Question

If $\overline{a},\overline{b},\overline{c}$ are non coplanar vectors and $\lambda$ is a real number, then the vectors $\overline{a}+2\overline{b}+3\overline{c},\lambda \overline{b}+4\overline{c}$ and $(2\lambda -1)\overline{c}$ are non-coplanar for

• A. All values of $\lambda$
• B. No value of $\lambda$
• C. All except two values of $\lambda$
• D. All except three values of $\lambda$

Answer 1

The correct option is C All except two values of $\lambda$

For three vectors to be non-coplanar, their scalar triple product should not be equal to 0

Let $\overrightarrow{A}=\overrightarrow{a}+2\overrightarrow{b}+3\overrightarrow{c},\overrightarrow{B}=\lambda \overrightarrow{b}+4\overrightarrow{c},\overrightarrow{C}=(2\lambda -1)\overrightarrow{c}$

for these to be non-coplanar

$\Rightarrow \overrightarrow{A}.\overrightarrow{B}\times \overrightarrow{C}\ne 0$

$\Rightarrow (\overrightarrow{a}+2\overrightarrow{b}+3\overrightarrow{c}).\left(\right(\lambda \overrightarrow{b}+4\overrightarrow{c})\times (2\lambda -1\left)\overrightarrow{c}\right)\ne 0$

$\Rightarrow (\overrightarrow{a}+2\overrightarrow{b}+3\overrightarrow{c}).\left(\lambda \right(2\lambda -1)b\times c)\ne 0$ $(\because \overrightarrow{c}\times \overrightarrow{c}=0)$

$\Rightarrow \lambda (2\lambda -1)\overrightarrow{a}.\overrightarrow{b}\times \overrightarrow{c}\ne 0$

Now as $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are non-coplanar,

$\Rightarrow \overrightarrow{a}.\overrightarrow{b}\times \overrightarrow{c}\ne 0$

$\Rightarrow$The above equation will satisfy for all $\lambda$ except for $\lambda =0,\lambda =\frac{1}{2}$

Hence, answer is option $C$