Quiz 1 for MATH 105 SECTION 205
January 14, 2015
Given Name
Family Name
Student Number
1. Let u = h3, −4i and v = h1, 2i, then
(a) (1 point) Compute u + 2v.
(a)
h5, 0i
Answer.
u + 2v = h3, −4i + 2h1, 2i
= h3, −4i + h2, 4i
= h3 + 2, −4 + 4i
= h5, 0i
(b) (1 point) Find the angle between u and v?
(b)
Answer.
Recall the formula:
cos(θ) =
u·v
.
|u||v|
Notice that
u · v = 3 · 1 + (−4) · 2
= 3−8
= −5
p
|u| =
32 + (−4)2
√
9 + 16
=
√
=
25
= 5
p
12 + 2 2
|v| =
√
=
1+4
√
=
5.
So we get
√
−5
1
5
√ = −√ = −
cos(θ) =
.
5
5· 5
5
Therefor, we have
√ !
5
θ = cos−1 −
5
√ !
5
= π − cos−1
.
5
√ √ cos−1 − 55 or π − cos−1 55
2. Let P be the plane which passes through the point (3, 0, 1) and is parallel to the plane 3x − 2y − z = 0, then
(a) (1 point) Write down the equation of the plane P .
(a)
3x − 2y − z = 8
Answer.
Since the plane P is parallel to the plane 3x − 2y − z = 0, then h3, −2, −1i is a normal vector
to the plane P . Since the plane P passes through the point (3, 0, 1), then the equation of P is
3(x − 3) − 2(y − 0) − (z − 1) = 0.
That is,
3x − 2y − z = 8.
(b) (1 point) Find the value of a such that the point (a, 2, 0) is on the plane P .
(b)
Answer.
4
By the result of part a, since (a, 2, 0) is on the plane P , then
3a − 2 · 2 − 0 = 8.
Then 3a = 12, that is,
a = 4.
(c) (1 point) Find the value of b such that the plane P is orthogonal to the plane 2x − by = 1.
(c)
−3
Answer.
By the result of part a, we know that h3, −2, −1i is a normal vector to the plane P . For the
plane 2x − by = 1, then h2, −b, 0i is a normal vector to the plane 2x − by = 1. By the assumption, we
have
h3, −2, −1i · h2, −b, 0i = 0.
That is,
3 · 2 − 2 · (−b) − 1 · 0 = 0.
That is, 6 + 2b = 0. So
b = −3.
3. Let f (x, y) = ex and g(x, y) = y sin(xy), then
(a) (1 point) Compute fx (x, y).
Answer.
(a)
ex
(b)
0
It’s easy to see that
fx (x, y) = ex .
(b) (1 point) Compute fy (x, y).
Answer.
For fy (x, y), we should think x is a constant, then ex is a constant. So
fy (x, y) = 0.
(c) (1 point) Compute gy (x, y).
(c)
Answer.
sin(xy) + xy cos(xy)
By the product rule and the chain rule, we have
gy (x, y) = sin(xy) + y cos(xy) · x
= sin(xy) + xy cos(xy)
4. (2 points) Sketch two level curves of the surface z =
Answer.
get
x2
y2
+ .
9
16
For the level curves, we just need to set z = z0 for some z0 . For example, we set z = z0 = 1, we
x2
y2
+
= 1,
9
16
which is an ellipse.
If we set z = z0 = 2, we get
y2
x2
+
= 2,
9
16
that is,
x2
y2
+
= 1,
18 32
which is also an ellipse.
Remark: You must draw these two level curves in the same xy-plane, do not separately.
Your Score:
/10