Question

The value of $c$ so that for all real $x$, then vectors $cx\hat{i}-6\hat{j}+3\hat{k},x\hat{i}+2\hat{j}+2cx\hat{k}$ make an obtuse angle are-

• A. $c<0$
• B. $0<c<4/3$
• C. $-4/3<c<0$
• D. $c>0$

Answer 1

The correct option is A $c<0$

$\overline{a}=cx\hat{i}-6\hat{j}+3\hat{k}$

$\overline{b}=x\hat{i}+2\hat{j}+2cx\hat{k}$

$\overline{a}$ & $\overline{b}$ contain an obtuse angle

$cos\theta =\frac{\overline{a}.\overline{b}}{\left|\overline{a}\right|\left|\overline{b}\right|}<0$

$\Rightarrow \frac{(c{x}^2-12+6cx)}{\sqrt{c^2{x}^2+36+9}\sqrt{x^2+4+4c^2{x}^2}}<0$

$\Rightarrow c{x}^2+6cx-12<0$

$\Rightarrow c(x^2+6x)<12$

$\Rightarrow c<\frac{12}{x^2+6x}=\frac{12}{x^2+6x+9-9}$ $=\frac{12}{(x+3{)}^2-9}$

$\Rightarrow c<min\left[\frac{12}{(x+3{)}^2-9}\right]$
$\Rightarrow c<\frac{12}{\infty -9}$

$\therefore c<0$