
Example
Write an equation for the plane containing points $P = (1, 1, -2),$ $Q = (0, 2, 1),$ and $R = (-1, -1, 0)$ in both standard and general forms.
Solution
We will find a normal vector to the plane containing the points $P = (1, 1, -2),$ $Q = (0, 2, 1),$ and $R = (-1, -1, 0)$ by computing the cross product of $\overrightarrow{PQ}$ and $\overrightarrow{PR}.$
Now $$\color{blue}{\overrightarrow{PQ}}=\langle 0-1,2-1,1-(-2) \rangle=\langle -1,1,3 \rangle$$ and $$\color{magenta}{\overrightarrow{PR}}=\langle -1-1,-1-1,0-(-2) \rangle=\langle -2,-2,2 \rangle.$$ We then have $$ \begin{array}{lll} \displaystyle \color{darkorange}{{\bf n}} &\displaystyle= \color{darkorange}{\overrightarrow{PQ} \times \overrightarrow{PR}} &\mbox{}\\ \displaystyle &\displaystyle= \langle -1,1,3 \rangle \times \langle -2,-2,2 \rangle &\mbox{}\\ \displaystyle &\displaystyle= \left|\begin{array}{ccc}{\bf i}&{\bf j}&{\bf k}\\ -1&1&3 \\ -2&-2&2 \\\end{array}\right|&\mbox{}\\ \displaystyle &\displaystyle= (2-(-6)){\bf i}-(-2-(-6)){\bf j}+(2-(-2)){\bf k}&\mbox{}\\ \displaystyle &\displaystyle= 8{\bf i}-4{\bf j}+4{\bf k}&\mbox{}\\ \displaystyle &\displaystyle= \langle 8,-4,4\rangle&\mbox{}\\ \end{array} $$ Taking the initial point $P = (1, 1, -2)$ of $\overrightarrow{PQ}$ and $\overrightarrow{PR}$ as our point in the plane, the equation of our plane is $$ \begin{array}{lll} \displaystyle {\bf n}\cdot \langle x-x_0,y-y_0,z-z_0 \rangle&\displaystyle= \langle 8,-4,4\rangle \cdot \langle x-1,y-1,z+2 \rangle &\mbox{}\\ \displaystyle &\displaystyle=8(x-1)-4(y-1)+4(z+2) &\mbox{}\\ \displaystyle &\displaystyle=8x-8-4y+4+4z+8 &\mbox{}\\ \displaystyle &\displaystyle=8x-4y+4z+4&\mbox{}\\ \displaystyle &\displaystyle=0 &\mbox{}\\ \end{array} $$ The equation of our plane is then $8x-4y+4z+4=0$ which can be more simply stated as $$ 2x-y+z+1=0 $$
Example
Find an equation of the plane that passes through point $(1, 4, 3)$ and contains the line given by $\displaystyle x=\frac{y-1}{2}=z+1.$
Solution
All we need is three points on the plane so that we can do what we did in the last example.
The point $P=(1, 4, 3)$ is handed to us on a velvet pillow, but the other two we need to work for.
Since the plane we seek contains the line $\displaystyle x=\frac{y-1}{2}=z+1,$ all we need are two solutions to this equation to get our other two points.
Since the symmetric equations are the result of eliminating the parameter, we simply uneliminate the parameter by writing $$ \displaystyle x=\frac{y-1}{2}=z+1=t $$ and solve for $x,$ $y,$ and $z.$ $$ \begin{array}{lll} \displaystyle x&\displaystyle=t &\mbox{}\\ \displaystyle y&\displaystyle=2t+1 &\mbox{}\\ \displaystyle z&\displaystyle= t-1.&\mbox{}\\ \end{array} $$ Since any value of $t$ will generate solutions, let's use a couple easy values, $t=0$ and $t=1.$
For $t=0,$ we have $x=0,$ $1,$ and $z=-1.$ Thus, $Q=(0,1,-1)$ will be our second point.
For $t=1,$ we have $x=1,$ $3,$ and $z=0.$ Thus, $R=(1,3,0)$ will be our third point.
As before, we now find a normal vector to the plane containing the points $P = (1, 4, 3),$ $Q = (0, 1, -1),$ and $R = (1,3,0)$ by computing the cross product of $\overrightarrow{PQ}$ and $\overrightarrow{PR}.$
Now $$\color{#0594f8}{\overrightarrow{PQ}}=\langle 0-1,1-4,-1-3 \rangle=\langle -1,-3,-4 \rangle$$ and $$\color{magenta}{\overrightarrow{PR}}=\langle 1-1,3-4,0-3 \rangle=\langle 0,-1,-3 \rangle.$$ We may now find our normal vector, $$ \begin{array}{lll} \displaystyle \color{darkorange}{{\bf n}} &\displaystyle= \color{darkorange}{\overrightarrow{PQ} \times \overrightarrow{PR}} &\mbox{}\\ \displaystyle &\displaystyle= \langle -1,-3,-4 \rangle \times \langle 0,-1,-3\rangle &\mbox{}\\ \displaystyle &\displaystyle= \left|\begin{array}{ccc}{\bf i}&{\bf j}&{\bf k}\\ -1&-3&-4 \\ 0&-1&-3 \\\end{array}\right|&\mbox{}\\ \displaystyle &\displaystyle= (9-4){\bf i}-(3-0){\bf j}+(1-0){\bf k}&\mbox{}\\ \displaystyle &\displaystyle= 5{\bf i}-3{\bf j}+{\bf k}&\mbox{}\\ \displaystyle &\displaystyle= \langle 5,-3,1\rangle&\mbox{}\\ \end{array} $$ Taking the initial point $P = (1, 4, 3)$ of $\overrightarrow{PQ}$ and $\overrightarrow{PR}$ as our point in the plane, the equation of our plane is $$ \begin{array}{lll} \displaystyle {\bf n}\cdot \langle x-x_0,y-y_0,z-z_0 \rangle&\displaystyle= \langle 5,-3,1\rangle \cdot \langle x-1,y-4,z-3 \rangle &\mbox{}\\ \displaystyle &\displaystyle=5(x-1)-3(y-4)+1\cdot(z-3) &\mbox{}\\ \displaystyle &\displaystyle=5x-5-3y+12+z-3 &\mbox{}\\ \displaystyle &\displaystyle=5x-3y+z+4&\mbox{}\\ \displaystyle &\displaystyle=0 &\mbox{}\\ \end{array} $$ The equation of our plane is then $5x-3y+z+4=0.$
Distance Between a Point and a Plane
Suppose a plane with normal vector ${\bf n}$ passes through point $Q.$ We would like to find the distance $D$ from the plane to a point $P$ not in the plane.
Dropping a perpendicular from $P$ to a point $R$ on the plane, we see that the vector $\overrightarrow{RP}$ is parallel to ${\bf n}.$
We notice two things: 1) $D=\Vert \overrightarrow{RP} \Vert,$ and 2) $\overrightarrow{RP}=\mbox{proj}_{{\bf n}}\overrightarrow{QP}$
Distance Between a Point and a Plane
From the above observations we have $D=\Vert \mbox{proj}_{{\bf n}}\overrightarrow{QP} \Vert$
Then $ \displaystyle D =\Vert \mbox{proj}_{{\bf n}}\overrightarrow{QP} \Vert =\left|\mbox{comp}_{{\bf n}}\overrightarrow{QP}\right| =\left|\frac{\overrightarrow{QP} \cdot {\bf n}}{\Vert{\bf n} \Vert}\right| =\frac{|\overrightarrow{QP} \cdot {\bf n}|}{|\Vert{\bf n} \Vert|} =\frac{|\overrightarrow{QP} \cdot {\bf n}|}{\Vert{\bf n} \Vert} $
Stating the result formally...
Distance Between a Point and a Plane
Suppose a plane with normal vector ${\bf n}$ passes through point $Q.$ Then the distance $d$ from the plane to a point $P$ not in the plane is given by $$ \displaystyle D =\frac{|\overrightarrow{QP} \cdot {\bf n}|}{\Vert{\bf n} \Vert} $$
Example
Find the distance between point $P = (3, 1, 2)$ and the plane given by $x - 2y + z = 5.$
Solution
The plane given by $x - 2y + z = 5$ has ${\bf n}=\langle 1,-2,1 \rangle$ as a normal vector.
All we need now is a point $Q$ on the plane. Keeping things easy, we let $x=0$ and $y=0$ which forces $z=5.$ (This is a $z$-intercept of the surface.) Thus, $Q=(0,0,5)$ is point on the plane.
Now $$ \overrightarrow{QP}=\langle 3-0,1-0,2-5 \rangle=\langle 3,1,-3 \rangle $$ The distance is then $$ \begin{array}{lll} \displaystyle D &\displaystyle= \frac{|\overrightarrow{QP} \cdot {\bf n}|}{\Vert{\bf n} \Vert} &\mbox{}\\ \displaystyle &\displaystyle= \frac{|\langle 3,1,-3 \rangle \cdot \langle 1,-2,1 \rangle|}{\Vert\langle 1,-2,1 \rangle \Vert}&\mbox{}\\ \displaystyle &\displaystyle= \frac{|3\cdot 1+1(-2)+(-3)\cdot 1|}{\sqrt{1^2+(-2)^2+1^2}}&\mbox{}\\ \displaystyle &\displaystyle= \frac{2}{\sqrt{6}}&\mbox{}\\ \displaystyle &\displaystyle= \frac{2\sqrt{6}}{6}&\mbox{}\\ \displaystyle &\displaystyle= \frac{\sqrt{6}}{3}&\mbox{}\\ \displaystyle &\displaystyle\approx 0.8164965809&\mbox{}\\ \end{array} $$
When Planes Intersect... or not
While lines admit several possibilities when they intersect, there are fewer when we consider planes.
Two planes can either
- intersect along a line in space
- be parallel to one another (and possibly equal)
Example
Find parametric and symmetric equations for the line formed by the intersection of the planes given by $x + y + z = 0$ and $2x - y + z = 0$
Solution
The intersection of the planes given by $x + y + z = 0$ and $2x - y + z = 0$ is the solution to the $2\times 3$ system of equations $$ \begin{cases} x + y + z = 0\\ 2x - y + z = 0 \end{cases} $$ The collection of solutions to this system is line of intersection.
Adding the two equations in the system together, $y$ is eliminated, and we have $3x+2z=0.$ Then, $\displaystyle z=-\frac{3}{2}x.$
Now, if we subtract the first equation from the second, $z$ is eliminated and we get $x-2y=0.$ Then $\displaystyle y=\frac{1}{2}x.$
The above shows that every solution to the above system has the form $\displaystyle \left(x,y,z\right)=\left(x,\frac{1}{2}x,-\frac{3}{2}x\right)$
We may now replace $x$ with a parameter $t$ to express the line in parametric form $$ \begin{array}{lll} \displaystyle x&\displaystyle=t &\mbox{}\\ \displaystyle y&\displaystyle=\frac{1}{2}t &\mbox{}\\ \displaystyle z&\displaystyle=-\frac{3}{2}t &\mbox{}\\ \end{array} $$ Eliminating the parameter we obtain the symmetric equation. $$ x=2y=-\frac{2}{3}z $$ If you dislike fractions, we may equivalently express this as $3x=6y=-2z.$
The Angle Between Two Intersecting Planes
Suppose we have two planes which intersect along a line.
From the equations of these planes, we may determine a normal vector ${\bf n}_1$ and ${\bf n}_2$ to each of these planes.
The Angle Between Two Intersecting Planes
From there we may simply find the acute angle between ${\bf n}_1$ and ${\bf n}_2$ since $$ \cos \theta =\frac{{\bf n}_1\cdot {\bf n}_2}{\Vert {\bf n}_1\Vert \Vert {\bf n}_2\Vert} $$
Example
Determine whether the planes $$x + y + z = 4$$ and $$x - 3y + 5z = 1$$ are parallel, orthogonal, or neither.
If the planes are intersecting, but not orthogonal, find the measure of the angle between them.
Solution
For the planes $x + y + z = 4$ and $x - 3y + 5z = 1,$ we will use the normal vectors ${\bf n}_1=\langle 1,1,1 \rangle$ and ${\bf n}_2=\langle 1,-3,5 \rangle.$
Now, ${\bf n}_1\cdot {\bf n}_2=\langle 1,1,1 \rangle \cdot \langle 1,-3,5 \rangle=1-3+5=3.$ So, the planes are not orthogonal.
Also, since $$ \begin{array}{lll} \displaystyle {\bf n}_1\times {\bf n}_2&\displaystyle= \langle 1,1,1 \rangle \times \langle 1,-3,5 \rangle&\mbox{}\\ \displaystyle &\displaystyle= \left|\begin{array}{ccc}{\bf i}&{\bf j}&{\bf k}\\ 1&1&1 \\ 1&-3&5 \\\end{array}\right|&\mbox{}\\ \displaystyle &\displaystyle= (5-(-3)){\bf i}-(5-1){\bf j}+(-3-1){\bf k}&\mbox{}\\ \displaystyle &\displaystyle= 8{\bf i}-4{\bf j}-4{\bf k}&\mbox{}\\ \displaystyle &\displaystyle= \langle 8,-4,-4\rangle &\mbox{}\\ \displaystyle &\displaystyle\neq {\bf 0} &\mbox{}\\ \end{array} $$ ${\bf n}_1$ and ${\bf n}_2$ are not parallel either.
We conclude that the two planes intersect non-orthogonally.
We may now find the acute angle of intersection. $$ \begin{array}{lll} \displaystyle \cos \theta &\displaystyle= \frac{{\bf n}_1\cdot {\bf n}_2}{\Vert {\bf n}_1\Vert \Vert {\bf n}_2\Vert}&\mbox{}\\ \displaystyle &\displaystyle= \frac{\langle 1,1,1 \rangle\cdot \langle 1,-3,5 \rangle}{\Vert \langle 1,1,1 \rangle \Vert \Vert \langle 1,-3,5 \rangle \Vert} &\mbox{}\\ \displaystyle &\displaystyle= \frac{3 \rangle}{\sqrt{1^2+1^2+1^2} \sqrt{1^2+(-3)^2+5^2}} &\mbox{}\\ \displaystyle &\displaystyle= \frac{3}{\sqrt{3} \sqrt{35}} &\mbox{}\\ \displaystyle &\displaystyle= \frac{\sqrt{3}}{\sqrt{35}} &\mbox{}\\ \end{array} $$ Then $$ \begin{array}{lll} \displaystyle \theta &\displaystyle= \cos^{-1}\left(\frac{\sqrt{3}}{\sqrt{35}}\right) &\mbox{}\\ \displaystyle &\displaystyle\approx 1.27367381 &\mbox{}\\ \end{array} $$ The angle between the two planes is about $1.27$ radians, or about $73^{\circ}.$
Distance Between a Point and a Plane (Redux)
If $Q=(x,y,z)$ is an arbitrary point in the plane and $P=(x_0,y_0,z_0)$ is some point off of the plane, and if the equation the plane is given by $ax+by+cz+d=0$ with normal vector ${\bf n}=\langle a,b,c \rangle,$ we have that the distance $D$ between these planes is given by $$ \begin{array}{lll} D&=\displaystyle\frac{|\overrightarrow{QP} \cdot {\bf n}|}{\Vert{\bf n} \Vert}&\\ &=\displaystyle \frac{|\langle x_0-x,y_0-y,z_0-z\rangle \cdot \langle a,b,c \rangle|}{\Vert \langle a,b,c \rangle \Vert}&\\ &=\displaystyle \frac{|a(x_0-x)+b(y_0-y)+c(z_0-z)|}{\sqrt{a^2+b^2+c^2}}&\\ &=\displaystyle \frac{|ax_0-ax+by_0-by+cz_0-bz|}{\sqrt{a^2+b^2+c^2}}&\\ &=\displaystyle \frac{|ax_0+by_0+cz_0-ax-by-cz|}{\sqrt{a^2+b^2+c^2}}&\\ &=\displaystyle \frac{|ax_0+by_0+cz_0+d|}{\sqrt{a^2+b^2+c^2}}&\mbox{since $ax+by+cz+d=0$}\\ %&=\displaystyle \frac{|ax_0+by_0+cz_0+d|}{\sqrt{a^2+b^2+c^2}}&\mbox{}\\ \end{array} $$ We have the following result...
Distance Between a Point and a Plane (Redux)
Let $P=(x_0, y_0, z_0)$ be a point. The distance from $P$ to plane $ax + by + cz + d = 0$ is given by $$ \begin{array}{lll} D&=\displaystyle \frac{|ax_0+by_0+cz_0+d|}{\sqrt{a^2+b^2+c^2}}&\\ \end{array} $$
Distance Between Two Planes
If two planes don't intersect, then there is a constant distance between them.
Using the above result, we may find the distance between the two planes.
Example
Find the distance between the two parallel planes given by $2x + y - z = 2$ and $2x + y - z = 8.$
Solution
Let's designate $2x + y - z = 2$ as $P_1$ and $2x + y - z = 8$ as $P_2.$
We'll find a point on $P_2$ and find its distance from $P_1$ using the formula for the distance between a point and plane.
Again, going for an easy solution, when $x=0$ and $y=0,$ then $z=-8$ on $P_2.$ Thus, we take $P=(0,0,8).$
The distance between $P=(0,0,8)$ and $P_1$ is $$ \begin{array}{lll} D&=\displaystyle \frac{|ax_0+by_0+cz_0+d|}{\sqrt{a^2+b^2+c^2}}&\\ &=\displaystyle \frac{|2x_0+y_0-z_0-2|}{\sqrt{2^2+1^2+(-1)^2}}&\\ &=\displaystyle \frac{|2\cdot 0+0-(-8)-2|}{\sqrt{2^2+1^2+(-1)^2}}&\\ &=\displaystyle \frac{6}{\sqrt{6}}&\\ &=\displaystyle \sqrt{6}&\\ \end{array} $$
Bonus Example
A solar panel is mounted on the roof of a house. The corners of the panel are positioned at the points with coordinates (in meters) $A=(0, 0, 0),$ $B=(18, 0, 0),$ $C=(18, 8, 8),$ and $D=(0, 8, 8).$
(a) Find the equation of the plane that contains the solar panel by using points $A,$ $B,$ and $D.$
(b) What is the angle of elevation of the roof?
Solution
(a) We first find $\overrightarrow{AB}$ and $\overrightarrow{AD}.$ $$\overrightarrow{AB}=\langle 18-0,0-0,0-0, \rangle=\langle 18,0,0\rangle$$ and $$\overrightarrow{AD}=\langle 0-0,8-0,8-0, \rangle=\langle 0,8,8\rangle$$ Then, a vector normal to the panel is $$ \begin{array}{lll} \displaystyle {\bf n} &\displaystyle= \overrightarrow{AB} \times \overrightarrow{AD}&\mbox{}\\ \displaystyle &\displaystyle= \left|\begin{array}{ccc}{\bf i}&{\bf j}&{\bf k}\\ 18&0&0 \\ 0&8&8 \\\end{array}\right|&\mbox{}\\ \displaystyle &\displaystyle= (0-0){\bf i}-(144-0){\bf j}+(144-0){\bf k}&\mbox{}\\ \displaystyle &\displaystyle= -144{\bf j}+144{\bf k}&\mbox{}\\ \displaystyle &\displaystyle= \langle 0,-144,144\rangle&\mbox{}\\ \end{array} $$ Taking the initial point $A = (0, 0, 0)$ of $\overrightarrow{AB}$ and $\overrightarrow{AD}$ as our point in the plane, the equation of our plane is $$ \begin{array}{lll} \displaystyle {\bf n}\cdot \langle x-x_0,y-y_0,z-z_0 \rangle&\displaystyle= \langle 8,-4,4\rangle \cdot \langle x-1,y-1,z+2 \rangle &\mbox{}\\ \displaystyle &\displaystyle=0(x-0)-144(y-0)+144(z-0) &\mbox{}\\ \displaystyle &\displaystyle=-144y+144z &\mbox{}\\ \displaystyle &\displaystyle=0 &\mbox{}\\ \end{array} $$ The equation of our plane is then $-144y+144z=0$ which can be more simply stated as $$ z=y $$ (b) We now find the angle of elevation of the roof. We mean this to be the angle $\theta$ seen below.
We note this is the angle between the plane $z=y$ with normal vector ${\bf n}=\langle 0,-144,144\rangle$ and the $xy$-plane, $z=0$ with normal vector ${\bf n}_2={\bf k}=\langle 0,0,1\rangle.$ Using the formula for the angle between two planes, $$ \begin{array}{lll} \displaystyle \cos \theta &\displaystyle= \frac{{\bf n}\cdot {\bf n}_2}{\Vert {\bf n}\Vert \Vert {\bf n}_2\Vert} &\mbox{}\\ \displaystyle &\displaystyle= \frac{\langle 0,-144,144\rangle\cdot \langle 0,0,1\rangle}{\Vert \langle 0,-144,144\rangle\Vert \Vert \langle 0,0,1\rangle\Vert}&\mbox{}\\ \displaystyle &\displaystyle= \frac{144}{\sqrt{0^2+(-144)^2+144^2} \cdot 1}&\mbox{}\\ \displaystyle &\displaystyle= \frac{144}{\sqrt{2\cdot 144^2}}&\mbox{}\\ \displaystyle &\displaystyle= \frac{144}{144\sqrt{2}}&\mbox{}\\ \displaystyle &\displaystyle= \frac{1}{\sqrt{2}}&\mbox{}\\ \displaystyle &\displaystyle= \frac{\sqrt{2}}{2}&\mbox{}\\ \end{array} $$ Thus, $$ \begin{array}{lll} \displaystyle \theta&\displaystyle=\cos^{-1}\left(\frac{\sqrt{2}}{2}\right) &\mbox{}\\ \displaystyle &\displaystyle= \frac{\pi}{4}&\mbox{}\\ \end{array} $$ Thus, the angle of elevation of the roof is $\displaystyle \frac{\pi}{4}$ radians, or $45^{\circ}.$