Question

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

• A. $\alpha$ = 1 , $\beta$ = 1
• B. $\alpha$ = 2 , $\beta$ = 1
• C. $\alpha$ = 3 , $\beta$ = 1
• D. $\alpha$ = 4 , $\beta$ = 1

Answer 1

Answer: (A)

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

From given, we have,

Since vectors are dependent, we have,

$\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]=0$

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

$1\cdot det\left(\begin{array}{cc}3& 4\\ \alpha & \beta \end{array}\right)-1\cdot det\left(\begin{array}{cc}4& 4\\ 1& \beta \end{array}\right)+1\cdot det\left(\begin{array}{cc}4& 3\\ 1& \alpha \end{array}\right)=0$

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

$1-\beta =0$

$\therefore \beta =1$

Since, $|c|=\sqrt{3}$

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

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

${\alpha }^2=1$

$\alpha =1$

$(\alpha ,\beta )=(1,1)$