Question

Find the value or range of value of m if:-
the vector 2i(cap) - mj(cap) +3mk(cap) and (1+m)i(cap) -2mj(cap) +k(cap) include an acute angle

Answer 1

Answer:

The range of m is

$\bf\,(-\infty,-2)\cup(\frac{-1}{2},\infty)$

Step-by-step explanation:

$\text{Concept:}$

$\text{The angle between the vectors }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ is }$

$\bf\,cos\theta=\frac{\overrightarrow{a}.\overrightarrow{b}}{|\overrightarrow{a}|\,|\overrightarrow{b}|}$

$\text{Let}$

$\overrightarrow{a}=2\overrightarrow{i}-m\overrightarrow{j}+3m\overrightarrow{k}$

$\overrightarrow{b}=(1+m)\overrightarrow{i}-2m\overrightarrow{j}+\overrightarrow{k}$

$\text{since }\theta\text{ is acute, }cos\theta>0$

$\implies\overrightarrow{a}.\overrightarrow{b}>0$

$\implies\,2(1+m)+2m^2+3m>0$

$\implies\,2+2m+2m^2+3m>0$

$\implies\,2m^2+5m+2>0$

$\implies\,(2m+1)(m+2)>0$

$\implies\bf\,m\in\,(-\infty,-2)\cup(\frac{-1}{2},\infty)$

Answer 2

The value of $m$ is $-2$ or $-\frac{1}{2}$

Step-by-step explanation:

Given,

Lets take two vector,

$\overrightarrow a=2\overrightarrow i-m\overrightarrow j+3m\overrightarrow k$ and $\overrightarrow b=(1+m)\overrightarrow i-2m\overrightarrow j+\overrightarrow k$

The angle between angle $\overrightarrow a\ and\ \overrightarrow b$ is given,

$Cos\theta=\frac{\overrightarrow a.\overrightarrow b}{\left | \overrightarrow a \right | \left | \overrightarrow b \right |}$

Since, $Cos\theta >0$

⇒ $\overrightarrow a.\overrightarrow b >0$

⇒$2(1+m)+2m^{2} +3m>0$

⇒$2+2m+2m^{2}+3m>0$

⇒$2m^{2}+5m+2>0$

⇒$(2m+1)(m+2)>0$

∴ $m=-2,-\frac{1}{2}$

So, The value of $m$ is $-2$ or $-\frac{1}{2}$