Q. If $\overrightarrow{\mathbf{a}},\ \overrightarrow{\mathbf{b}},\ \overrightarrow{\mathbf{c}}$ are non-coplanar vectors and λ be a real number, then the vectors $\overrightarrow{\mathbf{a}} + 2\overrightarrow{\mathbf{b}} + 3\overrightarrow{\mathbf{c}},\ \lambda\ \overrightarrow{\mathbf{b}} + 4\ \overrightarrow{\mathbf{c}}$ and $(2\lambda - 1)\ \overrightarrow{\mathbf{c}}$ are non-coplanar for

A) All values of λ
B) All except one value of λ
C) All except two values of λ
D) No value of λ
Explanation

(c)
Since, three vectors $\left( \overrightarrow{\mathbf{a}} + 2\overrightarrow{\mathbf{b}} + 3\overrightarrow{\mathbf{c}} \right),\left( \lambda\ \overrightarrow{\mathbf{b}} + 4\overrightarrow{\mathbf{c}} \right)\ and\ (2\lambda - 1)\overrightarrow{\mathbf{c}}$ are non-coplanar
$\therefore\ \left| \begin{matrix}
1 & 2 & 3 \\
0 & \lambda & 4 \\
0 & 0 & 2\lambda - 1
\end{matrix} \right| \neq 0$
$\Longrightarrow (2\lambda - 1)(\lambda) \neq 0$
$\Longrightarrow \ \lambda \neq 0,\frac{1}{2}$
Hence, these three vectors are non-coplanar for all except two values of λ.

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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