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 value of `lambda`
D. exactly one value of `lambda`

Answer

C

Explanation

Given that,
`[lambda(a+b)lambda^(2)blambdac]=["a b"+"c d"]`
`implies|{:(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),,a_(1),,a_(3)),(b_(1)+c_(1),,b_(2)+c_(2),,b_(3)+c_(3)),(b_(1),,b_(2),,b_(3)):}|`
`implieslambda^(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)):}|`
`implies" "lambda^(4)=-1`
`implies" "` So no real value of `lambda` exists.