Math 2210–001, Midterm 2, Study Guide
Summer 2003
(1) Find the equation of the plane that: (a) goes through (1, 2, 3); (b) and is parallel to the xz-plane.
Solution: A plane is desribed by the equation, Ax + By + Cz = D. Because j is perpendicular to the xz-plane, it is
also perpendicular to the plane that we seek. That is, the plane the we want has j for its normal vector; thus, it is
y = D. Since it goes through (1, 2, 3), plug y = 2 to get D = 2. That is the plane is y = 2. You may wish to find a
quicker solution to this problem.
(2) Find the equation of the plane that: (a) goes through (1, 2, 3); (b) and is parallel to the plane described by
2x − 2y − 5z = 0.
Solution: The perpendicular to the plane that we seek is, h2, −2, −5i. Thus, we want 2x − 2y − 5z = D to contain
(1, 2, 3). That is, 2(1) − 2(2) − 5(3) = D. So D = −17, and we get the plane, 2x − 2y − 5z = −17.
(3) Let x = h1, 2, 3i and y = h2, 3, 4i. Find two unit vectors that are perpendicular to both x and y.
Solution: The vector x × y is perpendicular
i
1
2
Note that |i × j × k| =
to both x and y but is not a unit vector. This vector is equal to
j k
2 3 = −i + 2j − k.
3 4
√
6. The two vectors that we seek are: ± √16 x × y.
(4) Consider the function f (x, y) = x3 −y 3 . If u = h1, 1i, then compute the directional derivative Du f at (x, y) = (−1, 9).
Solution: First and foremost, we need the gradient: ∇f (x, y) = 3x2 i − 3y 2 j. Therefore, Du f (x, y) = u · ∇f (x, y) =
3x2 − 3y 2 . Plug in x = −1 and y = 9 to get, Du f (−1, 9) = −240.
(5) Consider these two lines in 3-D:


 x = −2 + 2t
y = 1 + 4t

z =t


 x = 4 + 3t
y = 1 − 2t

 z = t.
Do they intersect? If no, then explain why not. If yes, then explain where they meet.
Solution: For them to meet, we must have −2 + 2t = 4 + 3t and 1 + 4t = 1 − 2t. The first identity implies that
t = −6 whereas the second implies that t = 0. Therefore, there is no t for which the two lines match; i.e., the lines
do not intersect.
(6) Consider the three-dimensional surface z = f (x, y) = y 2 + x2 .
(a) Plot this surface.
(b) Find the equation of the plane that is tangent to this surface at the point (x0 , y0 , z0 ) = ( √12 , √12 , 1).
Solution: The surface is a parabolic cone with round level-sets.
1
2
√
√
To find the tangent plane, first compute ∇f (x, y) = 2xi + 2yj. Thus, ∇f ( √12 , √12 ) = 2i + 2j. The tangent
plane, at ( √12 , √12 , 1) is described by all (x, y, z) such that z = f (p0 ) + ∇f ( √12 , √12 ) · (p − p0 ), where p = hx, yi and
p0 = h √12 , √12 i. That is,
√
√
1
1
+ 2 y− √
.
z = 1+ 2 x− √
2
2
Equivalently,
√
√
−z + 2x + 2y = 1.