Question

If $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k},\overrightarrow{b}=4\hat{i}+3\hat{j}+4\hat{k}$ and $\overrightarrow{c}=\hat{i}+\alpha \hat{j}+\beta \hat{k}$ are linearly dependent vectors and $|\overrightarrow{c}|=\sqrt{3},$ 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, $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are linearly dependent vectors.
$\Rightarrow \left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]=0\Rightarrow \left|\begin{array}{ccc}1& 1& 1\\ 4& 3& 4\\ 1& \alpha & \beta \end{array}\right|=0$

Applying $C_2\to C_2-C_1,C_3\to C_3-C_1$,
$\left|\begin{array}{ccc}1& 0& 0\\ 4& -1& 0\\ 1& \alpha -1& \beta -1\end{array}\right|=0\Rightarrow -(\beta -1)=0\Rightarrow \beta =1$

Also,
$|\overrightarrow{c}|=\sqrt{3}\Rightarrow 1+{\alpha }^2+{\beta }^2=3[\because \overrightarrow{c}=\hat{i}+\alpha \hat{j}+\beta \hat{k}]\Rightarrow 1+{\alpha }^2+1=3\Rightarrow {\alpha }^2=1\therefore \alpha =\pm 1$