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The distance of line $3y-2z-1=0=3x-z+4$ from the point $(2,-1,6)=?$

The distance of line $3y-2z-1=0=3x-z+4$ from the point $(2,-1,6)$ is:

(A) $2\sqrt 6$
(B) $\sqrt {26}$
(C) $2\sqrt 5 $
(D) $4\sqrt 2 $

Solution

We have, $3x-z+4=0$ or $z=3x+4$
& $3y-2z-1=0$ or $3y-2(3x+4)-1=0$ or $y=2x+3$

Any point $(x,y,z)$ on the line $ \equiv (t,2t + 3,3t + 4)$

Let $d$ be the distance between $(2,-1,6)$ & $ (t,2t + 3,3t + 4)$

Then, $d^2=(t-2)^2+(2t+3+1)^2+(3t+4-6)^2=14t^2+24$

Minimum $d$ = Required answer = $\sqrt {24}=2\sqrt 6$ when t = 0.

Answer: (A)

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$f(x)=x^6+2x^4+x^3+2x+3 $

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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$ $\Rightarrow n=7$