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Math 2243 Practice Test #
Name:
Eagle ID (last four digits only):
Instructor: Yi Lin Fall 2009
Show your work to receive credit, and box your final answer in every computation.
- [14 pts]
a) Find an equation of the largest (open) ball with center (1, 2 , 3) which does
not intersect the coordinate planes.
Solution: The distance from (1, 2 , 3) to the xy-plane is 3, to the xz-plane is
2, and to the yz-plane is 1. As a result, any open ball centered at (1, 2 , 3)
can have radius at most 1 if it does not intersect any coordinate planes. For
instance, if the radius is greater than 1, the ball will intersect the yz-plane.
(x โ 1)
2
2
2 < 1.
b) Given vectors a = 2i + 3j + k and b = i + j + k, find |a|, and the unit vector
in the direction of 2a โ 3 b
Solution:
|a| =
22 + 3^2 + 1^2 =
2 a โ 3 b = 4i + 6j + 2k โ 3 i โ 3 j โ 3 k = i + 2j + k;
| 2 a โ 3 b| =
2
2
2
The unit vector in the direction of 2a โ 3 b =
(i + 2j + k)
6 i
6 j
6 k
- [8pts] For what value of c the angle between the vectors (1, 2 , 3) and (1, 1 , c) is
ฯ
Solution: Note that (1, 2 , 3) ยท (1, 1 , c) = 1 + 2 + 3c = 3 + 3c. Suppose that the angle
between the vectors (1, 2 , 3) and (1, 1 , c) is
ฯ
. Then we have
cos
ฯ
(1, 2 , 3) ยท (1, 1 , c)
|(1, 2 , 3)| ยท |(1, 1 , c)|
3 + 3c โ 14
2 + c
2
It follows that
(3 + 3c)
2
ยท 14 ยท (2 + c
2 ).
3 c
2 โ 36 c + 24 = 0.
c
2 โ 12 c + 8 = 0.
c =
- [8 pts] Find a ร b if a = 2i + j + k and b = i + 3k.
Solution:
a ร b =
i j k
2 1 1
i โ
k +
j
= โ 3 i + k + 5j
- [10 pts] Find the distance from the point (1, 1, 1) to the plane x + 2 y + 3 z = 4.
Solution: Let S = (1, 1 , 1) and P = (4, 0 , 0). Note that P is a point on the plane.
We have that
P S = (โ 3 , 1 , 1) and that n = (1, 2 , 3) is the normal vector of the
plane. So the distance from S to the plane is
P S ยท
n
|n|
- [10 pts] On the xy-plane, find the equation of the line which passes through point
(โ 2 , 1) and which is perpendicular to the vector 3i โ 4 j.
Solution: Any plane which is perpendicular to the vector 3i โ 4 j must be of the
form 3x โ 4 y = c, where c is a constant. Since the plane passes through point
(โ 2 , 1), we must have
c = 3 ยท (โ2) โ 4 ยท (1) = โ 10.
So the plane equation is
3 x โ 4 y = โ 10.
Solve x and y in terms of t. We have that x = โ1 and y = โt. So the parametric
equations for the line is ๏ฃฑ ๏ฃด ๏ฃฒ
x = โ 1
y = โt
z = t.
- [12 pts] Find the linear equation of the plane that passes through the point P (1, 2 , 3)
and contains the line (^) ๏ฃฑ
๏ฃด ๏ฃฒ
x = 2t
y = t
z = 2t.
Solution: Note that the origin O(0, 0 , 0) and point Q(2, 1 , 2) lie on the line and so
lie on the plane as well. Now that
OP = (1, 2 , 3) and
OQ = (2, 1 , 2), their cross
product
OP ร
OQ =
i j k
1 2 3
i โ
j +
k
= i + 4j โ 3 k
must be perpendicular to the plane. We write down the equation for the plane
x + 4y โ 3 z = 0.
- [8 pts] Find the angle between the planes x + y + z + 2 = 0 and x + 2y + z = 0.
Solution: The angle ฮธ between these two planes coincides with the angle between
their normal vectors. So we have
cos ฮธ =
1 + 2^2 + 1
- [10 pts] Find an equation of the plane that passes through the line of intersection
of the planes x + y + 1 = 0 and 2x + y + z + 1 = 0 and is perpendicular to the
plane x + y + z = 1.
Solution: Let n 1 be normal vector of the plane x + y = 1 and n 2 the normal vector
of the plane 2x + y + z + 1 = 0. Then the line of intersection of these two planes
must be parallel to the vector
n 1 ร n 2 =
i j k
i โ
j +
k
= i โ j โ k
Note that the normal vector of the plane x + y + z) is (1, 1 , 1). By assumption, the
cross product
(i โ j โ k) ร (i + j + k) =
i j k
1 โ 1 โ 1
i โ
j +
k
= โ 2 j + 2k
must be perpendicular to the plane that we are looking for. Now let y = โ1. From
the equations {
x + y + 1 = 0
2 x + y + z = 1
we get that x = 0 and z = 2. Thus (0, โ 1 , 2) is a point on the line of intersection
of the planes x + y + 1 = 0 and 2x + y + z + 1 = 0. By assumption, it must also
be a point on the plane. We write down the equation for the plane
0(x โ 0) โ 2(y โ (โ1)) + 2(z โ 2) = 0,
i.e.,
โy + z โ 6 = 0.
