MCQ
If $\bar{a}, \bar{b}, \bar{c}$ are non-coplanar vectors and $\lambda$ is a real number, then the vectors $\bar{a}+2 \bar{b}+3 \bar{c}, \lambda \bar{b}+4 \bar{c}$ and $(2 \lambda-1) \bar{c}$ are non-coplanar for
  • A
    no value of $\lambda$
  • B
    all except one value of $\lambda$
  • all except two values of $\lambda$
  • D
    all values of $\lambda$

Answer

Correct option: C.
all except two values of $\lambda$
(C) Let $\bar{\alpha}, \bar{\beta}$ and $\bar{\gamma}$ be the given vectors $\bar{\alpha}, \bar{\beta}$ and $\bar{\gamma}$ are coplanar
$\therefore\left|\begin{array}{lcc}1 & 2 & 3 \\ 0 & \lambda & 4 \\ 0 & 0 & (2 \lambda-1)\end{array}\right|=0$
$\Rightarrow \lambda(2 \lambda-1)=0 \Rightarrow \lambda=0, \frac{1}{2}$
Hence, $\bar{\alpha}, \bar{\beta}, \bar{\gamma}$ are non-coplanar for all values of $\lambda$ except 0 and $\frac{1}{2}$.

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