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 $\left|\overrightarrow{c}\right|=\sqrt{3}$ 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 the three given vectors are linearly dependent, any one can be written in terms of the other two.
So, $\overrightarrow{c}=k\overrightarrow{a}+l\overrightarrow{b}$
Comparing the coefficients of $\hat{i},\hat{j},\hat{k}$ we get
$k+4l=1,k+3l=\alpha ,k+4l=\beta$
$\Rightarrow \beta =1$ and $\because |\overrightarrow{c}|=\sqrt{3},1+{\alpha }^2+{\beta }^2=3$
$\Rightarrow \alpha =\pm 1$