12.5: Lines and Planes

Lines and planes are perhaps the simplest of curves and surfaces in three dimensional space. They also will prove important as we seek to understand more complicated curves and surfaces.

The equation of a line in two dimensions is \(ax+by=d\), where not both of \(a\) and \(b\) are zero. By analogy, one might suppose that a line in three dimensions might be given by \(ax + by +cz = d\), but it turns out that (so long as not all three of \(a, b, c\) are zero) this is the equation of a plane.

A plane does not have an obvious "direction" as does a line. It is possible to associate a plane with a direction in a very useful way, however: there are exactly two directions perpendicular to a plane. Any vector with one of these two directions is called normal to the plane. So while there are many normal vectors to a given plane, they are all parallel or anti-parallel to each other.

Suppose two points \((v_1,v_2,v_3)\) and \((w_1,w_2,w_3)\) are in a plane; then the vector \(\langle w_1-v_1,w_2-v_2,w_3-v_3\rangle\) is parallel to the plane. In particular, if this vector is placed with its tail at \((v_1,v_2,v_3)\) then its head is at \((w_1,w_2,w_3)\) and it lies in the plane. As a result, any vector perpendicular to the plane is perpendicular to \(\langle w_1-v_1,w_2-v_2,w_3-v_3\rangle\). In fact, it is easy to see that the plane consists of precisely those points \((w_1,w_2,w_3)\) for which \(\langle w_1-v_1,w_2-v_2,w_3-v_3\rangle\) is perpendicular to a normal to the plane, as indicated in figure \(\PageIndex{1}\).

Turning this around, suppose we know that \(\langle a,b,c\rangle\) is normal to a plane containing the point \((v_1,v_2,v_3)\). Then \((x,y,z)\) is in the plane if and only if \(\langle a,b,c\rangle\) is perpendicular to \(\langle x-v_1,y-v_2,z-v_3\rangle\). In turn, we know that this is true precisely when \(\langle a,b,c\rangle \cdot \langle x-v_1,y-v_2,z-v_3\rangle=0\). That is, \((x,y,z)\) is in the plane if and only if \[\eqalign{\langle a,b,c\rangle \cdot \langle x-v_1,y-v_2,z-v_3\rangle&= 0\cr a(x-v_1)+b(y-v_2)+c(z-v_3)&= 0\cr ax+by+cz-av_1-bv_2-cv_3&=0\cr ax+by+cz&=av_1+bv_2+cv_3.\cr }\nonumber\]

Working backwards, note that if \((x,y,z)\) is a point satisfying \(ax+by+cz=d\), then \[\eqalign{ax+by+cz&= d\cr ax+by+cz-d&=0 \cr a(x-d/a)+b(y-0)+c(z-0)&=0\cr \langle a,b,c\rangle\cdot\langle x-d/a,y,z\rangle&= 0,\;\; \text{or} \;\;\langle a,b,c\rangle\cdot\langle x-d/a,y-0,z-0\rangle= 0.\cr}\nonumber\]

Notice that the point \((d/a,0,0)\) also satisfies \(ax+by+cz=d\), the equation of the plane. Which means that we have found a point, in the plane, (in fact, exactly where the \(x\)-axis pierces the plane) which can serve as our tail point, i.e. it corresponds to \((v_1, v_2, v_3)\) in the discussion above. And since the plane consists of exactly all those points \((x,y,z)\) satisfying the plane equation, the final dotted result just above shows that the \(\langle a,b,c\rangle\) vector is perpendicular to any vector \(\langle x-d/a,y-0,z-0\rangle\) lying in the plane which has its tail at \((d/a,0,0)\) and head at \((x,y,z)\). So the conclusion is, that \(\langle a,b,c\rangle\) is normal to the plane. (This argument fails if \(a=0\), but in that case, since at least one coefficient is non-zero, we can use \(b\) or \(c\) in the role of \(a\). Either \(a(x-0)+b(y-d/b)+c(z-0)= 0\), or \(a(x-0)+b(y-0)+c(z-d/c)= 0\) will suffice instead.)

Figure \(\PageIndex{1}\). A plane defined via vectors perpendicular to a normal.

Thus, given a vector \(\langle a,b,c\rangle\) we know that all planes perpendicular to this vector have the form \(ax+by+cz=d\), and any surface of this form is a plane perpendicular to \(\langle a,b,c\rangle\).

Example \(\PageIndex{1}\)

Find an equation for the plane perpendicular to \(\langle 1,2,3\rangle\) and containing the point \((5,0,7)\).

Solution

Using the derivation above, the plane is \(1x+2y+3z=1\cdot 5+2\cdot 0+3\cdot 7=26\). That is to say, we know that the plane is \(x+2y+3z=d\), and to find \(d\) we have substituted the known point on the plane to get \(1\cdot 5+2\cdot 0+3\cdot7=d\), so \(d=26\). We could also write this simply as \(1(x−5)+2(y-0)+3(z−7)=0\), which is for many purposes a fine representation; it can always be multiplied out to give \(x+2y+3z=26\).

Example \(\PageIndex{2}\)

Find a vector normal to the plane \(2x-3y+z=15\).

Solution

One example is \(\langle 2, -3, 1\rangle\). Any vector parallel or anti-parallel to this works as well, so for example \(-2\langle 2, -3,1\rangle= \langle -4,6,-2\rangle\) is also normal to the plane.

We will frequently need to find an equation for a plane given certain information about the plane. While there may occasionally be slightly shorter ways to get to the desired result, it is always possible, and usually advisable, to use the given information to find a normal to the plane and a point on the plane, and then to find the equation as above.

Example \(\PageIndex{3}\)

The planes \(x-z=1\) and \(y+2z=3\) intersect in a line. Find a third plane that contains this line and that is perpendicular to the plane \(x+y-2z=1\).

Solution

First, we note that two planes are perpendicular if and only if their normal vectors are perpendicular. Thus, we seek a vector \(\langle a,b,c\rangle\) that is perpendicular to \(\langle 1,1,-2\rangle\). In addition, since the desired plane with normal (\langle a,b,c\rangle\) is to contain another certain line yet to be determined, \(\langle a,b,c\rangle\) must also be perpendicular to any vector parallel to this line. Since \(\langle a,b,c\rangle\) must be perpendicular to two vectors, we may find it by computing the cross product of the two. So we have yet to find a vector parallel to the line of intersection of the given planes. For this, it suffices to know two points on the line.

To find two points on this line, we must find two different points each of which lies on both of the two planes, \(x-z=1\) and \(y+2z=3\). Now any point on both planes must satisfy \(x-z=1\) and also \(y+2z=3\), and so we are looking for two distinct solutions \(x,y,z\) to these two simultaneous equations in three unknowns. Since this gives us one degree of freedom, an easy way to work this out is to try picking two values for \(x\), say \(x=1\) and \(x=2\). Then solving for the corresponding values of \(y\) and \(z\) we get \(y=3\) and \(y=1\), and \(z=0\) and \(z=1\) respectively, so the points \((1,3,0)\) and \((2,1,1)\) are both on the line of intersection because both are on both planes. This choice is arbitrary, any other pair of points would serve equally as well. Now \(\langle 2-1,1-3,1-0\rangle=\langle 1,-2,1\rangle\) is a vector that is parallel to the line of intersection of the two given planes.

For the cross product, we may choose \(\langle a,b,c\rangle=\langle 1,1,-2\rangle\times \langle 1,-2,1\rangle=\langle -3,-3,-3\rangle\). While this vector would do perfectly well, any vector parallel or anti-parallel to it will work as well, so for example we might choose \(\langle 1,1,1\rangle\), which is anti-parallel to it, and so this vector is normal to the desired plane that we are looking for. And we also already know that \((2,1,1)\) is a point on the plane. Therefore an equation of the plane is \(1\cdot x+1\cdot y+1\cdot z=1\cdot 2+1\cdot 1+1\cdot 1\), i.e. \(x+y+z=4\). As a quick check, since \((1,3,0)\) is also on the line of intersection, it should be on the plane; since \(1+3+0=4\), we see that this is indeed the case.

Note that had we used \(\langle -3,-3,-3\rangle\) as the normal, we would have discovered the equation \(-3x-3y-3z=-12\), then we might well have noticed that we could divide both sides by \(-3\) to get the equivalent \(x+y+z=4\).

So we now understand equations of planes; let us turn to lines. Unfortunately, it turns out to be quite inconvenient to represent a typical line with a single equation; we need to approach lines in a different way.

Unlike a plane, a line in three dimensions does have an obvious direction, namely, the direction of any vector parallel to it. In fact a line can be defined and uniquely identified by providing one point on the line and a vector parallel to the line (in one of two possible directions). That is, the line consists of exactly those points we can reach by starting at the point and going for some distance in the direction of the vector. Let's see how we can translate this into more mathematical language.

Suppose a line contains the point \( (v_1,v_2,v_3)\) and is parallel to the vector \(\langle a,b,c\rangle\). We call \(\langle a,b,c\rangle\) a direction vector for the line. If we place the vector \( \langle v_1,v_2,v_3\rangle\) with its tail at the origin and its head at \( (v_1,v_2,v_3)\), and if we then place the vector \(\langle a,b,c\rangle\) with its tail at \( (v_1,v_2,v_3)\), then the head of \(\langle a,b,c\rangle\) is at a point on the line. We can get to any point on the line by doing the same thing, except using \(t\langle a,b,c\rangle\) in place of \(\langle a,b,c\rangle\), where \(t\) is some real number. Because of the way vector addition works, the point at the head of the vector \(t\langle a,b,c\rangle\) is the point at the head of the combined vector \( \langle v_1,v_2,v_3\rangle+t\langle a,b,c\rangle\), namely \( (v_1+ta, v_2+tb, v_3+tc)\); see figure \(\PageIndex{2}\).

Figure \(\PageIndex{2}\). Vector form of a line.

In other words, by running \(t\) through all possible real values, the vector \(\langle v_1,v_2,v_3\rangle+ t\langle a,b,c\rangle\) can be made to point to any point on the line when its tail has been placed at the origin. Another common way to write this is as a set of parametric equations: \[x= v_1+ta\qquad y= v_2+tb \qquad z= v_3+tc.\nonumber\]

It is occasionally useful to use this form of a line even in two dimensions; a vector form for a line in the \(x\)-\(y\) plane is \(\langle v_1,v_2\rangle+ t\langle a,b\rangle\), which is the same as \(\langle v_1,v_2,0\rangle+ t\langle a,b,0\rangle\).

Example \(\PageIndex{4}\)

Find a vector expression for the line through \((6,1,-3)\) and \((2,4,5)\).

Solution

To get a vector parallel to the line we subtract \(\langle 6,1,-3\rangle-\langle2,4,5\rangle=\langle 4,-3,-8\rangle\). The line is then given by \(\langle 2,4,5\rangle+t\langle 4,-3,-8\rangle\); there are of course many other possibilities, such as \(\langle 6,1,-3\rangle+t\langle 4,-3,-8\rangle\).

Example \(\PageIndex{5}\)

Determine whether the lines \(\langle 1,1,1\rangle+t\langle 1,2,-1\rangle\) and \(\langle 3,2,1\rangle+t\langle -1,-5,3\rangle\) are parallel, intersect, or neither.

Solution

In two dimensions, two lines either intersect or are parallel; in three dimensions, lines that do not intersect might not be parallel. In this case, since the direction vectors for the lines are not parallel or anti-parallel we know the lines are not parallel. If they intersect, there must be two values \(a\) and \(b\) so that \(\langle 1,1,1\rangle+a\langle 1,2,-1\rangle= \langle 3,2,1\rangle+b\langle -1,-5,3\rangle\), that is, \[\eqalign{1+a&= 3-b\cr 1+2a&= 2-5b\cr 1-a&= 1+3b.\cr}\nonumber\] This gives three equations in two unknowns, for which in general there may or may not be a solution. In this case, it is easy to discover that \(a=3\) and \(b=-1\) satisfies all three equations, so the lines do intersect at the point \((4,7,-2)\).

Example \(\PageIndex{6}\)

Find the distance from the point \((1,2,3)\) to the plane \(2x-y+3z=5\).

Solution

The distance from a point \(P\) to a plane is the shortest distance from \(P\) to any point on the plane; this is the distance measured from \(P\) perpendicular to the plane; see figure 12.5.3. This distance is the absolute value of the scalar projection of \(\overrightarrow{\strut QP}\) onto a normal vector of unit length \(\bf n\), where \(Q\) is any point on the plane. It is easy to find a point on the plane, say \((1,0,1)\). Thus the distance is \[\dfrac{\overrightarrow{\strut QP}\cdot{\bf n}}{|{\bf n}|}= \dfrac{ {\langle 0,2,2\rangle \cdot \langle 2,-1,3\rangle}} {|\langle 2,-1,3\rangle|}= \dfrac{4}{\sqrt{14}}.\nonumber\]

Figure \(\PageIndex{3}\). Distance from a point to a plane.

Example \(\PageIndex{7}\)

Find the distance from the point \((-1,2,1)\) to the line \(\langle 1,1,1\rangle + t\langle 2,3,-1\rangle\).

Solution

Again we want the distance measured perpendicular to the line, as indicated in figure \(\PageIndex{4}\). The desired distance is \[ |\overrightarrow{\strut QP}| \sin\theta= \dfrac { |\overrightarrow{\strut QP} \times {\bf A}|} {|{\bf A}|},\nonumber\]

where \(\bf A\) is any vector parallel to the line. From the equation of the line, we can use \(Q=(1,1,1)\) and \({\bf A}=\langle 2,3,-1\rangle\), so the distance is \[\dfrac{ |\langle -2,1,0\rangle \times \langle2,3,-1\rangle|}{\sqrt{14}}= \dfrac{ |\langle-1,-2,-8\rangle|} {\sqrt{14}}= \dfrac{\sqrt{69}}{\sqrt{14}}.\nonumber\]

Figure \(\PageIndex{4}\). Distance from a point to a line.

Exercises \(\PageIndex{}\)

Exercise \(\PageIndex{1}\)

Find an equation of the plane containing \((6,2,1)\) and perpendicular to \(\langle 1,1,1\rangle\).

Answer

\((x−6)+(y−2)+(z−1)=0\)

Exercise \(\PageIndex{2}\)

Find an equation of the plane containing \((-1,2,-3)\) and perpendicular to \(\langle 4,5,-1\rangle\).

Answer

\(4(x+1)+5(y−2)−(z+3)=0\)

Exercise \(\PageIndex{3}\)

Find an equation of the plane containing \((1,2,-3)\), \((0,1,-2)\) and \((1,2,-2)\).

Answer

\((x−1)−(y−2)=0\)

Exercise \(\PageIndex{4}\)

Find an equation of the plane containing \((1,0,0)\), \((4,2,0)\) and \((3,2,1)\).

Answer

\(−2(x−1)+3y−2z=0\)

Exercise \(\PageIndex{5}\)

Find an equation of the plane containing \((1,0,0)\) and the line \(\langle 1,0,2\rangle + t\langle 3,2,1\rangle\).

Answer

\(4(x−1)−6y=0\)

Exercise \(\PageIndex{6}\)

Find an equation of the plane containing the line of intersection of \(x+y+z=1\) and \(x-y+2z=2\), and perpendicular to the plane \(2x+3y−z=4\).

Answer

\(7x−y+11z=11\)

Exercise \(\PageIndex{7}\)

Find an equation of the plane containing the line of intersection of \(x+2y−z=3\) and \(3x−y+4z=7\), and perpendicular to the plane \(6x−y+3z=16\).

Answer

\(−68x−153y+85z=−1462/7\)

Exercise \(\PageIndex{8}\)

Find an equation of the plane containing the line of intersection of \(x+3y−z=6\) and \(2x+2y−3z=8\), and perpendicular to the plane \(3x+y−z=11\).

Answer

\(3x−19y−10z=−10\)

Exercise \(\PageIndex{9}\)

Find an equation of the line through \((1,0,3)\) and \((1,2,4)\).

Answer

\(\langle x,y,z\rangle = \langle 1,0,3\rangle+t\langle 0,2,1\rangle\)

Exercise \(\PageIndex{10}\)

Find an equation of the line through \((1,0,3)\) and perpendicular to the plane \(x+2y-z=1\).

Answer

\(\langle x,y,z\rangle = \langle 1,0,3\rangle+t\langle 1,2,-1\rangle\)

Exercise \(\PageIndex{11}\)

Find an equation of the line through the origin and perpendicular to the plane \(x+y-z=2\).

Answer

\(\langle x,y,z\rangle = t\langle 1,1,-1\rangle\)

Exercise \(\PageIndex{12}\)

Find \(a\) and \(c\) so that \((a,1,c)\) is on the line through \((0,2,3)\) and \((2,7,5)\).

Answer

\(a=−2/5, \quad c=13/5\)

Exercise \(\PageIndex{13}\)

Explain how to discover the solution in example \(\PageIndex{5}\).

Exercise \(\PageIndex{14}\)

Determine whether the lines \(\langle 1,3,-1\rangle+t\langle 1,1,0\rangle\) and \(\langle 0,0,0\rangle+t\langle 1,4,5\rangle\) are parallel, intersect, or neither.

Answer

Neither.

Exercise \(\PageIndex{15}\)

Determine whether the lines \(\langle 1,0,2\rangle+t\langle -1,-1,2\rangle\) and \(\langle 4,4,2\rangle+t\langle 2,2,-4\rangle\) are parallel, intersect, or neither.

Answer

Parallel.

Exercise \(\PageIndex{16}\)

Determine whether the lines \(\langle 1,2,-1\rangle+t\langle 1,2,3\rangle\) and \(\langle 1,0,1\rangle+t\langle 2/3,2,4/3\rangle\) are parallel, intersect, or neither.

Answer

Intersect at \((3,6,5)\).

Exercise \(\PageIndex{17}\)

Determine whether the lines \(\langle 1,1,2\rangle+t\langle 1,2,-3\rangle\) and \(\langle 2,3,-1\rangle+t\langle 2,4,-6\rangle\) are parallel, intersect, or neither.

Answer

They are the same line.

Exercise \(\PageIndex{18}\)

Find a unit normal vector to each of the coordinate planes.

Answer

\(\langle 1,0,0\rangle,\quad\langle 0,1,0\rangle,\quad\langle 0,0,1\rangle\)

Exercise \(\PageIndex{19}\)

Show that \(\langle 2,1,3 \rangle + t \langle 1,1,2 \rangle\) and \(\langle 3, 2, 5 \rangle + s \langle 2, 2, 4 \rangle\) are the same line.

Exercise \(\PageIndex{20}\)

Give a prose description for each of the following processes:

a. Given two distinct points, find the line that goes through them.
b. Given three points (not all on the same line), find the plane that goes through them. Why do we need the caveat that not all three points are on the same line?
c. Given a line and a point not on the line, find the plane that contains them both.
d. Given a plane and a point not on the plane, find the line that is perpendicular to the plane through the given point.

Exercise \(\PageIndex{21}\)

Find the distance from \((2,2,2)\) to \(x+y+z=-1\).

Answer

\(\dfrac{7}{\sqrt{3}}\)

Exercise \(\PageIndex{22}\)

Find the distance from \((2,-1,-1)\) to \(2x-3y+z=2\).

Answer

\(\dfrac{4}{\sqrt{14}}\)

Exercise \(\PageIndex{23}\)

Find the distance from \((2,-1,1)\) to the line \(\langle 2,2,0\rangle+t\langle 1,2,3\rangle\).

Answer

\(\sqrt{\dfrac{131}{14}}\)

Exercise \(\PageIndex{24}\)

Find the distance from the point \((1,0,1)\) to the line \(\langle 3,2,1\rangle+t\langle 2,-1,-2\rangle\).

Answer

\(\dfrac{\sqrt{68}}{3}\)

Exercise \(\PageIndex{25}\)

Find the distance between the lines \(\langle 5,3,1\rangle+t\langle 2,4,3\rangle\) and \(\langle 6,1,0\rangle+t\langle 3,5,7\rangle\).

Answer

\(\dfrac{25}{\sqrt{198}}\)

Exercise \(\PageIndex{26}\)

Find the distance between the lines \(\langle 2,1,3\rangle+t\langle -1,2,-3\rangle\) and \(\langle 1,-3,4\rangle+t\langle 4,-4,1\rangle\).

Answer

\(\dfrac{50}{\sqrt{237}}\)

Exercise \(\PageIndex{27}\)

Find the distance between the lines \(\langle 1,2,3\rangle+t\langle 2,-1,3\rangle\) and \(\langle 4,5,6\rangle+t\langle -4,2,-6\rangle\).

Answer

\(3\sqrt{\dfrac{13}{7}}\)

Exercise \(\PageIndex{28}\)

Find the distance between the lines \(\langle 3,2,1\rangle+t\langle 1,4,-1\rangle\) and \(\langle 3,1,3\rangle+t\langle 2,8,-2\rangle\).

Answer

\(\sqrt{3}\)

Exercise \(\PageIndex{29}\)

Find the cosine of the angle between the planes \(x+y+z=2\) and \(x+2y+3z=8\).

Answer

\(\sqrt{\dfrac{6}{7}}\)

Exercise \(\PageIndex{30}\)

Find the cosine of the angle between the planes \(x-y+2z=2\) and \(3x-2y+z=5\).

Answer

\(\dfrac{1}{2}\sqrt{\dfrac{7}{3}}\)

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