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Section 6.5 Lines and Planes

Lines and planes are maybe the most basic of curves and also surfaces in three dimensional space. They also will prove essential as we seek to understand more complicated curves and surfaces.

You are watching: Find an equation for the plane that contains the line and is perpendicular to the plane

You may recall that the equation the a heat in two dimensions is (ax+by=c ext;) it is reasonable to suppose that a heat in 3 dimensions is provided by (ax + by +cz = d ext.) however it transforms out the this is the equation that a plane. Us will revolve our attention to a research of planes and return to think about lines later on in this section.

A plane does not have actually an apparent “direction” as does a line. It is possible to associate a airplane with a direction in a an extremely useful way, however: there are specifically two direction perpendicular come a plane. Any type of vector with among these 2 directions is referred to as normal to the plane. While there are numerous normal vectors come a given plane, they room 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; climate the vector (langle w_1-v_1,w_2-v_2,w_3-v_3 angle) is parallel come the plane. In particular, if this vector is put with the tail in ~ ((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 come the aircraft is perpendicular come (langle w_1-v_1,w_2-v_2,w_3-v_3 angle ext.) In fact, that is simple to view that the aircraft consists that precisely those points ((w_1,w_2,w_3)) because that which (langle w_1-v_1,w_2-v_2,w_3-v_3 angle) is perpendicular to a regular to the plane, as shown in Figure 6.11. Transforming this around, suppose we know that (langle a,b,c angle) is common to a airplane containing the allude ((v_1,v_2,v_3) ext.) climate ((x,y,z)) is in the airplane if and also only if (langle a,b,c angle) is perpendicular come (langle x-v_1,y-v_2,z-v_3 angle ext.) In turn, we understand that this is true precisely when (langle a,b,c anglecdotlangle x-v_1,y-v_2,z-v_3 angle=0 ext.) the is, ((x,y,z)) is in the plane if and only if


eginalign*langle a,b,c anglecdotlangle x-v_1,y-v_2,z-v_3 angleamp =0\a(x-v_1)+b(y-v_2)+c(z-v_3)amp =0\ax+by+cz-av_1-bv_2-cv_3amp =0\ax+by+czamp =av_1+bv_2+cv_3 ext.endalign*
eginalign*ax+by+czamp =d\ax+by+cz-damp =0\a(x-d/a)+b(y-0)+c(z-0)amp =0\langle a,b,c anglecdotlangle x-d/a,y,z angleamp =0 ext.endalign*

Namely, (langle a,b,c angle) is perpendicular come the vector through tail in ~ ((d/a,0,0)) and head at ((x,y,z) ext.) This means that the point out ((x,y,z)) that satisfy the equation (ax+by+cz=d) type a airplane perpendicular come (langle a,b,c angle ext.) (This doesn"t job-related if (a=0 ext,) yet in that case we have the right to use (b) or (c) in the role of (a ext.) the 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 ext.))


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Figure 6.11. A aircraft defined via vectors perpendicular to a normal.

Thus, offered a vector (langle a,b,c angle) we understand that every planes perpendicular come this vector have actually the type (ax+by+cz=d ext,) and any surface ar of this form is a airplane perpendicular come (langle a,b,c angle ext.)

Definition 6.10. Scalar Equation that a Plane.

Any airplane can be composed in the form


where (a, b, c, d) room constants and also not all (a,b, c) space zero.

This plane is perpendicular come the vector (langle a,b,c angle ext.)

Example 6.11. Perpendicular Plane.

Find one equation because that the aircraft perpendicular to (langle 1,2,3 angle) and containing the point ((5,0,7) ext.)


Using the formula above, the airplane is (1x+2y+3z=d ext.) To uncover (d) we might substitute the known allude on the plane to acquire (5+2cdot0+3cdot7=d ext,) for this reason (d=26 ext.)


We will commonly need to find an equation for a plane given certain information about the plane. If there may occasionally it is in slightly shorter ways to acquire to the wanted result, the is constantly possible, and usually advisable, to usage the offered information to find a typical to the airplane and a point on the plane, and also then to discover the equation as above.

Example 6.13. Plane Perpendicular.

The plane (x-z=1) and (y+2z=3) crossing in a line. Discover a 3rd plane that contains this line and is perpendicular come the aircraft (x+y-2z=1 ext.)


First, we keep in mind that two planes space perpendicular if and only if their common vectors space perpendicular. Thus, we seek a vector (langle a,b,c angle) that is perpendicular to (langle 1,1,-2 angle ext.) In addition, because the desired airplane is to contain a certain line, (langle a,b,c angle) should be perpendicular to any kind of vector parallel to this line. Because (langle a,b,c angle) need to be perpendicular to 2 vectors, we may uncover it by computer the overcome product the the two.

Therefore we require a vector parallel to the line of intersection of the offered planes. Because that this, that suffices to know two clues on the line. To discover two clues on this line, we must uncover two points the are all at once on the 2 planes, (x-z=1) and also (y+2z=3 ext.) Any suggest on both planes will fulfill (x-z=1) and also (y+2z=3 ext.) it is easy to uncover values because that (x) and also (z) solve the first, such together (x=1, z=0) and (x=2, z=1 ext.) climate we can find matching values because that (y) making use of the second equation, namely (y=3) and (y=1 ext,) for this reason ((1,3,0)) and also ((2,1,1)) room two such points. They space both ~ above the heat of intersection because they are had in both planes.

Now (langle 2-1,1-3,1-0 angle=langle 1,-2,1 angle) is parallel come the line. Finally, us may choose (langle a,b,c angle=langle 1,1,-2 angle imes langle 1,-2,1 angle=langle -3,-3,-3 angle ext.) while this vector will carry out perfectly well, any vector parallel or anti-parallel to it will work as well. For example we might select (langle 1,1,1 angle) i beg your pardon is anti-parallel come it, and also easier to occupational with.

Now we understand that (langle 1,1,1 angle) is typical to the desired plane and ((2,1,1)) is a suggest on the plane. This offers an equation that ((1)x + (1)y + (1)z = d ext.) Substituting the worth of the suggest into the equation gives (d = 4 ext,) and therefore an equation the the airplane is (x+y+z=4 ext.) together a quick check, since ((1,3,0)) is also on the line, it should be on the plane; due to the fact that (1+3+0=4 ext,) we see that this is undoubtedly the case.

Note that had actually we offered (langle -3,-3,-3 angle) as the normal, we would have uncovered the equation (-3x-3y-3z=-12 ext.) then we could well have noticed that we might divide both sides by (-3) to gain the tantamount (x+y+z=4 ext.)


We will now turn our fist to a research of lines. Unfortunately, it transforms out to be quite inconvenient to represent a common line through a single equation; we need to strategy lines in a different way.

Unlike a plane, a line in three dimensions does have actually an obvious direction, namely, the direction of any type of vector parallel come it. In reality a line have the right to be defined and also uniquely identified by giving one suggest on the line and a vector parallel to the line (in one of two possible directions). That is, the line is composed of precisely those points we deserve to reach by beginning at the allude and walking for part distance in the direction the the vector. Let"s see how we deserve to translate this into much more mathematical language.

Suppose a line includes the suggest ((v_1,v_2,v_3)) and also is parallel to the vector (langle a,b,c angle ext.) If we place the vector (langle v_1,v_2,v_3 angle) v its tail at the origin and its head in ~ ((v_1,v_2,v_3) ext,) and if we location the vector (langle a,b,c angle) with its tail in ~ ((v_1,v_2,v_3) ext,) climate the head of (langle a,b,c angle) is in ~ a suggest on the line. Us can obtain to any suggest on the heat by act the very same thing, other than using (tlangle a,b,c angle) in place of (langle a,b,c angle ext,) where (t) is some actual number. Because of the means vector addition works, the point at the head of the vector (tlangle a,b,c angle) is the suggest at the head of the vector (langle v_1,v_2,v_3 angle+tlangle a,b,c angle ext,) namely ((v_1+ta,v_2+tb,v_3+tc) ext;) check out Figure 6.12.


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Figure 6.12. Vector kind of a line.

In various other words, together (t) runs with all possible real values, the vector (langle v_1,v_2,v_3 angle+tlangle a,b,c angle) points come every suggest on the line once its tail is put at the origin. The is occasionally beneficial to use this type of a line also in 2 dimensions; a vector type for a heat in the (x)-(y)-plane is (langle v_1,v_2 angle+tlangle a,b angle ext,) which is the same as (langle v_1,v_2,0 angle+tlangle a,b,0 angle ext.)

Definition 6.14.

Vector Equation the a Line an equation because that a heat passing through suggest ((v_1, v_2, v_3)) and parallel to the vector(langle a, b, c angle) is


whereby the vector (langle a, b, c angle) is dubbed the direction vector for the line.

Another common means to write this is together a set of parametric equations:


where ((v_1, v_2, v_3)) is a allude on the line and also (langle a, b, c angle) is parallel come the line.

Example 6.16. Vector Expression.

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


To obtain a vector parallel come the line us subtract (langle 6,1,-3 angle-langle2,4,5 angle=langle 4,-3,-8 angle ext.) The heat is then offered by (langle 2,4,5 angle+tlangle 4,-3,-8 angle ext;) there room of course numerous other possibilities, such as (langle 6,1,-3 angle+tlangle 4,-3,-8 angle ext.)


Example 6.17. Intersecting Lines.

Determine whether the currently (langle 1,1,1 angle+tlangle 1,2,-1 angle) and also (langle 3,2,1 angle+tlangle -1,-5,3 angle) room parallel, intersect, or neither.

See more: Meriwether Lewis And Clark Quotes, Lewis And Clark Expedition Quotes


In two dimensions, 2 lines either intersect or space parallel; in three dimensions, lines that do not intersect can not be parallel. In this case, due to the fact that the direction vectors because that the lines space not parallel or anti-parallel we recognize the lines are not parallel. If lock intersect, there should be two values (a) and also (b) so that (langle 1,1,1 angle+alangle 1,2,-1 angle= langle 3,2,1 angle+blangle -1,-5,3 angle ext.) that is, the adhering to must have actually a solution:


This gives three equations in 2 unknowns, for this reason there might or may not it is in a solution in general. In this case, that is basic to uncover that (a=3) and (b=-1) satisfies all 3 equations. Substituting these values right into (langle 1,1,1 angle+alangle 1,2,-1 angle= langle 3,2,1 angle+blangle -1,-5,3 angle ext,) the allude of intersection is ((4,7,-2) ext.)


Example 6.18. Distance from a suggest to a Plane.

Find the distance from the point ((1,2,3)) come the aircraft (2x-y+3z=5 ext.)


The street from a allude (P) to a aircraft is the shortest distance from (P) come any suggest on the plane; this is the distance measured indigenous (P) perpendicular to the plane; check out Figure 6.13. This distance is the absolute worth of the scalar forecast of (overrightarrowstrut QP) top top a regular vector (vectn ext,) whereby (Q) is any suggest on the plane. The is easy to discover a suggest on the plane, to speak ((1,0,1) ext.) thus the street is


eginequation*= 4oversqrt14 ext.endequation*