How to Draw a Line Going Through Two Planes
Lines and planes are perhaps the simplest of curves and surfaces in 3 dimensional space. They also will prove of import as we seek to understand more complicated curves and surfaces.
The equation of a line in 2 dimensions is $ax+past=c$; it is reasonable to look that a line in three dimensions is given by $ax + by +cz = d$; reasonable, only wrong—it turns out that this is the equation of a plane.
A plane does not have an obvious "management'' equally does a line. It is possible to associate a plane with a management in a very useful way, still: there are exactly two directions perpendicular to a plane. Whatever vector with one of these ii directions is called normal to the aeroplane. So while there are many normal vectors to a given aeroplane, they are all parallel or anti-parallel to each other.
Suppose two points $\ds (v_1,v_2,v_3)$ and $\ds (w_1,w_2,w_3)$ are in a plane; then 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. Every bit a event, 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, information technology 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 12.5.i. That is, suppose we know that $\langle a,b,c\rangle$ is normal to a plane containing the betoken $\ds (v_1,v_2,v_3)$. And then $(x,y,z)$ is in the aeroplane if and only if $\langle a,b,c\rangle$ is perpendicular to $\ds \langle x-v_1,y-v_2,z-v_3\rangle$. In turn, we know that this is true precisely when $\ds \langle a,b,c\rangle\cdot\langle ten-v_1,y-v_2,z-v_3\rangle=0$. Thus, $(ten,y,z)$ is in the airplane if and simply if $$\eqalign{ \langle a,b,c\rangle\cdot\langle x-v_1,y-v_2,z-v_3\rangle&=0\cr a(10-v_1)+b(y-v_2)+c(z-v_3)&=0\cr ax+by+cz-av_1-bv_2-cv_3&=0\cr ax+by+cz&=av_1+bv_2+cv_3.\cr }$$ Working backwards, note that if $(x,y,z)$ is a bespeak satisfying $ax+by+cz=d$ and then $$\eqalign{ ax+by+cz&=d\cr ax+by+cz-d&=0\cr a(x-d/a)+b(y-0)+c(z-0)&=0\cr \langle a,b,c\rangle\cdot\langle x-d/a,y,z\rangle&=0.\cr }$$ Namely, $\langle a,b,c\rangle$ is perpendicular to the vector with tail at $(d/a,0,0)$ and caput at $(x,y,z)$. This means that the points $(x,y,z)$ that satisfy the equation $ax+by+cz=d$ form a plane perpendicular to $\langle a,b,c\rangle$. (This doesn't work if $a=0$, but in that case we tin apply $b$ or $c$ in the office of $a$. That is, either $a(x-0)+b(y-d/b)+c(z-0)=0$ or $a(x-0)+b(y-0)+c(z-d/c)=0$.)
Effigy 12.5.1. 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 class is a plane perpendicular to $\langle a,b,c\rangle$.
Instance 12.5.i Find an equation for the plane perpendicular to $\langle 1,2,iii\rangle$ and containing the point $(5,0,7)$.
Using the derivation above, the plane is $1x+2y+3z=one\cdot5+2\cdot0+3\cdot7=26$. Alternately, nosotros know that the plane is $x+2y+3z=d$, and to find $d$ we may substitute the known bespeak on the aeroplane to become $v+2\cdot0+3\cdot7=d$, so $d=26$. We could as well write this simply as $(x-five)+2(y)+3(z-7)=0$, which is for many purposes a fine representation; information technology tin can always be multiplied out to give $x+2y+3z=26$. $\square$
Example 12.five.2 Find a vector normal to the plane $2x-3y+z=xv$.
One case is $\langle ii, -3,1\rangle$. Whatsoever vector parallel or anti-parallel to this works also, so for case $-2\langle 2, -3,1\rangle=\langle -iv,half dozen,-2\rangle$ is also normal to the plane. $\square$
We will oftentimes need to detect an equation for a aeroplane given certain information about the plane. While there may occasionally be slightly shorter ways to get to the desired result, information technology is ever possible, and usually advisable, to use the given information to find a normal to the plane and a point on the plane, and so to detect the equation as above.
Example 12.v.3 The planes $x-z=one$ and $y+2z=3$ intersect in a line. Find a tertiary plane that contains this line and is perpendicular to the airplane $x+y-2z=1$.
First, we notation that two planes are perpendicular if and just 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 is to contain 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 two vectors, we may detect it by computing the cross product of the ii. So we demand a vector parallel to the line of intersection of the given planes. For this, it suffices to know ii points on the line. To notice two points on this line, nosotros must find 2 points that are simultaneously on the two planes, $x-z=ane$ and $y+2z=3$. Any bespeak on both planes will satisfy $x-z=ane$ and $y+2z=3$. Information technology is easy to find values for $ten$ and $z$ satisfying the offset, such every bit $10=1, z=0$ and $x=2, z=one$. And so we can notice corresponding values for $y$ using the second equation, namely $y=3$ and $y=one$, so $(one,iii,0)$ and $(2,1,1)$ are both on the line of intersection considering both are on both planes. Now $\langle 2-i,ane-3,1-0\rangle=\langle one,-2,1\rangle$ is parallel to the line. Finally, we may choose $\langle a,b,c\rangle=\langle i,one,-2\rangle\times \langle 1,-2,1\rangle=\langle -3,-3,-3\rangle$. While this vector will do perfectly well, any vector parallel or anti-parallel to information technology will piece of work as well, so for example nosotros might cull $\langle 1,1,i\rangle$ which is anti-parallel to it.
Now we know that $\langle 1,i,ane\rangle$ is normal to the desired aeroplane and $(2,i,1)$ is a point on the plane. Therefore an equation of the plane is $x+y+z=4$. As a quick check, since $(i,three,0)$ is also on the line, it should exist on the plane; since $1+iii+0=4$, we see that this is indeed the case.
Notation that had we used $\langle -iii,-three,-3\rangle$ as the normal, nosotros would have discovered the equation $-3x-3y-3z=-12$, then we might well have noticed that we could separate both sides by $-iii$ to get the equivalent $x+y+z=four$. $\square$
So nosotros now understand equations of planes; let the states turn to lines. Unfortunately, it turns out to be quite inconvenient to stand for a typical line with a single equation; nosotros demand to approach lines in a dissimilar style.
Unlike a plane, a line in three dimensions does take an obvious direction, namely, the direction of any vector parallel to it. In fact a line can exist defined and uniquely identified by providing 1 point on the line and a vector parallel to the line (in 1 of two possible directions). That is, the line consists of exactly those points we can reach past starting at the point and going for some distance in the direction of the vector. Let's run across how nosotros can translate this into more mathematical language.
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 call $\langle a,b,c\rangle$ a management vector for the line. If we 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 $\ds (v_1,v_2,v_3)$, then the head of $\langle a,b,c\rangle$ is at a indicate on the line. We can get to any point on the line past doing the aforementioned thing, except using $t\langle a,b,c\rangle$ in identify of $\langle a,b,c\rangle$, where $t$ is some existent number. Because of the way vector improver works, the betoken at the head of the vector $t\langle a,b,c\rangle$ is the bespeak at the caput 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 effigy 12.v.2.
Figure 12.five.2. Vector form of a line.
In other words, every bit $t$ runs through all possible existent values, the vector $\ds \langle v_1,v_2,v_3\rangle+t\langle a,b,c\rangle$ points to every point on the line when its tail is placed at the origin. Another common way to write this is as a gear up of parametric equations: $$ x= v_1+ta\qquad y=v_2+tb \qquad z=v_3+tc.$$ Information technology is occasionally useful to utilise this form of a line even in two dimensions; a vector form for a line in the $ten$-$y$ aeroplane is $\ds \langle v_1,v_2\rangle+t\langle a,b\rangle$, which is the aforementioned as $\ds \langle v_1,v_2,0\rangle+t\langle a,b,0\rangle$.
Case 12.5.4 Find a vector expression for the line through $(6,1,-3)$ and $(2,4,5)$. To become a vector parallel to the line we decrease $\langle 6,1,-three\rangle-\langle2,iv,five\rangle=\langle four,-3,-8\rangle$. The line is then given by $\langle 2,4,5\rangle+t\langle 4,-three,-8\rangle$; at that place are of course many other possibilities, such as $\langle vi,i,-3\rangle+t\langle 4,-three,-8\rangle$. $\square$
Example 12.five.v Determine whether the lines $\langle 1,1,1\rangle+t\langle 1,ii,-1\rangle$ and $\langle iii,two,1\rangle+t\langle -1,-5,three\rangle$ are parallel, intersect, or neither.
In two dimensions, two lines either intersect or are parallel; in 3 dimensions, lines that do non intersect might not be parallel. In this case, since the management vectors for the lines are not parallel or anti-parallel nosotros know the lines are non parallel. If they intersect, in that location must be two values $a$ and $b$ and so that $\langle 1,1,1\rangle+a\langle 1,ii,-1\rangle= \langle 3,2,one\rangle+b\langle -1,-v,3\rangle$, that is, $$\eqalign{ 1+a&=iii-b\cr 1+2a&=2-5b\cr i-a&=1+3b\cr }$$ This gives 3 equations in ii unknowns, so there may or may not be a solution in general. In this case, it is like shooting fish in a barrel to discover that $a=iii$ and $b=-1$ satisfies all iii equations, so the lines do intersect at the point $(4,7,-2)$. $\foursquare$
Instance 12.5.6 Discover the distance from the point $(1,2,3)$ to the aeroplane $2x-y+3z=v$. The altitude from a point $P$ to a plane is the shortest distance from $P$ to any betoken on the plane; this is the distance measured from $P$ perpendicular to the plane; come across figure 12.v.3. This distance is the accented value of the scalar projection of $\ds \overrightarrow{\strut QP}$ onto a normal vector $\bf northward$, where $Q$ is any signal on the plane. It is like shooting fish in a barrel to observe a point on the plane, say $(i,0,ane)$. Thus the distance is $$ {\overrightarrow{\strut QP}\cdot {\bf north}\over|{\bf due north}|}= {\langle 0,2,2\rangle\cdot\langle two,-ane,3\rangle\over|\langle 2,-ane,3\rangle|}= {four\over\sqrt{14}}. $$ $\square$
Figure 12.5.three. Altitude from a point to a plane.
Case 12.5.vii Notice the distance from the point $(-ane,2,i)$ to the line $\langle ane,1,one\rangle + t\langle 2,3,-ane\rangle$. Once more we want the distance measured perpendicular to the line, as indicated in figure 12.five.4. The desired altitude is $$ |\overrightarrow{\strut QP}|\sin\theta= {|\overrightarrow{\strut QP}\times{\bf A}|\over|{\bf A}|}, $$ where $\bf A$ is whatever vector parallel to the line. From the equation of the line, we can utilise $Q=(i,one,1)$ and ${\bf A}=\langle 2,3,-1\rangle$, and so the altitude is $$ {|\langle -ii,ane,0\rangle\times\langle2,three,-1\rangle|\over\sqrt{xiv}}= {|\langle-1,-2,-viii\rangle|\over\sqrt{14}}={\sqrt{69}\over\sqrt{14}}. $$ $\square$
Figure 12.5.4. Distance from a point to a line.
Exercises 12.5
You tin can use Sage to compute distances to lines and planes, since this just involves vector arithmetic that nosotros have already seen. Of course, you can also apply Sage to do some of the computations involved in finding equations of planes and lines.
Ex 12.5.1 Find an equation of the airplane containing $(6,ii,one)$ and perpendicular to $\langle one,i,1\rangle$. (answer)
Ex 12.5.2 Detect an equation of the airplane containing $(-one,2,-three)$ and perpendicular to $\langle 4,five,-ane\rangle$. (answer)
Ex 12.5.3 Find an equation of the plane containing $(1,2,-3)$, $(0,1,-2)$ and $(1,ii,-ii)$. (reply)
Ex 12.5.4 Find an equation of the plane containing $(1,0,0)$, $(iv,2,0)$ and $(3,ii,1)$. (answer)
Ex 12.v.5 Observe an equation of the plane containing $(ane,0,0)$ and the line $\langle 1,0,ii\rangle + t\langle 3,two,1\rangle$. (reply)
Ex 12.five.half dozen Detect 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)
Ex 12.five.7 Notice an equation of the plane containing the line of intersection of $x+2y-z=3$ and $3x-y+4z=vii$, and perpendicular to the aeroplane $6x-y+3z=16$. (answer)
Ex 12.5.eight Observe an equation of the plane containing the line of intersection of $x+3y-z=6$ and $2x+2y-3z=eight$, and perpendicular to the plane $3x+y-z=11$. (reply)
Ex 12.5.9 Find an equation of the line through $(ane,0,3)$ and $(1,2,4)$. (answer)
Ex 12.5.10 Find an equation of the line through $(ane,0,three)$ and perpendicular to the plane $x+2y-z=1$. (answer)
Ex 12.v.11 Find an equation of the line through the origin and perpendicular to the airplane $ten+y-z=ii$. (respond)
Ex 12.5.12 Find $a$ and $c$ so that $(a,i,c)$ is on the line through $(0,2,three)$ and $(2,7,five)$. (answer)
Ex 12.5.xiii Explain how to discover the solution in example 12.5.5.
Ex 12.5.fourteen Make up one's mind whether the lines $\langle ane,3,-1\rangle+t\langle 1,1,0\rangle$ and $\langle 0,0,0\rangle+t\langle 1,iv,5\rangle$ are parallel, intersect, or neither. (answer)
Ex 12.5.15 Determine whether the lines $\langle 1,0,2\rangle+t\langle -1,-1,ii\rangle$ and $\langle iv,iv,2\rangle+t\langle two,2,-four\rangle$ are parallel, intersect, or neither. (answer)
Ex 12.5.16 Determine whether the lines $\langle 1,ii,-1\rangle+t\langle 1,two,3\rangle$ and $\langle one,0,i\rangle+t\langle two/3,ii,iv/3\rangle$ are parallel, intersect, or neither. (answer)
Ex 12.5.17 Determine whether the lines $\langle i,1,two\rangle+t\langle 1,2,-3\rangle$ and $\langle two,three,-1\rangle+t\langle 2,4,-half dozen\rangle$ are parallel, intersect, or neither. (answer)
Ex 12.five.xviii Discover a unit normal vector to each of the coordinate planes.
Ex 12.5.nineteen Testify that $\langle ii,1,iii \rangle + t \langle 1,ane,2 \rangle$ and $\langle 3, 2, 5 \rangle + s \langle two, ii, 4 \rangle$ are the same line.
Ex 12.v.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 (non all on the same line), find the aeroplane that goes through them. Why exercise we need the caveat that not all points be on the same line?
c. Given a line and a point not on the line, discover the plane that contains them both.
d. Given a airplane and a indicate not on the plane, observe the line that is perpendicular to the plane through the given point.
Ex 12.5.21 Detect the distance from $(two,2,2)$ to $x+y+z=-one$. (reply)
Ex 12.5.22 Find the altitude from $(2,-1,-1)$ to $2x-3y+z=2$. (respond)
Ex 12.five.23 Find the distance from $(2,-ane,1)$ to $\langle ii,ii,0\rangle+t\langle one,2,3\rangle$. (answer)
Ex 12.5.24 Find the distance from $(i,0,1)$ to $\langle three,2,i\rangle+t\langle 2,-one,-2\rangle$. (answer)
Ex 12.5.25 Discover the distance betwixt the lines $\langle 5,three,1\rangle+t\langle ii,4,3\rangle$ and $\langle 6,1,0\rangle+t\langle 3,v,7\rangle$. (answer)
Ex 12.five.26 Find the distance between the lines $\langle 2,i,3\rangle+t\langle -1,2,-three\rangle$ and $\langle 1,-3,iv\rangle+t\langle iv,-4,1\rangle$. (answer)
Ex 12.5.27 Discover the distance between the lines $\langle one,2,3\rangle+t\langle ii,-1,3\rangle$ and $\langle iv,5,6\rangle+t\langle -4,2,-half-dozen\rangle$. (reply)
Ex 12.five.28 Discover the distance between the lines $\langle 3,2,one\rangle+t\langle 1,four,-ane\rangle$ and $\langle three,1,3\rangle+t\langle 2,8,-2\rangle$. (reply)
Ex 12.5.29 Find the cosine of the angle betwixt the planes $x+y+z=ii$ and $x+2y+3z=8$. (answer)
Ex 12.5.30 Observe the cosine of the angle between the planes $x-y+2z=two$ and $3x-2y+z=5$. (answer)
Source: https://www.whitman.edu/mathematics/calculus_online/section12.05.html
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