122 4. Equations
~22 + bzy + c2z = 0.
Suppose that neither equation is a constant multiple of the other.
(a) Manipulate the first equation multiplied by c2 and the second
multiplied by cr to obtain a single equation relating the variables
x and y alone.
(b) Show that, for any solution x, y, z,
x : y : z = (blc2 - b2cl) : (CIQ - alc2) : (alb2 - a2bl).
(c) Prove that the set of solutions of the system is given by
x = (blc2 - b2cl)k
Y = (ClU2 - a1c2p
z = (alb2 - a2bl)k
where k is any number.
For convenience we can use the display
X Y z
‘xclxalxbl
b, c2 a2 b,
as a mnemonic. In vectorial terms, the solution set of the two equa-
tions consists of the set of vectors (x, y,z) which are orthogonal to
the vectors (al, bl, cl) and (02, ba, ~2). Such vectors must be multiples
of the cross product of these two vectors.
- Solve the system
x+2y+3z = 0
x-y+z = 0
x2 + y2 + z2 = 152. - Consider the pair of equations, where up # 0:
at2 + bt + c = 0
pt2 + qt + T = 0.
(a) Show that, if the equations have a root u in common, then
u2 : u : 1 = (br - cq) : (cp - ar) : (aq - bp).