If `bar a,bar b,bar c` are non coplanar vectors and `lambda` is a real number then `[lambda(bar a+bar b)lambda^2 bar b lambda bar c]=[bar a bar b+bar c bar b]` for

Options

A. exactly two values of `lambda`
B. exactly three values of `lambda`
C. no volue of `lambda`
D. exactly one value of `lambda`

Answer

C

Explanation

Given , `[lambda(a+b) lambda^(2)b lambdac]= [a(b+c)b]`
`|{:(lambda(a_(1)+b_(1)),lambda(a_(2)+b_(2)),lambda(a_(3) + b_(3))),(lambda^(2)b_(1),lambda^(2)b_(2),lambda^(2)b_(3)),(lambdac_(1),lambdac_(2),lambdac_(3)):}|=|{:(a_(1),b_(1)+c_(1),b_(1)),(a_(2),b_(2)+c_(2) ,b_(2)),(a_(3),b_(3)+c_(3),b_(3)):}| = |{:(a_(1),a_(2),a_(3)),(b_(1)+c_(1),b_(2)+c_(2),b_(3)+c_(3)),(b_(1),b_(2),b_(3)):}|`
`rArr lambda^(4)|{:(a_(1),a_(2),a_(3)),(b_(1),b_(2),b_(3)),(c_(1),c_(2),c_(3)):}|= - |{:(a_(1),a_(2),a_(3)),(b_(1),b_(2),b_(3)),(c_(1),c_(2),c_(3)):}|`
`rArr lambda^(4) = - 1`
So, no value of `lambda` exists.