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Unformatted text preview: An Alternate Proof of the Vector Triple Product Formula
William C. Schulz, Northern Arizona University, Flagstaff, AZ Here is a proof of the vector triple product formula
which is based on the scalar triple product formula
and the observation that
n X (c X n) = |n|2c for cJ_n. (3) (Note that cJ_n implies |n X (c X n)| = |n|2|c| and, by the right-hand rule,
n x (c x n) points in the same direction as c.) The proof of (l) is trivial if b and c are linearly dependent. Assume they are not,
in which case b x c = n # O and the perpendicularity property of the cross product
implies that a X (b X c) is in the plane of b and c. Thus, a><(b><c)=>\b+uc (4) for some scalars A, ,u. To determine A and u, take the dot product of both sides with
(c X n)/[bcn] (where n = b X c) and note that [bcn] = (b X c) ~ 11 = n O n = |n|2. This
yields (an)_ b‘(c><n) c-(an)
ax(bxc)l [bcn] _>\ [bcn] +M [hen]
lnlz _>\ a-c=}\. Observe that, since c X b = — (b X c),
Now begin with (4), and proceed anew with b and c interchanged. This yields
a - b = — ll: (5) completing the proof of (1). Alternatively, one may obtain (5) by taking the dot
product of (4) with (b X n)/[bcn]. It is possible to make the entire proof “thumb free” by proving (3) as follows.
The perpendicularity property implies n><(c><n)=ozc+,8n. (6) Taking the dot product of (6) with 11 gives 0 = a0 + ,8n - n, yielding ,8 = 0. Taking
73 the dot product with c and using (2) yields
ac-c=c~n><(an)=(c><n)-(c><n)=|c><n|2= |c|2|n|2; so a = |n|2, proving (3). This shows that (3), and thus (1), would remain valid if the cross product were defined by a left-hand rule instead of a right-hand rule.
0 A Nonstandard Solution to a Standard Problem
Florence S. Gordon, New York Institute of Technology, Old Westbury, NY Sometimes, the usual solution to a standard problem becomes so routine that one
never thinks to look beyond it for an alternate solution. A case in point is the old
standby: Find the equation of the circle, given the coordinates P1(al,bl) and
P2(a2,b2) of the endpoints of a diameter. The standard approach leads us to locate the center, calculate the radius, and
then substitute into the formula for the equation of a circle to obtain a1 + a2 2 b1 + b2 2 (a1 — a2)2 +(b1— b2)2
(x — ) + y — = —— .
2 2 4
Expanding and simplifying this, we obtain xz—(a1 +£12)x+y2—(bl +b2)y+ala2+b1b2=0. (l) A simpler and more elegant approach can be based on geometric principles. If
Q(x, y) is any point on the circle different from P1 and P2, then the angle P1 QP2 is
a right angle. Thus, the slopes of P1 Q and P2 Q are negative reciprocals, and we
have y_b1 X_a2 x—a1 y—b2. This, upon cross-multiplying, immediately leads to the same solution as in (1)
above. This method also provides a different geometric insight to this standard
problem. This approach can be extended to handle the comparable problem in
three dimensions: find the equation of the sphere, given the endpoints of any diameter.
The standard approach is to parallel the initial argument mentioned above, using
analytic geometry. A much simpler approach is to utilize some elementary vector
analysis. If Pl(al,bl,c1) and P2(a2,b2,c2) are the two given points and Q(x, y, z) is
any other point on the surface of the sphere, then the vectors P1 Q and P2 Q must
be perpendicular, and so their dot product will be zero. This leads directly to the
solution x2 — (a1 + a2)x + ala2 +y2 — (bl + b2)y + blb2 + 22 — (c1 + c2)z + clc2 = 0. Editor’s Note: The same type of argument [see, for example, page 112 of C. H. Lehmann’s Analytic
Geometry, John Wiley & Sons, 1942] can be used to prove that any angle Pl QP2 inscribed in a circle is a
right angle. Pl(— r, 0) and P2(r, 0) determine the circle x2 + y2 = r2. For any point Q(x, y) on this circle,
Pl Q has slope m1=y/(x + r) and P2Q has slope m2 =y/(x — r). Hence, mlmz =y2/(x — r)2 = — l. O 74 ...
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- Summer '11