Calculus- Early Transcendentals, 2021a

434 Three Dimensions
w 1 c 1 + w 2 c 2 + w 3 c 3 = 0.
Multiply the first equation by w 3 and the second by v 3 and subtract to get
w 3 v 1 c 1 + w 3 v 2 c 2 + w 3 v 3 c 3 = 0
v 3 w 1 c 1 + v 3 w 2 c 2 + v 3 w 3 c 3 = 0
(v 1 w 3 − w 1 v 3 )c 1 +(v 2 w 3 − w 2 v 3 )c 2 = 0
Of course, this equation in two variables has many solutions; a particularly easy one to see is c 1 = v 2 w 3 −
w 2 v 3 , c 2 = w 1 v 3 − v 1 w 3 . Substituting back into either of the original equations and solving for c 3 gives
c 3 = v 1 w 2 − w 1 v 2 .
This particular answer to the problem turns out to have some nice properties, and it is dignified with a
name: the cross product:
v × w = 〈v 2 w 3 − w 2 v 3 ,w 1 v 3 − v 1 w 3 ,v 1 w 2 − w 1 v 2 〉.
While there is a nice pattern to this vector, it can be a bit difficult to memorize; here is a convenient
mnemonic. The determinant of a two by two matrix is
∣ a b
c d∣ = ad − cb.
This is extended to the determinant of a three by three matrix:
∣ x y z ∣∣∣∣∣ v 1 v 2 v 3 = x
∣ v ∣ 2 v 3 ∣∣∣ − y
∣
w
w 1 w 2 w 2 w 3
∣ v ∣ 1 v 3 ∣∣∣ + z
w 1 w 3
∣ v ∣
1 v 2 ∣∣∣
w 1 w 2
3
= x(v 2 w 3 − w 2 v 3 ) − y(v 1 w 3 − w 1 v 3 )+z(v 1 w 2 − w 1 v 2 )
= x(v 2 w 3 − w 2 v 3 )+y(w 1 v 3 − v 1 w 3 )+z(v 1 w 2 − w 1 v 2 ).
Each of the two by two matrices is formed by deleting the top row and one column of the three by three
matrix; the subtraction of the middle term must also be memorized. This is not the place to extol the uses of
the determinant; suffice it to say that determinants are extraordinarily useful and important. Here we want
to use it merely as a mnemonic device. You will have noticed that the three expressions in parentheses on
the last line are precisely the three coordinates of the cross product; replacing x, y, z by i, j, k gives us
∣ i j k ∣∣∣∣∣ v 1 v 2 v 3 =(v 2 w 3 − w 2 v 3 )i − (v 1 w 3 − w 1 v 3 )j +(v 1 w 2 − w 1 v 2 )k
∣w 1 w 2 w 3
=(v 2 w 3 − w 2 v 3 )i +(w 1 v 3 − v 1 w 3 )j +(v 1 w 2 − w 1 v 2 )k
= 〈v 2 w 3 − w 2 v 3 ,w 1 v 3 − v 1 w 3 ,v 1 w 2 − w 1 v 2 〉
= v × w.
Given v and w, there are typically two possible directions and an infinite number of magnitudes that
will give a vector perpendicular to both v and w. As we have picked a particular one, we should investigate
the magnitude and direction.