Question

The vectors $a=2{\lambda }^{2}i+4\lambda j+k$ and $b=7i-2j+\lambda k$ make an obtuse angle whereas the angle between b and k is acute and less then $\frac{\pi }{6},$ then determine the domain of $\lambda .$

• A. $0<\lambda <\frac{1}{2}$
• B. $\lambda >\frac{1}{2}$
• C. $\lambda =$null set
• D. $\lambda >\sqrt{159}$

Answer 1

The correct option is C $\lambda =$null set

I. $a.b=ab\cos \theta <0$ as $\theta$ is obtuse
$\therefore 14{\lambda }^2-8\lambda +\lambda <0$

$\Rightarrow 7\lambda (2\lambda -1)<0$$\Rightarrow 0<\lambda <\frac{1}{2}=D_1$ ...(1)

II. $b.k=\left|b\right|.1.\cos \theta ,\lambda =\sqrt{53+{\lambda }^2}\cos \theta$

$\therefore \cos \theta =\frac{\lambda }{\sqrt{53+{\lambda }^2}}$

Since $\theta <\frac{\pi }{6}$

$\therefore \sin \theta <\frac{1}{2}$ and $\cos \theta >\frac{\sqrt{3}}{2}$

$\therefore \frac{\lambda }{\sqrt{53+{\lambda }^2}}>\frac{\sqrt{3}}{2}$

$\Rightarrow 4{\lambda }^2>3(53+{\lambda }^2)$

$\therefore$ ${\lambda }^2>159$ or $\lambda >\sqrt{159}=D_2$

$\left(1\right)$ and $\left(2\right)$ are separate independent conditions.

But if they are to hold simultaneously it can't be possible.
Hence the domain of $\lambda$

i.e.intersection $D_1\bigcap D_2$ ...(2) is a null set.