Question

The vectors $\overrightarrow{a}=x\hat{i}+(x+1)\hat{j}+(x+2)\hat{k},\overrightarrow{b}=(x+3)\hat{i}+(x+4)\hat{j}+(x+5)\hat{k}$ and $\overrightarrow{c}=(x+6)\hat{i}+(x+7)\hat{j}+(x+8)\hat{k}$ are coplanar for

• A. all values of x
• B. x < 0
• C. x > 0
• D. None of these

Answer 1

The correct option is A all values of x

$\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are coplanar, iff $\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]=0$
We have, $\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]=\left|\begin{array}{ccc}x& x+1& x+2\\ x+3& x+4& x+5\\ x+6& x+7& x+8\end{array}\right|$
$=\left|\begin{array}{ccc}x& x+1& x+2\\ 3& 3& 3\\ 6& 6& 6\end{array}\right|\left[\begin{array}{ccc}Applying{R}_2\to & R_2-& R_1,\\ R_3\to & R_3-& R_1\end{array}\right]$
= 0 for all x [$\because R_1$ and $R_2$ are proportional]