Question

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

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

Answer 1

Answer: (C)

All except two values of $\lambda$

Answer 2

As $\mathbf { a } , \mathbf { b } , \mathbf { c }$ are non-coplanar vectors. $\therefore$

Now, $\mathbf { a } + 2 \mathbf { b } + 3 \mathbf { c } , \lambda \mathbf { b } + 4 \mathbf { c }$ and $( 2 \lambda - 1 ) \mathbf { c }$ will be non-coplanar$( \mathbf { a } + 2 \mathbf { b } + 3 \mathbf { c } ) \cdot \{ ( \lambda \mathbf { b } + 4 \mathbf { c } ) \times ( 2 \lambda - 1 ) \mathbf { c } \} \neq 0$i.e., i.e., $\lambda ( 2 \lambda - 1 ) [ \mathbf { a } \mathbf { b } \mathbf { c } ] \neq 0$

$\therefore$ $\lambda \neq 0 , \frac { 1 } { 2 }$

Thus, given vectors will be non-coplanar for all values of $\lambda$ except two values: $\lambda = 0$ and $\frac { 1 } { 2 }$