How to Draw Normal to a Plane

Lines and planes are possibly the simplest of curves and surfaces in three dimensional space. They also volition prove important equally nosotros seek to understand more complicated curves and surfaces.

The equation of a line in two dimensions is $ax+by=c$; it is reasonable to expect that a line in three dimensions is given by $ax + past +cz = d$; reasonable, only wrong—information technology turns out that 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 manner, however: there are exactly ii directions perpendicular to a plane. Whatever vector with ane of these two directions is called normal to the airplane. So while in that location are many normal vectors to a given plane, they are all parallel or anti-parallel to each other.

Suppose ii points $\ds (v_1,v_2,v_3)$ and $\ds (w_1,w_2,w_3)$ are in a aeroplane; and so the vector $\ds \langle w_1-v_1,w_2-v_2,w_3-v_3\rangle$ is parallel to the plane; in particular, if this vector is placed with its tail at $\ds (v_1,v_2,v_3)$ then its head is at $\ds (w_1,w_2,w_3)$ and it lies in the plane. Equally a result, any vector perpendicular to the plane is perpendicular to $\ds \langle w_1-v_1,w_2-v_2,w_3-v_3\rangle$. In fact, it is easy to see that the plane consists of precisely those points $\ds (w_1,w_2,w_3)$ for which $\ds \langle w_1-v_1,w_2-v_2,w_3-v_3\rangle$ is perpendicular to a normal to the plane, as indicated in effigy 12.v.1. That is, suppose we know that $\langle a,b,c\rangle$ is normal to a plane containing the point $\ds (v_1,v_2,v_3)$. And so $(x,y,z)$ is in the aeroplane if and only if $\langle a,b,c\rangle$ is perpendicular to $\ds \langle ten-v_1,y-v_2,z-v_3\rangle$. In turn, we know that this is truthful precisely when $\ds \langle a,b,c\rangle\cdot\langle x-v_1,y-v_2,z-v_3\rangle=0$. Thus, $(x,y,z)$ is in the plane if and but if $$\eqalign{ \langle a,b,c\rangle\cdot\langle 10-v_1,y-v_2,z-v_3\rangle&=0\cr a(10-v_1)+b(y-v_2)+c(z-v_3)&=0\cr ax+by+cz-av_1-bv_2-cv_3&=0\cr ax+by+cz&=av_1+bv_2+cv_3.\cr }$$ Working backwards, note that if $(10,y,z)$ is a signal satisfying $ax+by+cz=d$ then $$\eqalign{ ax+by+cz&=d\cr ax+by+cz-d&=0\cr a(ten-d/a)+b(y-0)+c(z-0)&=0\cr \langle a,b,c\rangle\cdot\langle 10-d/a,y,z\rangle&=0.\cr }$$ Namely, $\langle a,b,c\rangle$ is perpendicular to the vector with tail at $(d/a,0,0)$ and head at $(x,y,z)$. This means that the points $(x,y,z)$ that satisfy the equation $ax+by+cz=d$ form a airplane perpendicular to $\langle a,b,c\rangle$. (This doesn't work if $a=0$, but in that case we tin use $b$ or $c$ in the office of $a$. That is, either $a(10-0)+b(y-d/b)+c(z-0)=0$ or $a(x-0)+b(y-0)+c(z-d/c)=0$.)

Figure 12.5.one. 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 $ax+past+cz=d$, and whatever surface of this form is a plane perpendicular to $\langle a,b,c\rangle$.

Example 12.5.ane Find an equation for the plane perpendicular to $\langle 1,2,three\rangle$ and containing the indicate $(5,0,7)$.

Using the derivation above, the plane is $1x+2y+3z=1\cdot5+2\cdot0+3\cdot7=26$. Alternately, nosotros know that the plane is $x+2y+3z=d$, and to find $d$ we may substitute the known point on the plane to get $5+2\cdot0+3\cdot7=d$, and so $d=26$. Nosotros could also write this simply as $(x-5)+2(y)+three(z-seven)=0$, which is for many purposes a fine representation; it can ever be multiplied out to give $x+2y+3z=26$. $\foursquare$

Example 12.5.two Find a vector normal to the plane $2x-3y+z=15$.

Ane example is $\langle ii, -3,ane\rangle$. Any vector parallel or anti-parallel to this works likewise, so for example $-2\langle 2, -3,1\rangle=\langle -4,6,-two\rangle$ is also normal to the plane. $\square$

We volition 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 upshot, it is always possible, and usually appropriate, to use the given information to find a normal to the aeroplane and a bespeak on the airplane, and then to discover the equation as above.

Example 12.5.3 The planes $x-z=1$ and $y+2z=3$ intersect in a line. Find a third airplane that contains this line and is perpendicular to the plane $ten+y-2z=i$.

Get-go, nosotros 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 airplane is to incorporate a certain line, $\langle a,b,c\rangle$ must be perpendicular to any vector parallel to this line. Since $\langle a,b,c\rangle$ must be perpendicular to ii vectors, we may find it by computing the cantankerous product of the two. So we need a vector parallel to the line of intersection of the given planes. For this, information technology suffices to know two points on the line. To notice two points on this line, we must detect two points that are simultaneously on the two planes, $ten-z=1$ and $y+2z=3$. Any point on both planes will satisfy $ten-z=1$ and $y+2z=3$. Information technology is easy to detect values for $x$ and $z$ satisfying the first, such as $ten=1, z=0$ and $x=2, z=1$. And then we can find corresponding values for $y$ using the 2d equation, namely $y=3$ and $y=1$, so $(ane,3,0)$ and $(2,1,i)$ are both on the line of intersection considering both are on both planes. At present $\langle 2-ane,1-three,1-0\rangle=\langle 1,-2,i\rangle$ is parallel to the line. Finally, we may cull $\langle a,b,c\rangle=\langle one,1,-2\rangle\times \langle 1,-ii,1\rangle=\langle -3,-3,-3\rangle$. While this vector volition do perfectly well, any vector parallel or anti-parallel to it volition piece of work too, so for instance nosotros might choose $\langle ane,one,1\rangle$ which is anti-parallel to information technology.

At present nosotros know that $\langle 1,1,1\rangle$ is normal to the desired aeroplane and $(2,ane,1)$ is a indicate on the plane. Therefore an equation of the plane is $10+y+z=four$. As a quick check, since $(i,3,0)$ is besides on the line, it should be on the plane; since $1+3+0=iv$, we see that this is indeed the example.

Notation that had nosotros used $\langle -iii,-3,-3\rangle$ equally 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 go the equivalent $x+y+z=4$. $\square$

And so we now understand equations of planes; allow us plough 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 manner.

Different a plane, a line in iii dimensions does have an obvious management, namely, the direction of any vector parallel to it. In fact a line can be defined and uniquely identified past providing one bespeak on the line and a vector parallel to the line (in i of two possible directions). That is, the line consists of exactly those points we tin can attain past starting at the point and going for some altitude in the management of the vector. Let'south encounter how we can translate this into more mathematical linguistic communication.

Suppose a line contains the point $\ds (v_1,v_2,v_3)$ and is parallel to the vector $\langle a,b,c\rangle$; we phone call $\langle a,b,c\rangle$ a direction vector for the line. If nosotros place the vector $\ds \langle v_1,v_2,v_3\rangle$ with its tail at the origin and its head at $\ds (v_1,v_2,v_3)$, and if we place the vector $\langle a,b,c\rangle$ with its tail at $\ds (v_1,v_2,v_3)$, so the head of $\langle a,b,c\rangle$ is at a betoken on the line. Nosotros can go to any point on the line by doing the same thing, except using $t\langle a,b,c\rangle$ in identify of $\langle a,b,c\rangle$, where $t$ is some real number. Because of the style 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 vector $\ds \langle v_1,v_2,v_3\rangle+t\langle a,b,c\rangle$, namely $\ds (v_1+ta,v_2+tb,v_3+tc)$; see figure 12.5.2.

Figure 12.5.2. Vector form of a line.

In other words, every bit $t$ runs through all possible real values, the vector $\ds \langle v_1,v_2,v_3\rangle+t\langle a,b,c\rangle$ points to every indicate on the line when its tail is placed at the origin. Some other common way to write this is as a ready of parametric equations: $$ x= v_1+ta\qquad y=v_2+tb \qquad z=v_3+tc.$$ It is occasionally useful to use this grade of a line even in ii dimensions; a vector form for a line in the $x$-$y$ plane is $\ds \langle v_1,v_2\rangle+t\langle a,b\rangle$, which is the same every bit $\ds \langle v_1,v_2,0\rangle+t\langle a,b,0\rangle$.

Instance 12.five.iv Notice a vector expression for the line through $(6,1,-3)$ and $(2,four,five)$. To become a vector parallel to the line we decrease $\langle 6,ane,-3\rangle-\langle2,4,5\rangle=\langle 4,-3,-8\rangle$. The line is then given by $\langle ii,four,five\rangle+t\langle 4,-3,-8\rangle$; in that location are of course many other possibilities, such as $\langle 6,1,-3\rangle+t\langle four,-3,-viii\rangle$. $\square$

Case 12.five.v Determine whether the lines $\langle one,1,i\rangle+t\langle 1,ii,-one\rangle$ and $\langle 3,2,1\rangle+t\langle -one,-five,iii\rangle$ are parallel, intersect, or neither.

In ii dimensions, two lines either intersect or are parallel; in 3 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 ii values $a$ and $b$ and so that $\langle 1,i,1\rangle+a\langle 1,2,-1\rangle= \langle 3,2,1\rangle+b\langle -1,-5,3\rangle$, that is, $$\eqalign{ 1+a&=3-b\cr 1+2a&=2-5b\cr i-a&=one+3b\cr }$$ This gives iii equations in ii unknowns, so in that location may or may non exist a solution in general. In this case, it is easy to discover that $a=3$ and $b=-1$ satisfies all three equations, and then the lines do intersect at the point $(4,7,-two)$. $\square$

Example 12.5.6 Find the distance from the point $(1,ii,3)$ to the aeroplane $2x-y+3z=five$. The altitude from a bespeak $P$ to a plane is the shortest altitude from $P$ to any point on the airplane; this is the distance measured from $P$ perpendicular to the plane; see figure 12.5.three. This distance is the absolute value of the scalar projection of $\ds \overrightarrow{\strut QP}$ onto a normal vector $\bf n$, where $Q$ is any point on the plane. It is easy to detect a point on the plane, say $(1,0,1)$. Thus the distance is $$ {\overrightarrow{\strut QP}\cdot {\bf n}\over|{\bf n}|}= {\langle 0,2,two\rangle\cdot\langle 2,-ane,3\rangle\over|\langle 2,-1,three\rangle|}= {iv\over\sqrt{14}}. $$ $\square$

Figure 12.5.3. Distance from a betoken to a plane.

Example 12.5.7 Find the distance from the betoken $(-1,2,1)$ to the line $\langle 1,1,ane\rangle + t\langle 2,3,-ane\rangle$. Again we want the distance measured perpendicular to the line, as indicated in figure 12.5.4. The desired distance is $$ |\overrightarrow{\strut QP}|\sin\theta= {|\overrightarrow{\strut QP}\times{\bf A}|\over|{\bf A}|}, $$ where $\bf A$ is any vector parallel to the line. From the equation of the line, nosotros can use $Q=(ane,one,1)$ and ${\bf A}=\langle ii,3,-1\rangle$, and then the distance is $$ {|\langle -2,1,0\rangle\times\langle2,iii,-one\rangle|\over\sqrt{fourteen}}= {|\langle-1,-2,-8\rangle|\over\sqrt{fourteen}}={\sqrt{69}\over\sqrt{14}}. $$ $\square$

Effigy 12.five.4. Altitude from a betoken to a line.

Exercises 12.five

Yous can use Sage to compute distances to lines and planes, since this just involves vector arithmetic that nosotros accept already seen. Of course, you can likewise use Sage to exercise some of the computations involved in finding equations of planes and lines.

Ex 12.five.1 Discover an equation of the plane containing $(6,ii,1)$ and perpendicular to $\langle i,one,1\rangle$. (answer)

Ex 12.5.2 Observe an equation of the airplane containing $(-1,two,-three)$ and perpendicular to $\langle 4,five,-1\rangle$. (answer)

Ex 12.v.3 Find an equation of the plane containing $(ane,two,-3)$, $(0,one,-two)$ and $(1,2,-2)$. (answer)

Ex 12.5.four Discover an equation of the airplane containing $(1,0,0)$, $(four,2,0)$ and $(3,2,1)$. (reply)

Ex 12.5.5 Discover an equation of the airplane containing $(1,0,0)$ and the line $\langle one,0,2\rangle + t\langle 3,2,1\rangle$. (answer)

Ex 12.5.6 Discover an equation of the plane containing the line of intersection of $x+y+z=1$ and $x-y+2z=two$, and perpendicular to the airplane $2x+3y-z=4$. (answer)

Ex 12.5.7 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)

Ex 12.5.8 Find an equation of the airplane containing the line of intersection of $x+3y-z=6$ and $2x+2y-3z=8$, and perpendicular to the plane $3x+y-z=eleven$. (reply)

Ex 12.5.9 Find an equation of the line through $(one,0,three)$ and $(1,two,4)$. (answer)

Ex 12.5.10 Detect an equation of the line through $(1,0,3)$ and perpendicular to the airplane $x+2y-z=1$. (answer)

Ex 12.5.eleven Detect an equation of the line through the origin and perpendicular to the plane $x+y-z=2$. (answer)

Ex 12.five.12 Discover $a$ and $c$ so that $(a,ane,c)$ is on the line through $(0,2,3)$ and $(2,7,5)$. (reply)

Ex 12.5.thirteen Explain how to find the solution in instance 12.5.five.

Ex 12.5.fourteen Decide whether the lines $\langle 1,3,-1\rangle+t\langle one,one,0\rangle$ and $\langle 0,0,0\rangle+t\langle i,4,five\rangle$ are parallel, intersect, or neither. (answer)

Ex 12.5.xv Determine whether the lines $\langle 1,0,2\rangle+t\langle -1,-1,2\rangle$ and $\langle 4,iv,ii\rangle+t\langle 2,2,-4\rangle$ are parallel, intersect, or neither. (answer)

Ex 12.5.16 Determine whether the lines $\langle 1,2,-ane\rangle+t\langle 1,two,3\rangle$ and $\langle 1,0,1\rangle+t\langle 2/3,2,4/iii\rangle$ are parallel, intersect, or neither. (reply)

Ex 12.5.17 Determine whether the lines $\langle one,i,2\rangle+t\langle ane,two,-iii\rangle$ and $\langle two,3,-1\rangle+t\langle two,4,-6\rangle$ are parallel, intersect, or neither. (answer)

Ex 12.5.eighteen Find a unit of measurement normal vector to each of the coordinate planes.

Ex 12.v.19 Show that $\langle 2,1,3 \rangle + t \langle 1,1,2 \rangle$ and $\langle iii, ii, five \rangle + south \langle 2, 2, iv \rangle$ are the same line.

Ex 12.v.xx Give a prose description for each of the post-obit processes:

a. Given ii distinct points, find the line that goes through them.

b. Given three points (not all on the same line), find the aeroplane that goes through them. Why do we need the caveat that not all points be on the same line?

c. Given a line and a point not on the line, notice the plane that contains them both.

d. Given a plane and a indicate not on the airplane, find the line that is perpendicular to the airplane through the given point.

Ex 12.5.21 Detect the distance from $(2,ii,2)$ to $x+y+z=-1$. (answer)

Ex 12.5.22 Find the altitude from $(2,-one,-1)$ to $2x-3y+z=2$. (answer)

Ex 12.five.23 Discover the distance from $(two,-1,one)$ to $\langle 2,2,0\rangle+t\langle 1,two,3\rangle$. (answer)

Ex 12.5.24 Observe the distance from $(1,0,one)$ to $\langle iii,two,1\rangle+t\langle 2,-1,-ii\rangle$. (respond)

Ex 12.5.25 Find the distance between the lines $\langle v,iii,one\rangle+t\langle two,4,3\rangle$ and $\langle half-dozen,one,0\rangle+t\langle three,five,7\rangle$. (answer)

Ex 12.5.26 Find the distance between the lines $\langle 2,ane,3\rangle+t\langle -ane,ii,-3\rangle$ and $\langle ane,-3,four\rangle+t\langle 4,-4,one\rangle$. (reply)

Ex 12.5.27 Find the distance betwixt the lines $\langle 1,ii,3\rangle+t\langle two,-1,three\rangle$ and $\langle 4,5,6\rangle+t\langle -4,ii,-6\rangle$. (answer)

Ex 12.v.28 Find the distance betwixt the lines $\langle three,ii,1\rangle+t\langle 1,iv,-1\rangle$ and $\langle iii,1,iii\rangle+t\langle 2,8,-two\rangle$. (respond)

Ex 12.5.29 Find the cosine of the angle betwixt the planes $10+y+z=2$ and $ten+2y+3z=8$. (answer)

Ex 12.5.30 Find the cosine of the bending between the planes $x-y+2z=2$ and $3x-2y+z=5$. (answer)

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Source: https://www.whitman.edu/mathematics/calculus_online/section12.05.html

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