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

If a, b, c are non-coplanar vectors and is a real number then [ (…

MCQ

If $\bar{a}, \bar{b}, \bar{c}$ are non-coplanar vectors and $\lambda$ is a real number then $\left[\lambda(\bar{a}+\bar{b}) \ \lambda^2 \bar{b} \ \lambda \bar{c}\right]=[\bar{a} \bar{b}+\bar{c} \bar{b}]$ for

A
exactly three values of $\lambda$
B
exactly two values of $\lambda$
C
exactly one value of $\lambda$
✓
no value of $\lambda$

✓

Answer

Correct option: D.

no value of $\lambda$

(D) $\left[\lambda(\overline{ a }+\overline{ b }) \lambda^2 \overline{b} \quad \lambda \overline{ c }\right]=\left[\begin{array}{lll}\overline{ a } & \overline{ b }+\overline{ c } & \overline{ b }\end{array}\right]$
$\Rightarrow \lambda^4[\overline{ a }+\overline{ b } \overline{ b } \overline{ c }]=[\overline{ a } \overline{ b }+\overline{ c } \overline{ b }]$
$\begin{array}{l}\Rightarrow \lambda^4\{[\overline{ abc }]+[\overline{ b } \overline{ bc }]\}=\{[\overline{ a } \overline{ bc }]+[\overline{ ac } \overline{ b }]\} \\ \Rightarrow \lambda^4[\overline{ a } \overline{ bc }]=-[\overline{ a } \overline{ b } \overline{ c }]\end{array}$
$\Rightarrow\left(\lambda^4+1\right)[\overline{ a } \overline{ bc }]=0$
But, $[\overline{ a } \overline{ b } \overline{ c }] \neq 0$.
$\therefore \quad \lambda^4+1=0$
This is not true for any real value of $\lambda$.