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### Planes $x-2y-2z+1=0$ & $2x-3y-6z+1=0$ ....

Let the acute angle bisector of the two planes $x-2y-2z+1=0$ and $2x-3y-6z+1=0$ be the plane P. Then which of the following points lies on P?

(A) $(0,2,-4)$
(B) $(4,0,-2)$
(C) $(-2,0,-\frac {1}{2})$
(D) $(3,1,-\frac {1}{2})$

Solution

Bisectors are given by,

$\frac{{x - 2y - 2z + 1}}{3} = \pm \frac{{2x - 3y - 6z + 1}}{7}$

$\Rightarrow 7x - 14y - 14z + 7 = \pm (6x - 9y - 18z + 3)$

Hence, $x-5y+4z+4=0$ & $13x-23y-32z+10=0$

Let $\theta$ be the angle between $x-2y-2z+1=0$ & $x-5y+4z+4=0$

$\cos \theta = \frac{{|1 + 10 - 8|}}{{3 \times \sqrt {42} }} = \frac{1}{{\sqrt {42} }} < \frac{1}{{\sqrt 2 }}$

$\theta > 45^\circ$

So, $x-5y+4z+4=0$ is the obtuse angle bisector.

$\therefore$ Acute angle bisector $P \equiv 13x - 23y - 32z + 10 = 0$

The point $(-2,0,-\frac {1}{2})$ satisfies P.

### $f(x)=x^6+2x^4+x^3+2x+3 $$\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$$n=?$
Let $f(x)=x^6+2x^4+x^3+2x+3,x \in R$. Then the natural number n for which $\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$ is _ _ _ _ . Solution Since the limit has $\left[ {\frac{0}{0}} \right]$ form, L.H. Rule is applicable. Thus, $\mathop {\lim }\limits_{x \to 1} n{x^{n - 1}}f(1) - f'(x) = 44$ $\therefore nf(1) - f'(1) = 44$ $\therefore n.9 - ({6.1^5} + {8.1^3} + {3.1^2} + 2.1) = 44$ $\Rightarrow 9n - 19 = 44$ $\Rightarrow n=7$