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  • Let the points on the plane P be equidistant from the points  (-4,2,1)$ and (2,-2,3). Then the acute angle between the plane <img alt="\mathrm{p}" src="https://entrancecorner.oncodec

Let the points on the plane P be equidistant from the points  (-4,2,1)$ and (2,-2,3). Then the acute angle between the plane \mathrm{p} and the plane \mathrm{2x+y+3z} is

Option: 1

\mathrm{\frac{\pi}{6}}


Option: 2

\mathrm{\frac{\pi}{4}}


Option: 3

\mathrm{\frac{\pi}{3}}


Option: 4

\mathrm{\frac{5\pi}{12}}


Answers (1)

best_answer

\mathrm{A:(-4,2,1) \quad B:(2,-2,3) }

mid point of A & B=\mathrm{(-1,0,2) }

Normal vector : AB=\mathrm{(6,-4,2) }\mathrm{\therefore \quad P^{'}: \quad 6(x+1)+(-4)(y-0)+2(z-2)=0}

\mathrm{3 x-2 y+z+1=0 }

\mathrm{P: \quad 2 x+y+3 z-1=0 }

Angle between \mathrm{ p \& P^{\prime}=(\theta)=\cos ^{2}\left(\frac{6-2+3}{14}\right)}

\mathrm{ =\frac{\pi}{3}}

Posted by

seema garhwal

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