Question

If $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are non coplanar vectors and $\lambda$ is a real number, then $\left[\begin{array}{ccc}\lambda (\overrightarrow{a}+\overrightarrow{b})& {\lambda }^2\overrightarrow{b}& \lambda \overrightarrow{c}\end{array}\right]=\left[\begin{array}{ccc}\overrightarrow{a}& \overrightarrow{b}+\overrightarrow{c}& \overrightarrow{b}\end{array}\right],\lambda \in R$ for:

• A. exactly one value of $\lambda$
• B. no value of $\lambda$
• C. exactly three values of $\lambda$
• D. exactly two values of $\lambda$

Answer 1

The correct option is B no value of $\lambda$

Given:
$\left[\begin{array}{ccc}\lambda (\overrightarrow{a}+\overrightarrow{b})& {\lambda }^2\overrightarrow{b}& \lambda \overrightarrow{c}\end{array}\right]=\left[\begin{array}{ccc}\overrightarrow{a}& \overrightarrow{b}+\overrightarrow{c}& \overrightarrow{b}\end{array}\right]$
$\Rightarrow \left|\begin{array}{ccc}\lambda & \lambda & 0\\ 0& {\lambda }^2& 0\\ 0& 0& \lambda \end{array}\right|\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]=\left|\begin{array}{ccc}1& 0& 0\\ 0& 1& 1\\ 0& 1& 0\end{array}\right|\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]$
$\Rightarrow {\lambda }^4=-1$
Clearly, $\lambda$ has no real value.