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= 7(2,2,0) − 6(1,3,0) = (14,14,0) − (6,18,0)
= (8,−4,0)
Note that v and w lie in the x y-plane, and that u × (v × w) also lies in that plane. Also,
u × (v × w) is perpendicular to both u and v × w = (0,0,4) (see Figure 1.4.8).
z
v × w
u
y
0
w
v
u × (v × w)
x
Figure 1.4.8
For vectors v = v i
j
k and w
i
j
k in component form, the cross product
1
+ v 2 + v 3
= w 1 + w 2 + w 3
is written as: v×w = ( v w
w )i
w
w )j
w
w )k. It is often easier to use the
2
3 − v 3
2
+( v 3 1− v 1 3 +( v 1 2− v 2 1
component form for the cross product, because it can be represented as a determinant. We
will not go too deeply into the theory of determinants7; we will just cover what is essential
for our purposes.
7See ANTON and RORRES for a fuller development.
1.4 Cross Product
27
A 2 × 2 matrix is an array of two rows and two columns of scalars, written as
a
b
a
b
or
c
d
c
d
where a, b, c, d are scalars. The determinant of such a matrix, written as
a
b
a
b
or
det
,
c
d
c
d
is the scalar defined by the following formula:
a
b = ad− bc
c
d
It may help to remember this formula as being the product of the scalars on the downward
diagonal minus the product of the scalars on the upward diagonal.
Example 1.14.
1
2 = (1)(4)−(2)(3) = 4−6 = −2
3
4
A 3 × 3 matrix is an array of three rows and three columns of scalars, written as
a
a
a
a
a
a
1
2
3
1
2
3
b
b
b
b
b
b
1
2
3
or
1
2
3 ,
c
c
c
c
c
c
1
2
3
1
2
3
and its determinant is given by the formula:
a
a
a
1
2
3
b
b
b
b
b
b
b
b
b
2
3
1
3
1
2
(1.14)
1
2
3
= a 1
− a
+ a
c
c
2
c
c
3
c
c
c
c
c
2
3
1
3
1
2
1
2
3
One way to remember the above formula is the following: multiply each scalar in the first
row by the determinant of the 2 ×2 matrix that remains after removing the row and column
that contain that scalar, then sum those products up, putting alternating plus and minus
signs in front of each (starting with a plus).
Example 1.15.
1
0
2
−1 3
4
3
4
−1
4
−1 3 = 1
− 0
+ 2
= 1(−2 − 0) − 0(8 − 3) + 2(0 + 1) = 0
0
2
1
2
1
0
1
0
2
28
CHAPTER 1. VECTORS IN EUCLIDEAN SPACE
We defined the determinant as a scalar, derived from algebraic operations on scalar en-
tries in a matrix. However, if we put three vectors in the first row of a 3 × 3 matrix, then
the definition still makes sense, since we would be performing scalar multiplication on those
three vectors (they would be multiplied by the 2×2 scalar determinants as before). This gives
us a determinant that is now a vector, and lets us write the cross product of v = v i
j
k
1
+ v 2 + v 3
and w = w i
j
k as a determinant:
1
+ w 2 + w 3
i
j
k
v
v
v
v
v
v
v × w = v
v
v
2
3
i
1
3
j
1
2
k
1
2
3
=
−
+
w
w
w
w
w
w
w
w
w
2
3
1
3
1
2
1
2
3
= ( v w
w )i
w
w )j
w
w )k
2
3 − v 3
2
+ ( v 3 1 − v 1 3 + ( v 1 2 − v 2 1
Example 1.16. Let v = 4i − j + 3k and w = i + 2k. Then
i
j
k
−1 3
4
3
4
−1
v × w = 4 −1 3 =
i −
j +
k = −2i − 5j + k
0
2
1
2
1
0
1
0
2
The scalar triple product can also be written as a determinant. In fact, by Example 1.12,
the following theorem provides an alternate definition of the determinant of a 3 × 3 matrix
as the volume of a parallelepiped whose adjacent sides are the rows of the matrix and form
a right-handed system (a left-handed system would give the negative volume).
Theorem 1.17. For any vectors u = ( u , u , u ), v
, v , v ), w
, w , w ) in
1
2
3
= ( v 1 2 3
= ( w 1
2
3
R3:
u
u
u
1
2
3
u · (v × w) = v
v
v
(1.15)
1
2
3
w
w
w
1
2
3
Example 1.17. Find the volume of the parallelepiped with adjacent sides u = (2,1,3), v =
(−1,3,2), w = (1,1,−2) (see Figure 1.4.9).
z
Solution: By Theorem 1.15, the volume vol( P) of the parallelepiped
P is the absolute value of the scalar triple product of the three
adjacent sides (in any order). By Theorem 1.17,
v
u
2
1
3
y
u · (v × w) = −1 3
2
0
1
1
−2
x
w
3
2
−1
2
−1 3
= 2
− 1
+ 3
1
−2
1
−2
1
1
Figure 1.4.9
P
= 2(−8) − 1(0) + 3(−4) = −28, so
vol( P) = |−28| = 28.
1.4 Cross Product
29
Interchanging the dot and cross products can be useful in proving vector identities:
u · w u · z
Example 1.18. Prove: (u × v) · (w × z) =
for all vectors u, v, w, z in R3.
v · w v · z
Solution: Let x = u × v. Then
(u × v) · (w × z) = x · (w × z)
= w · (z × x) (by formula (1.12))
= w · (z × (u × v))
= w · ((z · v)u − (z · u)v) (by Theorem 1.16)
= (z · v)(w · u) − (z · u)(w · v)
= (u · w)(v · z) − (u · z)(v · w) (by commutativity of the dot product).
u · w u · z
= v·w v·z
Exercises
A
For Exercises 1-6, calculate v × w.
1. v = (5,1,−2), w = (4,−4,3)
2. v = (7,2,−10), w = (2,6,4)
3. v = (2,1,4), w = (1,−2,0)
4. v = (1,3,2), w = (7,2,−10)
5. v = −i + 2j + k, w = −3i + 6j + 3k
6. v = i, w = 3i + 2j + 4k
For Exercises 7-8, calculate the area of the triangle △ PQR.
7. P = (5,1,−2), Q = (4,−4,3), R = (2,4,0)
8. P = (4,0,2), Q = (2,1,5), R = (−1,0,−1)
For Exercises 9-10, calculate the area of the parallelogram PQRS.
9. P = (2,1,3), Q = (1,4,5), R = (2,5,3), S = (3,2,1)
10. P = (−2,−2), Q = (1,4), R = (6,6), S = (3,0)
For Exercises 11-12, find the volume of the parallelepiped with adjacent sides u, v, w.
11. u = (1,1,3), v = (2,1,4), w = (5,1,−2)
12. u = (1,3,2), v = (7,2,−10), w = (1,0,1)
For Exercises 13-14, calculate u · (v × w) and u × (v × w).
13. u = (1,1,1), v = (3,0,2), w = (2,2,2)
14. u = (1,0,2), v = (−1,0,3), w = (2,0,−2)
15. Calculate (u × v) · (w × z) for u = (1,1,1), v = (3,0,2), w = (2,2,2), z = (2,1,4).
30
CHAPTER 1. VECTORS IN EUCLIDEAN SPACE
B
16. If v and w are unit vectors in R3, under what condition(s) would v × w also be a unit
vector in R3 ? Justify your answer.
17. Show that if v × w = 0 for all w in R3, then v = 0.
18. Prove Theorem 1.14(b).
19. Prove Theorem 1.14(c).
20. Prove Theorem 1.14(d).
21. Prove Theorem 1.14(e).
22. Prove Theorem 1.14(f).
23. Prove Theorem 1.16.
24. Prove Theorem 1.17. ( Hint: Expand both sides of the equation. )
25. Prove the following for all vectors v, w in R3:
(a)
v × w 2 + |v · w|2 = v 2 w 2
(b) If v · w = 0 and v × w = 0, then v = 0 or w = 0.
C
26. Prove that in Example 1.8 the formula for the area of the triangle △ PQR yields the
same value no matter which two adjacent sides are chosen. To do this, show that 1 u
2
×
(−w) = 1 v
2
× w , where u = PR, −w = PQ, and v = QR, w = QP as before. Similarly,
show that 1 (
v
2
−u) × (−v) = 12
× w , where −u = RP and −v = RQ.
27. Consider the vector equation a × x = b in R3, where a = 0. Show that:
(a) a · b = 0
b × a
(b) x =
+ ka is a solution to the equation, for any scalar k
a 2
28. Prove the Jacobi identity: u × (v × w) + v × (w × u) + w × (u × v) = 0
29. Show that u, v, w lie in the same plane in R3 if and only if u · (v × w) = 0.
30. For all vectors u, v, w, z in R3, show that
(u × v) × (w × z) = (z · (u × v))w − (w · (u × v))z
and that
(u × v) × (w × z) = (u · (w × z))v − (v · (w × z))u
Why do both equations make sense geometrically?
1.5 Lines and Planes
31
1.5 Lines and Planes
Now that we know how to perform some operations on vectors, we can start to deal with some
familiar geometric objects, like lines and planes, in the language of vectors. The reason
for doing this is simple: using vectors makes it easier to study objects in 3-dimensional
Euclidean space. We will first consider lines.
Line through a point, parallel to a vector
Let P = ( x , y , z ) be a point in
0
0
0
R3, let v = ( a, b, c) be a nonzero vector, and let L be the line
through P which is parallel to v (see Figure 1.5.1).
z
L
P( x
tv
0, y 0, z 0)
r + tv
r
t > 0
r + tv