(a) Let p = a i + j + k , q = i + b j+ k and r = i + j + ck If these vectors are coplanar, then [p q r] = 0 |
a
1
1
1
b
1
1
1
c
|=0 using operation R2→R2+R1,R3→R3−R1 |
a
1
1
1−a
b−1
0
1−a
0
c−1
|=0 Expanding along R1, a(b−1)(c−1)−(1−a)(c−1)−(1−a)(b−1)=0 ⇒[(a−1)+1](b−1)(c−1)+(1−a)(1−c)+(1−a)(1−b)=0 ⇒[(a−1)(b−1)(c−1)+(b−1)(c−1)+(1−a)(1−c)+(1−a)(1−b)=0] ⇒−(1−a)(1−b)(1−c)+(−b+1)(−c+1)+(1−a)(1−c)+(1−a)(1−b)=0 On dividing both sides by (1 - a) (1 - b) (1 - c), we get −1+