Question

If $\overline{a}=\hat{i}+\hat{j}+\hat{k},\overline{b}=4\hat{i}+3\hat{j}+4\hat{k}$ and $\overline{c}=\hat{i}+\alpha \hat{j}+\beta \hat{k}$ are linearly dependent vectors & $\left|\overline{c}\right|=\sqrt{3}$,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 $\alpha =\pm 1,\beta =1$

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

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

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

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

if $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are lineraly dependent then $\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]=0$

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

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

$3\beta -4\alpha -4\beta +4+4\alpha -3=0$

$-3\beta =-3$

$\beta =1$

put $\beta in\overrightarrow{c}$

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

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

$\sqrt{1^2+{\alpha }^2+1^2}=\sqrt{3}$

$1+{\alpha }^2+1=3$

${\alpha }^2=3-2$

${\alpha }^2=1$

$\alpha =\sqrt{1}$

$\alpha =\pm 1$

and

$\beta =1$