Question

If $a=i+j+k,b=4i+3j+4k$ and $c=i+\alpha j+\beta k$ are linearly dependent vectors and $\left|c\right|=\sqrt{3}$, 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

The correct option is D $\alpha =\pm 1,\beta =1$

Since $a,b,c$ are linearly dependent.

$\left[abc\right]=0$

$\Rightarrow \left|\begin{array}{ccc}1& 1& 1\\ 4& 3& 4\\ 1& \alpha & \beta \end{array}\right|=0$

$\Rightarrow \left|\begin{array}{ccc}1& 0& 0\\ 4& -1& 0\\ 1& \alpha -1& \beta -1\end{array}\right|=0$

$\Rightarrow -(\beta -1)=0$

$\Rightarrow \beta =1$

Also, $\left|c\right|=\sqrt{2}\Rightarrow 1+{\alpha }^2+{\beta }^2=3$

$\Rightarrow {\alpha }^2=1\Rightarrow \alpha =\pm 1$