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

Answer: (D)

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

$\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}$

$\overrightarrow{b}=\widehat{4i}+\widehat{3j}+\widehat{4k}$

$\overrightarrow{c}=\hat{i}+\alpha \hat{j}+\beta \hat{k}$

As they are linearly independent

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

$\Rightarrow 1(4-3)+\alpha (4-4)+\beta (3-4)=0$

$\Rightarrow 1-\beta =0$

$\Rightarrow \beta =1⟶\left(1\right)$

AS,

$\left|c\right|=\sqrt{3}$

$\sqrt{{\left(1\right)}^2+(\alpha {)}^2+{\left(\beta \right)}^2}=\sqrt{3}$

Squaring both side, and put $\beta =1$ from (1)

${\left(1\right)}^2+{\left(\alpha \right)}^2+{\left(1\right)}^2=3$

$\alpha =\pm 1⟶\left(2\right)$

$\therefore \alpha =\pm 1and\beta =1$