Example.1
If a, b
and g are such that a + b + g
= 0, then prove that
.
Solution
Operating C2
→ C2 –
cos g C1, C3
→ C3 –
cosb C1,
D = 
D = 
= 
Example.2
Solution
Since 
are coplanar there must exist k1,
k2 , k3 not all zero (say k1 > 0) such that 
Now by operating C1→ 
,
Example.3
Prove that 
= –64(a -b) (a -g)
(a -d) (b -g) (b -d)
(g- d).
Solution
Let (b + g -a - d)2 = A, (g + a -b -d)2 = B and (a +b
-g -d)2 = C
then A – B = (2g - 2d) (2b
-2a) = - 4(a - b) (g
- d)
and D = 
=
( A – B) ( B –C) ( C – A)
= - 64 ( a - b ) ( a
- g) ( a - d) ( b-
g) ( b - d) ( g
- d) .
Example.4
In D ABC, prove that 
.
Solution
D = 
=
=
= 
= 
= 0 [Q
C2 and C3 are identical when (a2 – b2)
and (a2– c2) are taken common]
Second Method
D
= 
= 
= 0
Example.5
If x, y, z are not all zero and if ax+by+cz= 0, bz+cy+az
= 0, cx+ay+bz =0 prove that x:y:z = 1:1:1 or 1: w :w2 or 1: w2:w where w is the complex cube roots of unity.
Solution
For non-trival
solution 
= 0
= ( a+ b+ c) (a2
+b2 +c2 – ab – bc – ca) =0 . . . . (1).
= ( a+ b+c ) ( a+ wb +w2c)( a+ w2b + wc)
= 0 . . . . (2).
Now, by using cramer
rule find x, y, z and using ( a+b+c) = 0 ,
( a+ wb +w2c)=0 and ( a+ w2b + wc) = 0, prove the desired result.
