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 coplanar and $$\left| {\overrightarrow c } \right| = \sqrt 3 ,$$ then :
A.
$$\alpha = \sqrt 2 ,\,\beta = 1$$
B.
$$\alpha = 1,\,\beta = \pm 1$$
C.
$$\alpha = \pm 1,\,\beta = 1$$
D.
$$\alpha = \pm 1,\,\beta = - 1$$
Answer :
$$\alpha = \pm 1,\,\beta = 1$$
Solution :
Since $$\overrightarrow a ,\,\overrightarrow b $$ and $$\overrightarrow c $$ are coplanar therefore
\[\begin{array}{l}
\left| \begin{array}{l}
\,1\,\,\,1\,\,\,1\\
4\,\,\,3\,\,\,4\\
1\,\,\,\alpha \,\,\beta
\end{array} \right| = 0\\
\Rightarrow \beta = 1\,;\,\left| {\overrightarrow c } \right| = \sqrt {1 + {\alpha ^2} + {\beta ^2}} = \sqrt 3 \\
\Rightarrow {\alpha ^2} + {\beta ^2} = 2\\
\Rightarrow {\alpha ^2} = 1\\
\therefore \,\alpha = \pm 1
\end{array}\]