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DEPARTMENT OF MATHEMATICS AND STATISTICS
UNIVERSITY OF MASSACHUSETTS
MATH 233
EXAM 1
Spring 2007
NAME:
Section Number: Instructor’s Name:
In problems that require reasoning or algebraic calculation, it is not sufficient just to write the answers. You must explain how you arrived at your answers, and show your algebraic calculations.
You can leave answers in terms of fractions
and square roots, but if approximate numerical answers are used,
they should be rounded off to 4 significant figures.
〈x, y, z〉, [x, y, z],
x y z
,
x y z
; are all permissible notations for
the vector xi + yj + zk.
Total
Perfect Paper −→ 100 Points.
There are seven pages, including this one, in this exam and five problems. Make sure you have them all before you begin!
FORMULAS
a · b = a 1 b 1 + a 2 b 2 + a 3 b 3 = |a||b| cos θ
|a| =
a · a =
√ a^21 + a^22 + a^23.
a × b =
∣∣ ∣∣ ∣∣ ∣
i j k a 1 a 2 a 3 b 1 b 2 b 3
∣∣ ∣∣ ∣∣ ∣
∣∣ ∣∣ ∣
a 2 a 3 b 2 b 3
∣∣ ∣∣ ∣ i^ −
∣∣ ∣∣ ∣
a 1 a 3 b 1 b 3
∣∣ ∣∣ ∣ j^ +
∣∣ ∣∣ ∣
a 1 a 2 b 1 b 2
∣∣ ∣∣ ∣ k
where
∣∣ ∣∣ ∣
a b c d
∣∣ ∣∣ ∣ =^ ad^ −^ bc.
|a × b| = |a||b| sin θ = area of parallelogram
with sides a, b.
The vector projection of b onto (in the
direction of)
a is proj (^) ab =
( a · b a · a
) a
The scalar projection (component) of b
onto
a is compab =
a · b |a|
The volume of the “box” determined by a, b and
c is |a · (b × c)|
Arc length of parametrized curve:
L =
∫ (^) b
a
|r′(t)| dt.
- a) (8 points) Find an equation of the plane which contains the points P (− 1 , 0 , 2), Q(1, − 2 , 1) and R(2, 0 , −1).
A normal vector to the plane can be found by taking the cross product of any two vectors that lie in the plane. Two vectors that lie in the plane are P Q〈 2 , − 2 , − 1 〉 and P R〈 3 , 0 , − 3 〉. So the normal vector is
n = 〈 2 , − 2 , − 1 〉 × 〈 3 , 0 , − 3 〉 =
∣∣ ∣∣ ∣∣ ∣
i j k 2 − 2 − 1 3 0 − 3
∣∣ ∣∣ ∣∣ ∣
∣∣ ∣∣ ∣
∣∣ ∣∣ ∣ i^ −
∣∣ ∣∣ ∣
∣∣ ∣∣ ∣ j^ +
∣∣ ∣∣ ∣
∣∣ ∣∣ ∣ k = 6 i − (−3)j + 6k = 〈 6 , 3 , 6 〉
A point on the plane is P (− 1 , 0 , 2). Therefore an equation of the plane is
6(x − (−1)) + 3(y − 0) + 6(z − 2) = 0 (1)
or, simplified,
2 x + y + 2z − 2 = 0 (2)
b) (6 points) Find the distance from the point (1, 0 , −1) to the plane 2x + y − 2 z = 1.
Using the distance formula for the distance between a point (x 0 , y 0 , z 0 ) and a plane ax + by + cz + d = 0,
D =
|ax 0 + by 0 + cz 0 + d| a^2 + b^2 + c^2
√ 22 + 1^2 + (−2)^2
c) (6 points) Find the point P in the plane 2x + y − 2 z = 1 which is closest to the point (1, 0 , −1). (Hint: You can use part b) of this problem to help find P or first find the equation of the line passing through P and the point (1, 0 , −1) and then solve for P .)
First, get the equation of the line that goes through the point (1, 0 , −1) that is normal to the plane:
x = 1 + 2t y = t z = − 1 − 2 t
Now, the point in the plane which is closest to (1, 0 , −1) is the intersection of this line and the plane. To find this intersection, substitute the line equations into the plane equation:
2(1 + 2t) + (t) − 2(− 1 − 2 t) = 1
Simplifying and solving for t,
9 t + 4 = 1 ⇒ t = −
Plugging this t-value into the line equation, we get the coordinates of the point of intersection:
x = 1 + 2(−
y = −
z = − 1 − 2(−
So, the point on the plane closest to (1, 0 , −1) is (^13 , −^13 , −^13 ).
- The velocity vector of a particle moving in space equals
v(t) = 2 ti − 2 tj + tk at any time t ≥ 0 (3)
. a)(6 points) At the time t = 4, this particle is at the point (0, 5 , 4). Find an equation of the tangent line to the curve at the time t = 4.
This line goes through the point (0, 5 , 4) and has parallel tangent vector v(4) = 〈 8 , − 8 , 4 〉, so has parametric equations
x = 8 t y = 5 − 8 t z = 4 + 4t
b) (6 points) Find the length of the arc traveled from time t = 2 to time t = 4.
Using the arclength formula, ∫ (^4)
2
|v(t)| dt =
∫ (^4)
2
√ (2t)^2 + (− 2 t)^2 + t^2 dt
∫ (^4)
2
9 t^2 dt
∫ (^4)
2
3 tdt
t^2
∣∣ ∣∣
4
2
c) (8 points) Find a vector function which represents the curve of intersection of the cylinder x^2 + y^2 = 1 and the plane 3x − 2 y + z = 2.
Since the first equation is the equation of a circular cylinder, set x = cos(t) and y = sin(t). Next, use the second equation to solve for z, so z = 2 − 3 x + 2y = 2 − 3 cos(t) + 2 sin(t). Therefore
r(t) = 〈cos(t), sin(t), 2 − 3 cos(t) + 2 sin(t)〉
- a) (10 points) Consider the points A(3, 1 , 0), B(1, 0 , 3) and C(0, 3 , 1). Find the area of the triangle ABC. (Hint: If you know how to find the area of a parallelogram spanned by 2 vectors, then you should be able to solve this problem.)
The area of the parallelogram is
|AB × AC| =
∣∣ ∣∣ ∣∣ ∣
i j k − 2 − 1 3 − 3 2 1
∣∣ ∣∣ ∣∣ ∣
=
∣∣ ∣∣ ∣
∣∣ ∣∣ ∣
∣∣ ∣∣ ∣ i^ −
∣∣ ∣∣ ∣
∣∣ ∣∣ ∣ j^ +
∣∣ ∣∣ ∣
∣∣ ∣∣ ∣ k
∣∣ ∣∣ ∣ = |〈− 7 , 7 , − 7 〉| =
So, the area of the triangle is
b) (10 points) Suppose a particle moving in space has velocity
v(t) = 〈cos t, sin 3t, e^2 t〉
and initial position r(0) = 〈 1 , 2 , 0 〉. Find the position vector function r(t).
Integrating the velocity vector, we get position:
r(t) = 〈sin t + c 1 , −
cos(3t) + c 2 ,
e^2 t^ + c 3 〉
Now, we insist that when t = 0 is plugged into the position equation, the position must be 〈 1 , 2 , 0 〉:
r(0) = 〈c 1 , −
Therefore c 1 = 1, c 2 = 73 and c 3 = −^12 , so the position vector is
r(t) = 〈sin t + 1, −
cos(3t) +
e^2 t^ −
