WRITTEN HOMEWORK #2 SOLUTIONS
(1) (Problem #62, Chapter 12.5) Find an equation for the plane consisting of all
points that are equidistant from the points (2, 5, 5) and (−6, 3, 1).
Solution. You can write down formulas for the distance of an arbitrary point
(x, y, z) from (2, 5, 5), (−6, 3, 1) and set them equal (just like a problem from
last week), but here I’ll show a slightly different solution.
Notice that the plane in question is orthogonal to the segment connecting
(2, 5, 5) and (−6, 3, 1). In other words, the plane we want has normal vector
n = h8, 2, 4i. Also, the midpoint of this segment is (−2, 4, 3), so the plane we
want has equation
8x + 2y + 4z = 8(−2) + 2(4) + 4(3) = 4.
You can divide this by 2 to get 4x + y + 2z = 2. (2) (Problem #75, Chapter 12.5) Show that the distance between the parallel
planes ax + by + cz + d1 = 0 and ax + by + cz + d2 = 0 is
|d1 − d2 |
D=√
.
a2 + b 2 + c 2
(Not all a, b, c are equal to 0.)
Solution. Select two arbitrary points P1 = (x1 , y1 , z1 ) and P2 = (x2 , y2 , z2 ),
on ax + by + cz + d1 = 0, ax + by + cz + d2 = 0, respectively. Then the vector
v = hx2 − x1 , y2 − y1 , z2 − z1 i is the vector starting at P1 and ending at P2 .
On the other hand, the projection of v onto the normal vector n = ha, b, ci is
the distance between the two planes. This length can be obtained by taking
the absolute value of the scalar projection of v onto n; this value is
v · n |a(x2 − x1 ) + b(y2 − y1 ) + c(z2 − z1 )|
√
.
|n| =
a2 + b 2 + c 2
Use the fact that axi + byi + czi = −di to see that this expression is equal to
| − d2 + d1 |
√
,
a2 + b 2 + c 2
as desired. (3) (Problem #28, Chapter 13.1) Show that the curve with parametric equations
x = sin t, y = cos t, z = sin2 t is the curve of intersection of the surfaces z = x2
and x2 + y 2 = 1. Use this fact (or not!) to help sketch the curve.
Solution. Let C be the curve of intersection between z = x2 and x2 + y 2 = 1.
On the one hand, the x, y coordinates of points on C satisfy x2 + y 2 = 1.
1
2
WRITTEN HOMEWORK #2 SOLUTIONS
Therefore, we can find t such that x = sin t, y = cos t. (This isn’t the usual
parameterization for the unit circle, but one can check that this really is a
genuine parameterization of the unit circle.) On the other hand, since points
on C also satisfy z = x2 , we must have z = x2 = sin2 t.
We’ve shown that every point on C can be written in the form (sin t, cos t, sin2 t).
We should also check that every point of the form (sin t, cos t, sin2 t) is on C.
This is clear, since sin2 t + cos2 t = 1, and sin2 t = (sin t)2 .
The graph below was drawn by computer, but you can get a good approximation to this by hand by noting that
(4) (Problem #48, Chapter 13.1) Two particles travel along the space curves
r1 (t) = ht, t2 , t3 i, r2 (t) = h1 + 2t, 1 + 6t, 1 + 14ti. Do the particles collide? Do
their paths intersect? If so (to either of the questions), at which points?
Solution. The particles collide if there exists some value of t where r1 (t) =
r2 (t). We claim that no such value of t exists. Indeed, setting the xcoordinates equal, we see that we would need t = 1 + 2t, or, solving for
t, that t = −1. On the other hand, when t = −1, the y-coordinates are 1, −5,
which are not equal. Therefore no value of t will make r1 (t) = r2 (t) true.
The particles cross paths if there exist two values t, s such that r1 (t) = r2 (s).
Setting x-coordinates equals yields t = 1 + 2s. Setting y-coordinates equal,
keeping in mind that t = 1 + 2s, yields
t2 = (1 + 2s)2 = 1 + 4s + 4s2 = 1 + 6s,
or, solving for s in the last equation, 4s2 − 2s = 0, or s = 0, 1/2. The
corresponding values of t are t = 1, 2, respectively. We now check whether
these (s, t) = (0, 1), (1/2, 2) pairs yield points with equal z-coordinates; since
13 = 1 + 14 · 0, and 23 = 1 + 14 · (1/2), we find that both these (s, t) pairs
yield points at which these two paths intersect; namely, (1, 1, 1) and (2, 4, 8).
(5) (Problem #34, Chapter 13.2) At what point do the curves r1 (t) = ht, 1 −
t, 3 + t2 i and r2 (s) = h3 − s, s − 2, s2 i intersect? Find the cosine of the angle
of their intersection. (That is, compute the cosine of the angle between the
tangent vectors to the two curves at that point.)
Solution.
(6) (Problem #54, Chapter 13.2) If a curve in R3 has the property that the
position vector r(t) is always perpendicular to the tangent vector r0 (t), show
that the curve lies on a sphere with center at the origin.