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posted by sophie on Thursday, May 12, 2011 at 9:25pm.

Given the plane 2x-3y+7z=4, a) Find the equation for the line perpendicular to the plane that intersects the plane at the point P=(6,5,1). b) How far is it from the point (4,-2,3) to the plane?

a)(x-6)/2=(y-5)/(-3)=(z-1)/7 b)The normal equation of the plane (2/sqrt(62))x-(3/sqrt(62))y+(7/sqrt(62))z- -4/sqrt(62)=0 62=2^2+(-3)^2+7^2 The distance=(2*4-3*(-2)+7*3-4)/sqrt(62)= sqrt(31/2)

why do you have = signs in part a?

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