Question

If $\mathbf { a } = \mathbf { i } + \mathbf { j } + \mathbf { k }$ $\mathbf { b } = 4 \mathbf { i } + 3 \mathbf { j } + 4 \mathbf { k }$ and $\mathbf { c } = \mathbf { i } + \alpha \mathbf { j } + \beta \mathbf { k }$ are linearly dependent vectors and , then

• A. $\alpha = 1 , \beta = - 1$
• B. $\alpha = 1 , \beta = \pm 1$
• C. $\alpha = - 1 , \beta = \pm 1$
• D. $\alpha = \pm 1 , \beta = 1$

Answer 1

Answer: (D)

$\alpha = \pm 1 , \beta = 1$

Answer 2

The given vectors are linearly dependent hence, there exist scalars $x , y , z$ not all zero, such that $x \mathbf { a } + y \mathbf { b } + z \mathbf { c } = \mathbf { 0 }$

i.e., $x ( \mathbf { i } + \mathbf { j } + \mathbf { k } ) + y ( 4 \mathbf { i } + 3 \mathbf { j } + 4 \mathbf { k } ) + z ( \mathbf { i } + \alpha \mathbf { j } + \beta \mathbf { k } ) = \mathbf { 0 }$,

i.e.,$( x + 4 y + z ) \mathbf { i } + ( x + 3 y + \alpha z ) \mathbf { j } + ( x + 4 y + \beta z ) \mathbf { k } = \mathbf { 0 }$

̃ $x + 4 y + z = 0$ , $x + 3 y + \alpha z = 0$, $x + 4 y + \beta z = 0$

For non-trivial solution, ̃ $\beta = 1$

$| c | ^ { 2 } = 3$ ̃ $1 + \alpha ^ { 2 } + \beta ^ { 2 } = 3$ ̃ $\alpha ^ { 2 } = 2 - \beta ^ { 2 } = 2 - 1 = 1$;

$\therefore$ $\alpha = \pm 1$

Trick : $| \mathbf { c } | = \sqrt { 1 + \alpha ^ { 2 } + \beta ^ { 2 } } = \sqrt { 3 }$ ̃ $\alpha ^ { 2 } + \beta ^ { 2 } = 2$

$\bullet \bullet$ $\mathbf { a } , \mathbf { b } , \mathbf { c }$ are linearly dependent, hence

̃ $\beta = 1$.

$\therefore$ $\alpha ^ { 2 } = 1$ ̃ $\alpha = \pm 1$ .