Unknown

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.

3. Solve the system
   x+2y+3z = 0
   x-y+z = 0
   x2 + y2 + z2 = 152.


4. 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).