PROBLEMS, SECTION 3
5. LINES AND PLANES
A great deal of analytic geometry can be simplified by the use of vector notation.
Such things as equations of lines and planes, and distances between points or be-tween lines and planes often occur in physics and it is very useful to be able to find them quickly. We shall talk about three-dimensional space most of the time although the ideas apply also to two dimensions. In analytic geometry a point is a set of three coordinates (x, y, z); we shall think of this point as the head of a vector r = ix + jy + kz with tail at the origin. Most of the time the vector will be in the background of our minds and we shall not draw it; we shall just plot the point (x, y, z) which is the head of the vector. In other words, the point (x, y, z) and the vectorr will be synonymous. We shall also use vectors joining two points.
In Figure 5.1 the vectorA from (1, 2, 3) to (x, y, z) is
A = r − C = (x, y, z) − (1, 2, 3) = (x − 1, y − 2, z − 3) or A = ix + jy + kz − (i + 2j + 3k) = i(x − 1) + j(y − 2) + k(z − 3).
Figure 5.1
Thus we have two ways of writing vector equations; we may choose the one we prefer.
Note the possible advantage of writing (1, 0,−2) for i−2k; since the zero is explicitly written, there is less chance of accidentally confusingi − 2k with i − 2j = (1, −2, 0).
On the other hand, 5j is simpler than (0, 5, 0).
In two dimensions, we write the equation of a straight line through (x0, y0) with slope m as
(5.1) y− y0
x− x0 = m.
Section 5 Lines and Planes 107
Figure 5.2
Suppose, instead of the slope, we are given a vector in the direction of the line, sayA = ia+jb (Figure 5.2). Then the line through (x0, y0) and in the directionA is determined and we should be able to write its equation. The directed line segment from (x0, y0) to any point (x, y) on the line is the vectorr − r0 with components x− x0 and y− y0:
(5.2) r − r0=i(x − x0) +j(y − y0).
This vector is parallel toA = ia + jb. Now if two vectors are parallel, their compo-nents are proportional. Thus we can write (for a, b= 0)
(5.3) x− x0
a = y− y0
b or y− y0
x− x0 = b a.
This is the equation of the given straight line. As a check we see that the slope of the line is m = b/a, so (5.3) is the same as (5.1).
Another way to write this equation is to say that if r − r0 andA are parallel vectors, one is some scalar multiple of the other, that is,
(5.4) r − r0=At, or r = r0+At,
where t is the scalar multiple. We can think of t as a parameter; the component form of (5.4) is a set of parametric equations of the line, namely
(5.5) x− x0= at,
y− y0= bt, or x = x0+ at, y = y0+ bt.
Eliminating t yields the equation of the line in (5.3).
In three dimensions, the same ideas can be used. We want the equations of a straight line through a given point (x0, y0, z0) and parallel to a given vector A = ai + bj + ck. If (x, y, z) is any point on the line, the vector joining (x0, y0, z0) and (x, y, z) is parallel to A. Then its components x − x0, y− y0, z− z0 are proportional to the components a, b, c ofA and we have
(5.6) x− x0
a =y− y0
b = z− z0
c
(symmetric equations of a straight line, a, b, c= 0).
If c, for instance, happens to be zero, we would have to write (5.6) in the form
(5.7) x− x0
a =y− y0
b , z = z0 (symmetric equations of a straight line when c = 0).
As in the two-dimensional case, equations (5.6) and (5.7) could be written
(5.8) r = r0+At, or
x = x0+ at, y = y0+ bt, z = z0+ ct,
(parametric equations of a straight line).
The parametric equations (5.8) have a partic-ularly useful interpretation when the parameter t means time. Consider a particle m (electron, billiard ball, or star) moving along the straight line L in Figure 5.3. Position yourself at the ori-gin and watch m move from P0 to P along L.
Your line of sight is the vectorr; it swings from r0 at t = 0 to r = r0+At at time t. Note that the velocity of m is dr/dt = A; A is a vector
along the line of motion. Figure 5.3
Going back to two dimensions, suppose we want the equation of a straight line L through the point (x0, y0) and perpendicular to a given vectorN = ai+bj. As above, the vector
r − r0= (x− x0)i + (y − y0)j
lies along the line. This time we want this vector perpendicular to N; recall that two vectors are perpendicular if their dot product is zero. Setting the dot product ofN and r − r0 equal to zero gives
(5.9) a(x− x0) + b(y− y0) = 0 or y− y0 x− x0 =−a
b.
This is the desired equation of the straight line L perpendicular to N. As a check, note from Figure 5.4 that the slope of the line L is
tan θ =− cot φ = −a/b.
Figure 5.4
Figure 5.5
In three dimensions, we use this method to write the equation of a plane. If (x0, y0, z0) is a given point in the plane and (x, y, z) is any other point in the plane,
Section 5 Lines and Planes 109
the vector (Figure 5.5)
r − r0= (x− x0)i + (y − y0)j + (z − z0)k
is in the plane. IfN = ai + bj + ck is normal (perpendicular) to the plane, then N andr − r0are perpendicular, so the equation of the plane isN · (r − r0) = 0, or
(5.10) a(x− x0) + b(y− y0) + c(z− z0) = 0,
or ax + by + cz = d, (equation of a plane) where d = ax0+ by0+ cz0.
If we are given equations like the ones above, we can read backwards to find A or N. Thus we can say that the equations (5.6), (5.7), and (5.8) are the equations of a straight line which is parallel to the vectorA = ai+bj+ck, and either equation in (5.10) is the equation of a plane perpendicular to the vectorN = ai + bj + ck.
Example 1. Find the equation of the plane through the three points A(−1, 1, 1), B(2, 3, 0), C(0, 1,−2).
A vector joining any pair of the given points lies in the plane. Two such vectors are−−→
AB = (2, 3, 0)− (−1, 1, 1) = (3, 2, −1) and−→
AC = (1, 0,−3). The cross product of these two vectors is perpendicular to the plane. This is
N = (−−→
AB)× (−→
AC) =
i j k
3 2 −1 1 0 −3
=−6i + 8j − 2k.
Now we write the equation of the plane with normal directionN through one of the given points, say B, using (5.10):
−6(x − 2) + 8(y − 3) − 2z = 0 or 3x − 4y + z + 6 = 0.
(Note that we could have dividedN by −2 to save arithmetic.)
Example 2. Find the equations of a line through (1, 0,−2) and perpendicular to the plane of Example 1.
The vector 3i − 4j + k is perpendicular to the plane of Example 1 and so parallel to the desired line. Thus by (5.6) the symmetric equations of the line are
(x− 1)
3 = y
−4 =(z + 2) 1 .
By (5.8) the parametric equations of the line arer = i − 2k + (3i − 4j + k)t or, if you like,r = (1, 0, −2) + (3, −4, 1)t.
Vectors give us a very convenient way of finding distances between points and lines or planes. Suppose we want to find the (perpendicular) distance from a point P
Figure 5.6
to the plane (5.10). (See Figure 5.6.) We pick any point Q we like in the plane (just by looking at the equation of the plane and thinking of some simple numbers x, y, z that satisfy it). The distance P R is what we want. Since P R and RQ are perpendicular (because P R is perpendicular to the plane), we have from Figure 5.6
(5.11) P R = P Q cos θ.
From the equation of the plane, we can find a vectorN normal to the plane. If we divide N by its magnitude, we have a unit vector normal to the plane; we denote this unit vector byn. Then |−−→
P Q· n| = (P Q) cos θ, which is what we need in (5.11) to find P R. (We have put in absolute value signs because−−→
P Q·n might be negative, whereas (P Q) cos θ, with θ acute as in Figure 5.6, is positive.)
Example 3. Find the distance from the point P (1,−2, 3) to the plane 3x− 2y + z + 1 = 0.
One point in the plane is (1, 2, 0); call this point Q. Then the vector from P to Q is
−−→P Q = (1, 2, 0)− (1, −2, 3) = (0, 4, −3) = 4j − 3k.
From the equation of the plane we get the normal vector N = 3i − 2j + k.
We getn by dividing N by |N| =√
14. Then we have
|P R| =−−→
P Q· n =(4j − 3k) · (3i − 2j + k)/√ 14
=(−8 − 3)/√
14 = 11/√ 14.
Figure 5.7 We can find the distance from a point P to a line in a similar way. In Figure 5.7 we want the perpendicular distance P R. We select any point on the line [that is, we pick any (x, y, z) satisfying the equations of the line]; call
this point Q. Then (see Figure 5.7) P R = P Q sin θ. Let A be a vector along the line and u a unit vector along the line (obtained by dividingA by its magnitude). Then
−−→
P Q× u = |P Q| sin θ, so we get
|P R| =−−→P Q× u .
Section 5 Lines and Planes 111
Example 4. Find the distance from P (1, 2,−1) to the line joining P1(0, 0, 0) and P2(−1, 0, 2).
Let A = −−−→P1P2 =−i + 2k; this is a vector along the line. Then a unit vector along the line is u = (1/√
5)(−i + 2k). Let us take Q to be P1(0, 0, 0). Then
−−→P Q =−i − 2j + k, so we get for the distance |P R|:
|P R| = 1
√5|(−i − 2j + k) × (−i + 2k)| = 1
√5| − 4i + j − 2k| = 21/5.
It is also straightforward to find the distance between two skew lines (and if you really want to appreciate vectors, just look up this calculation in an analytic geometry book that doesn’t use vectors!). Pick two points P and Q, one on each line (Figure 5.8). Then |−−→P Q· n|, where n is a unit vector perpendicular to both lines, is the distance we want. Now ifA and B are vectors along the two lines, then A × B is perpendicular to both, and n is just A × B divided by |A × B|.
Figure 5.8
Example 5. Find the distance between the linesr = i−2j+(i−k)t and r = 2j−k+(j−i)t.
If we write the first line asr = r0+At, then (the head of) r0is a simple choice for P , so we have
P = (1,−2, 0) and A = i − k.
Similarly, from the second line we find
Q = (0, 2,−1) and B = j − i.
ThenA × B = i + j + k and n = 1/√
3
(i + j + k). Also
−−→P Q = (0, 2,−1) − (1, −2, 0) = (−1, 4, −1) = −i + 4j − k.
Thus we get for the distance between the lines
−−→
P Q· n =(−i + 4j − k) · (i + j + k)/√
3 = |−1 + 4 − 1| /√
3 = 2/√ 3 .
Example 6. Find the direction of the line of intersection of the planes x− 2y + 3z = 4 and 2x + y − z = 5.
The desired line lies in both planes, and so is perpendicular to the two normal vectors to the planes, namely i − 2j + 3k and 2i + j − k. Then the direction of the line is that of the cross product of these normal vectors; this is−i + 7j + 5k.
Example 7. Find the cosine of the angle between the planes of Example 6.
The angle between the planes is the same as the angle between the normals to the planes. Thus our problem is to find the angle between the vectorsA = i−2j+3k andB = 2i + j − k. Since A · B = |A| |B| cos θ, we have −3 =√
14√
6 cos θ, and so cos θ =−
3/28. This gives the obtuse angle between the planes; the corresponding acute angle is π− θ, or arc cos
3/28.
PROBLEMS, SECTION 5
In Problems 1 to 5, all lines are in the (x, y) plane.
1. Write the equation of the straight line through (2, −3) with slope 3/4, in the para-metric formr = r0+At.