Question

Let ${\overrightarrow{p}}_1=6x\hat{i}+2m\hat{j}-\hat{k}$ and ${\overrightarrow{p}}_2=-m^{2}x\hat{i}+3x\hat{j}+2\hat{k}$, if the angle between them is obtuse, then $m$ belongs to

• A. $R$
• B. $R^{+}$
• C. $R^{-}$
• D. $R-\left\{0\right\}$

Answer 1

The correct option is A $R$

${\overrightarrow{p}}_1\cdot {\overrightarrow{p}}_2=|{\overrightarrow{p}}_1||{\overrightarrow{p}}_2|\cos \theta$
For angle to be obtuse $\cos \theta <0$
As $|{\overrightarrow{p}}_1|>0,|{\overrightarrow{p}}_2|>0$, so
${\overrightarrow{p}}_1\cdot {\overrightarrow{p}}_2<0\Rightarrow (6x\hat{i}+2m\hat{j}-\hat{k})\cdot (-m^{2}x\hat{i}+3x\hat{j}+2\hat{k})<0\Rightarrow -6m^2{x}^2+6mx-2<0\Rightarrow 3m^2{x}^2-3mx+1>0$

Here,
$3m^2>0\Rightarrow m\in R-\left\{0\right\}\cdots \left(1\right)D<0\Rightarrow 9m^2-12m^2<0\Rightarrow m^2>0\Rightarrow m\in R-\left\{0\right\}\cdots \left(2\right)$

From equation $\left(1\right)$ and $\left(2\right)$, we get
$m\in R-\left\{0\right\}$
When $m=0$, we get
$0-0+1>0$
Which is true

Hence, $m\in R$