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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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${\log _{\sqrt 5 }}\left[ {3 + \cos \left( {\frac{{3\pi }}{4} + x} \right) + \cos \left( {\frac{\pi }{4} + x} \right) + \cos \left( {\frac{\pi }{4} - x} \right) - \cos \left( {\frac{{3\pi }}{4} - x} \right)} \right]$

The range of the function $f(x) = {\log _{\sqrt 5 }}\left[ {3 + \cos \left( {\frac{{3\pi }}{4} + x} \right) + \cos \left( {\frac{\pi }{4} + x} \right) + \cos \left( {\frac{\pi }{4} - x} \right) - \cos \left( {\frac{{3\pi }}{4} - x} \right)} \right]$ is: (A) $[ - 2,2]$ (B) $\left[ {\frac{1}{{\sqrt 5 }},\sqrt 5 } \right]$ (C) $(0,\sqrt 5 )$ (D) $[ 0,2]$ Solution We have, $f(x) = {\log _{\sqrt 5 }}\left( {3 - 2\sin \frac{{3\pi }}{4}\sin x + 2\cos \frac{\pi }{4}\cos x} \right)$ $ \Rightarrow f(x) = {\log _{\sqrt 5 }}\left[ {3 + \sqrt 2 (\cos x - \sin x)} \right]$ Now, $ - \sqrt 2  \le \cos x - \sin x \le \sqrt 2 $ $\therefore - 2 \le \sqrt 2 (\cos x - \sin x) \le 2$ $\therefore 1 \le 3 + \sqrt 2 (\cos x - \sin x) \le 5$ $\therefore{\log _{\sqrt 5 }}1 \le {\log _{\sqrt 5 }}[3 + \sqrt 2 (\cos x - \sin x)] \le {\log _{\sqrt 5 }}5$ $ \Rightarrow 0 \le f(x) \le 2$ Answer: (D)
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