MAT 211 Quiz 1. Spring 2015. Jennings
1. (10 points) Find equation(s) for the line that passes through the point (2, 4, 6) in the direction of the
vector < 3, 5, 7 >.
Solution: Anything equivalent to this
(x(t), y(t), z(t)) = (2, 4, 6) + t < 3, 5, 7 >= (2 + 3t, 4 + 5t, 6 + 7t)
will do.
2. (10 points) Find a vector whose length is 5 and points in the same direction as the vector < 2, 3, 6 >.
Solution: The length of the vector is
|< 2, 3, 6 >| =
√
4 + 9 + 36 = 7
so
1
< 2, 3, 6 >
7
is a unit vector. Multiply by 5 to obtain the anwer
5
< 2, 3, 6 >=< 10/7, 15/7, 30/7 >
7
3. (10 points) Find the angle between the vectors < 1, 3, 5 > and < 6, 4, 2 >.
Solution: Let θ be the angle. Using the dot product,
< 1, 3, 5 > · < 6, 4, 2 >= |< 1, 3, 5 >||< 6, 4, 2 >| cos(θ)
√ √
28 = 35 56 cos(θ)
28
√ √ = cos(θ)
35 56
28
arccos √ √
=θ
35 56
so θ ≈ 0.88608 radians.
4. (10 points) Find an equation for the plane that contains the points (3, 4, 5), (6, 4, 2) and (3, 0, 2).
Solution: The vector that points from (3, 4, 5) to (6, 4, 2) is
(6, 4, 2) − (3, 4, 5) =< 3, 0 − 3 >
The vector that points from (3, 4, 5) to (3, 0, 2) is
(3, 0, 2) − (3, 4, 5) =< 0, −4, −3 >
Their cross product is a normal vector


ı̂
̂
k̂
det 3 0 −3 = −12ı̂ + 9̂ − 12k̂
0 −4 −3
so
< −12, 9, −12 > · < x − 3, y − 4, z − 5 >= 0
is one equation for the plane. Multiply out to obtain another
−12(x − 3) + 9(y − 4) − 12(z − 5) = 0
If you don’t like that one play around with the algebra to generate many more equivalent equations.
5. Consider the surface x2 + y 2 − z 2 = 1.
(a) (5 points) Sketch the intersection of the surface with each of the horizontal planes
z = 0, z = ±1, z = ±2.
Solution: The intersection with the plane z = c is the set of points satisfying x2 + y 2 − c2 = 1.
Add c2 to both sides to obtain
x2 + y 2 = 1 + c2
√
a circle of radius c, centered at (0, 0). Here they are
-2
2
2
2
1
1
1
-1
1
2
-2
-1
1
2
-2
-1
1
-1
-1
-1
-2
-2
-2
z=0
z = ±1
x2 + y 2 = 0
x2 + y 2 = 2
In each case the axes point in the x and y directions.
2
z = ±2
x2 + y 2 = 5
(b) (5 points) Sketch the intersection of the surface with the x, z-plane and the intersection with the
y, z-plane.
Solution: Both intersections look like this, with the z axis pointing up.
2
1
-2
-1
1
-1
-2
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(c) (5 points) Sketch the surface.
Solution: It’s a “cooling tower” (hyperboloid of one sheet) symmetric around the z axis with
the z axis pointing upward.
6. A point moves through space so that its position at time t is
(x(t), y(t), z(t)) = (t2 , cos(t), e5t )
Find its
(a) (5 points) velocity at time t
Solution:
< 2t, − sin(t), 5e5t >
(b) (5 points) acceleration at time t
Solution:
< 2, − cos(t), 25e5t >
7. (10 points) Find the volume of a “box” whose adjacent edges are < 1, 1, 0 >, < 1, 0, 1 >, < 0, 1, 1 >.
Solution:
volume = ± < 1, 1, 0 > × < 1, 0, 1 > · < 0, 1, 1 >
= ± < 1, −1, −1 > · < 0, 1, 1 >
= ±(−2)
so the volume is 2 cubic units.
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