Question

If $a$ is a non-zero scalar, then the vectors $\overrightarrow{\alpha }=a\hat{i}+2a\hat{j}-3a\hat{k},\overrightarrow{p}=(2a+1)\hat{i}$ $+(2a+3)\hat{j}+(a+1)\hat{k}$ , $\overrightarrow{r}=(3a+5)\hat{i}+(a+5)\hat{j}+(a+2)\hat{k}$

• A. Coplanar if $a<0$
• B. Coplanar if $a>0$
• C. Always coplanar
• D. Never coplanar

Answer 1

The correct option is D Never coplanar

Let they be coplanar
$\left|\begin{array}{ccc}a& 2a& -3a\\ 2a+1& 2a+3& a+1\\ 3a+5& a+5& a+2\end{array}\right|=0$
$\left(\begin{array}{cc}{R}_3\to & R_3-R_1\\ R_2\to & R_2-R_1\end{array}\right)$
$a\left|\begin{array}{ccc}1& 2& -3\\ a+1& 3& 4a+1\\ 2a+4& 5-a& 4a+2\end{array}\right|=0$
$R_2-R_2-R_1$
$R_3=R_3-R_1$
$a\left|\begin{array}{ccc}1& 2& -3\\ a& 1& 4(a+1)\\ 2a+3& 3-a& 4a+5\end{array}\right|=0$
$C_2\to C_2-C_1$
$C_3\to C_3-C_1$
$a\left|\begin{array}{ccc}1& 1& -4\\ a& 1-a& 3a+4\\ 2a+3& -3a& 2a+2\end{array}\right|=0$
$a\left(2\right(1-a^2)-1\left(\right(a^2+a)2-(2a+3\left)\right(3a+4\left)\right)-3a^2-(2a+3\left)\right(1-a\left)\right)=0$
$a\left(2\right(1-a^2)+9a^2+12a-1(2a^2+2a-6a^2-17a-12)-3a^2+2a^2+a-3)$
$a(11+10a^2+2a)=0$ but $10a^2+28a+11\ne 0$