Question

The value of $\lambda$ for which the four points $2 \mathbf { i } + 3 \mathbf { j } - \mathbf { k } , \mathbf { i } + 2 \mathbf { j } + 3 \mathbf { k } , 3 \mathbf { i } + 4 \mathbf { j } - 2 \mathbf { k } , \mathbf { i } - \lambda \mathbf { j } + 6 \mathbf { k }$ are coplanar

• A. 8
• B. 0
• C. – 2
• D. 6

Answer 1

Answer: (C)

– 2

Answer 2

The given four points are coplanar

$\therefore$ $x ( 2 \mathbf { i } + 3 \mathbf { j } - \mathbf { k } ) + y ( \mathbf { i } + 2 \mathbf { j } + 3 \mathbf { k } ) + z ( 3 \mathbf { i } + 4 \mathbf { j } - 2 \mathbf { k } ) + w ( \mathbf { i } - \lambda \mathbf { j } + 6 \mathbf { k } ) = \mathbf { 0 }$ and $x + y + z + w = 0$,

where $x , y , z$ , w are not all zero.

⇒$( 2 x + y + 3 z + w ) \mathbf { i } + ( 3 x + 2 y + 4 z - \lambda w ) \mathbf { j }$+ $( - x + 3 y - 2 z + 6 w ) \mathbf { k } = 0$ and $x + y + z + w = 0$

⇒ $2 x + y + 3 z + w = 0$ ,$3 x + 2 y + 4 z - \lambda w = 0$, $- x + 3 y - 2 z + 6 w = 0$ and $x + y + z + w = 0$

For non-trivial solution, $\left| \begin{array} { c c c c } 2 & 1 & 3 & 1 \\ 3 & 2 & 4 & - \lambda \\ - 1 & 3 & - 2 & 6 \\ 1 & 1 & 1 & 1 \end{array} \right| = 0$

⇒ $\left| \begin{array} { c c c c } 2 & 1 & 3 & 1 \\ 0 & 0 & 0 & - ( \lambda + 2 ) \\ - 1 & 3 & - 2 & 6 \\ 1 & 1 & 1 & 1 \end{array} \right| = 0$ , Operating $\left[ R _ { 2 } \rightarrow R _ { 2 } - R _ { 1 } - R _ { 4 } \right]$

⇒ $- ( \lambda + 2 )$ $\left| \begin{array} { c c c } 2 & 1 & 3 \\ - 1 & 3 & - 2 \\ 1 & 1 & 1 \end{array} \right| = 0$ ⇒ $\lambda = - 2$