Q. If the vectors $\overrightarrow{\mathbf{a}} = \left( c\ \log_{2}x \right)\widehat{\dot{\mathbf{i}}} - 6\widehat{\dot{\mathbf{j}}} + 3\widehat{\mathbf{k}}$ and $\overrightarrow{\mathbf{b}} = \left( \log_{2}x \right)\widehat{\dot{\mathbf{i}}} + 2\widehat{\dot{\mathbf{j}}} + (2c\ {log\ }_{2}x)\widehat{\mathbf{k}}$ make an abtuse angle for any $x\ \in (0,\infty)$, then the interval of which $c$ belongs
A) $\left( \frac{4}{3},0 \right)$
B) $\left( - \infty, - \frac{4}{3} \right)$
C) $\left( \frac{3}{4},0 \right)$
D) $\left( - \frac{3}{4},0 \right)$
Explanation
(b)
$Since,\cos{\theta = \frac{\overrightarrow{\mathbf{a}} \bullet \overrightarrow{\mathbf{b}}}{\overrightarrow{\mathbf{|a}}||\overrightarrow{\mathbf{b}}|}}$
$= \frac{c(\log_{2}x)^{2} - 12 + 6c\ \log_{2}x}{\lbrack\sqrt{(\ c\ \log_{2}x)^{2} + 36 + 9} \times \sqrt{(\log_{2}x)^{2} + 4 + 4(c\ \log_{2}x)^{2}}}$
For obtuse angle,
$\cos{\theta < 0}$
$\Longrightarrow \ c(\log_{2}x)^{2} - 12 + 6c\ \log_{2}x < 0$
$\Longrightarrow c < 0\ and\ D < 0$
$\Longrightarrow c < 0\ and\ (6c)^{2} + 48c < 0$
$\Longrightarrow c < 0\ and\ c < - \frac{4}{3}$
$\therefore c\epsilon\left( - \frac{4}{3},0 \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.