lie in the plane. The cross product
n = b×c
= (−1, 1, 3) × (−2, −1, 2)
= (5, −4, 3)
Section 1.5 selected answers
Math 131 Multivariate Calculus
D Joyce, Spring 2014
is ⊥ to the plane, so, as in problem 1, a generic
point x in the plane satisfies the equation
Exercises 1, 3, 5, 13, 23, 25, 27, 31.
n · (x − a) = 0.
1. Calculate an equation for the plane containing
the point (3, −1, 2) and ⊥ to i − j + 2k.
Here’s one way to solve the exercise that uses the
property that a generic point x in a plane passing
through the point a and ⊥ to the normal vector n
satisfies the equation
In this exercise, that’s (5−4, 3)·(x−3, y+1, z−2) =
0, which you can simplify, if you like.
5. Give an equation for the plane that is parallel
to the plane 5x−4y+z = 1 and that passes through
the point (2, −1, −2).
The coefficients (5, −2, 1) in the equation of the
n · (x − a) = 0.
plane give a vector normal to the plane, so the equaIn this exercise, that equation becomes (1, −1, 2) · tion of the parallel plane has the same coefficients.
(x − 3, y + 1, z − 2) = 0, which you can simplify, if Thus, the parallel plane has an equation of the form
you like, to x − y + 2z = 8.
5x − 4y + z = D
3. Find an equation for the plane containing the
where D is some constant. To see which conthree points (3, −1, 2), (2, 0, 5), and (1, −2, 4).
stant D is, evaluate the equation at the given point
There are several ways you could do this. For (2, −1, −2) on the plane:
instance, you could determine what values A, B,
C, and D have to have so that the equation
5(2) − 4(−1) + (−2) = 12.
Ax + By + Cz = D
Thus, D is 12, and the equation is 5x − 4y + z = 12.
is satisfied for those three points. You’ll need to
solve a system of 3 linear equations in 4 unknowns.
That’s not hard. It just involves row reducing a
matrix.
Here’s another way that uses cross and dot products of vectors.
The point a = (3, −1, 2) lies in the plane, and
the displacement vectors
13. Give a set of parametric equations for the line
of intersection of the planes x + 2y − 3z = 5 and
5x + 5y − z = 1.
There are many ways to find an answer. Here’s
one. Solve the system of two given equations simultaneously.
x + 2y − 3z = 5
5x + 5y − z = 1
b = (2, 0, 5) − (3, −1, 2) = (−1, 1, 3)
For instance, you might reduce this system to
13
z = − 23
x +
5
5
y − 14
z = 24
5
5
and
c = (1, −2, 4) − (3, −1, 2) = (−2, −1, 2)
1
From that, you can read off the general solution
(x, y, z) = (− 13
t−
5
23 14
, t
5 5
+
length of d is not the distance between a and the
line because it’s not ⊥ to the line. We need to
subtract the projection of d on the line from d to
get a perpendicular vector. Compute the projection
c·d
5
c = − 34
c
projc d =
c·c
24
, t),
5
which is, in vector notation,
, 14
, 1)t + (− 23
, 24 , 0).
x = ( −13
5
5
5 5
23. Find a single equation of the form Az + By + Then the vector e = d − proj d = 39 c is the vector
c
34
Cz = D that describes the plane given parametri- from the point a drawn at right angles to the line
cally as
l, so its length is the required distance.
x = 3s − t + 2
y = 4s + t
z = s + 5t + 3
27. Determine the distance between the two lines
l1 (t) = t(8, −1, 0) + (−1, 3, 5)
(Hint: Begin by writing the parametric equations and
in vector form and then find a vector normal to the
l2 (t) = t(0, 3, 1) + (0, 3, 4).
plane.)
Let’s use a notation for the lines. Let l1 (t) be the
Those three equations become the single vector
line ta1 + b1 , that is, the line with base point b1 =
equation
(−1, 3, 5) in the direction a1 = (8, −1, 0). Likewise,
let l2 (t) be the line ta2 +b2 , the line with base point
x = (x, y, z) = (3, 4, 1)s + (−1, 1, 5)t + (2, 0, 3).
b2 = (0, 3, 4) in the direction a2 = (0, 3, 1). First
That means that the two vectors (3, 4, 1) and note that they’re not parallel since their directions
(−1, 1, 5) lie in the plane, and their cross product a1 and a2 aren’t parallel. So they’re skew lines.
Let c = b2 − b1 = (1, 0, −1) be the vector ben = (3, 4, 1) × (−1, 1, 5) = (19, −16, 7)
tween the base points. Let n = a1 × a2 so that
n is ⊥ to each line. Then the projection of c on
is ⊥ the plane. Thus, the equation of the plane is
n, namely, projn c is a vector between the two lines
of the form 19x − 16y + 7z = D for some constant
and perpendicular to each. Therefore its length is
D. Find D by evaluating the equation at the point
the distance between the planes.
(2, 0, 3) which is known to lie in the plane. You’ll
get D = 59.
31. Find the distance between the two planes
given by the equations x − 3y + 2z = 1 and
25. Find the distance between the point (2, −1)
x − 3y + 2z = 8.
and the line l: x = 3t + 7, y = 5t − 3.
Note that these are parallel since their normal
A couple of different methods are outlined in the
vectors are both n = (1, −3, 2). Choose a point on
text.
each plane, say b1 = (4, 1, 0) on the first, and b2 =
Here’s one way to find the distance between a
(11, 1, 0) on the second. Let c = b2 − b1 = (7, 0, 0)
point a = (2, −1) and a line
be the vector between these points. Then, as in
exercise 23, the projection of c on n, namely, projn c
x = b + ct = (7, −3) + (3, 5)t.
is a vector between the two lines and perpendicular
Let d = a − b = (−5, 2). It’s the vector from the to each, so its length is the distance between the
base point b on the line to the given point a. The planes.
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