Question

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

• A. $c<0$
• B. $0<c<\frac{4}{3}$
• C. $-\frac{4}{3}<c<0$
• D. $c>0$

Answer 1

Answer: (C)

For an obtuse angle
$(cx\hat{i}-6\hat{j}+3\hat{k})\cdot (x\hat{i}+2\hat{j}+2cx\hat{k})<0$
$\Rightarrow c{x}^2-12+6cx<0$
$\Rightarrow c{x}^2+6cx-12<0$
We know that, if
$a{x}^2+bx+c>$ or $<0,\forall x$
Then, $b^2-4ac<0$
$\therefore (6c{)}^2-4c(-12)<0$
$\Rightarrow 3c^2+4c<0$
$\Rightarrow 3c^2\left(c+\frac{4}{3}\right)<0$
$\Rightarrow -\frac{4}{3}<c<0$