Question

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

• A. $\alpha =\sqrt{2},\beta =1$
• B. $\alpha =1,\beta =\pm 1$
• C. $\alpha =\pm 1,\beta =1$
• D. $\alpha =\pm 1,\beta =-1$
• E. $\alpha =-1,\beta =\pm 1$

Answer 1

The correct option is C $\alpha =\pm 1,\beta =1$

Given that, $a=\hat{i}+\hat{j}+\hat{k}$,
$b=4\hat{i}+3\hat{j}+4\hat{k}$
and $c=\hat{i}+\alpha \hat{j}+\beta \hat{k}$
Since, $a,b$ and $c$ are coplanar.
Then, $\left|\begin{array}{ccc}1& 1& 1\\ 4& 3& 4\\ 1& \alpha & \beta \end{array}\right|=0$
$\Rightarrow 1(3\beta -4\alpha )-1(4\beta -4)+1(4\alpha -3)=0$
$\Rightarrow 3\beta -4\alpha -4\beta +4+4\alpha -3=0$
$\Rightarrow \beta =1$
Also, $\left|c\right|=\sqrt{3}$
$\Rightarrow \sqrt{1+{\alpha }^2+{\beta }^2}=\sqrt{3}$
$\Rightarrow {\alpha }^2=1\Rightarrow \alpha =\pm 1$