Calculus III
Practice Problems 2: Answers
1. Find a unit vector orthogonal to the vectors V 6I
the parallelopiped determined by these three vectors?
Answer. First, calculate
7J K W I 2J
3K. What is the volume of
21 2 I 1 18 J 12 7 K
25 98, so the answer is U
Now, the unit vector in the direction of V W has length 19 17 5
731I 654I 192K. Since U is a unit vector perpendicular to the plane of V and W, the volume is the area
of the base, V W 25 98.
2. Find a vector which makes an angle of 30 with the plane determined by the vectors V 6I 7J K W
I 2J 3K.
Answer. Since the vectors V and W are the same vectors as in the above problem, the unit normal to the
plane of the vectors V and W is the vector U found there; U 731I 654I 192K.The vector X we seek
makes an angle of 60 with U. There is such a vector in the plane determined by U and V (see the figure).
So, if M is the unit vector in the direction of V, we can take X cos 60 U cos 30 M. First we calculate
V
6I 7J K
M
V 36 49 1 647I 755J 108K V
W
det
I
6
1
J
7
2
K
1
3
2
Thus
X
cos 60 U
2
2
5 731I 654I 192K 866 647I 755J 108K
926I 327J 190K cos 30 M
Of course there is a cone (as shown in the figure) of vectors solving this problem; we chose the unit vector in
the plane of V and V W.
U
X
W
PSfrag replacements
V
3. Find a vector normal to the plane given by the equation
3x
2y z
14
Answer. 3I
2J K.
W PQ PR I K J K I J K 4. Find a vector normal to the plane through P : 0 0 0 Q : 1 0 1 R : 0 1 1
Answer. The vectors PQ I K PR J K lie on the plane, so the vector we seek is
5. Find the equation of the plane through the origin which is normal to the line given parametrically by
3I 2J K t I J 2K
Answer. The vector L I J 2K is parallel to the line, so can be taken as the normal to the desired
plane. Thus the equation of the plane is x y 2z 0.
X
V where
6. Find V1 V2
3
V1
Answer.
V1
I 2J K V V 2
V2
V 2I
1
det
3
V2
V3
2J 3K 1
det
2
1
V3
2
2
0
I
1
3
2
2K
12
U V W , then the three vectors are mutually orthogonal.
7. Show that if det U V W
Answer. Let β be the angle between U and the normal to the plane of V and W, and α the angle between V
and W. Then
det U V W
U V W cos β
U V W cos β sin α
If this is U V W , then cos β sin α
three vectors must be orthogonal.
1, so β
0 and α
π 2 or β
Answer. The cross product 2I J 3K ! I J K 2I J 2 " 1 3 is on the line, the symmetric equations are
x 2 y 1 z 3
1
2
1
π and α
π 2. In both cases the
8. Find the symmetric equations of the line through the point (2,-1,3) which is perpendicular to the vectors
V 2I J 3K and W I J K.
K has the direction of the line. Since
9. Find the parametric equations of the line through the point (2,-1,3) which is parallel to the two planes given
by the equations
3x z 4
x 2y 5z 1
Answer. The line is parallel to the two planes, so is perpendicular to their normals. The normals are N 1
3I 0J K N2 I 2J 5K, so N1 N2 is in the direction of the line. We find N 1 N2 2I 14J 6K
so the parametric equations of the line are:
x
2
2t y
1
14t
z
3
6t 10. Find the equation of the plane through the point (2,-1,3) which is parallel to the vectors I
3I 2J K.
2J 3K and
4I 8J 4K.
Answer. The normal to the plane is the cross product of the two given vectors. This is N
Taking X0 2I J 3K, the equation is X X0 N 0, which comes to 4x 8y 4z 12.
2 y 1 z 1 0
3
4
2
Answer. Represent the given point by the vector X 2I K. The point X
2I J K is on the given
plane, and the normal to the plane is N
1 3 I 1 4 J 1 2 K. The distance then is
X X N J # 1 3 I 1 4 J 1 2 K 384 d
N
651
11. Find the distance of the point 2 0 1 from the plane given by the equation
x
0
0
12. Let L1 L2 be two lines in three dimensions which do not intersect. There are points P1 P2 on L1 L2
respectively such that the line joining P1 and P2 intersects each line at a right angle. These are the points on
the two lines which are closest together. Find a formula for the distance between P1 and P2 , in terms of the
equations of the lines
L1 : X Q1 tN1
L2 : X Q2 sN2
Answer. Let Π be the plane containing the line L 1 and parallel to the line L 2 . Then the distance between
P1 and P2 is the same as the distance of any point on L 1 from the plane Π (see the figure below). Now, the
normal to Π is N1 N2 , and Q1 is a point on L1 . Thus, using the distance formula, where we take Q 2 as a
point on L2 we get
Q2 Q1 N1 N2
d
N1 N2
Π
PSfrag replacements
U
V
W
X
P2
V
d
d
L1
P1
L2
If the geometry isn’t clear, we can argue as follows: Since the line joining P 1 and P2 is perpendicular to
L1 and L2 , it has the direction N 1 N2 . Now, the distance from Q 2 to P1 is the projection of Q 2 P1 in this
direction, so is
d
Q P N N $ N N 2
1
1
1
2
2
But we don’t know P1 , so we can’t use this formula. However, since Q 1 lies on the line L1 , we must have
Q1 P1 tN1 . Then, we can replace P 1 by Q1 , as this computation shows:
Q 2
N N % Q P tN N N % Q P N N is perpendicular to N N .
Q1
since N1
1
2
2
1
1
1
2
1
2
2
1
1
2
N % Q P N N tN1 N1
2
2
1
1
2