Let P be the plane containing the point(-5,-5,5) and with normal vector &lt;3,2,1&gt;. Find thepoint of P closest to the point (6,1,2).

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Let P be the plane containing the point(-5,-5,5) and with normal vector &amp;lt;3,2,1&amp;gt;. Find thepoint of P closest to the point (6,1,2).

grocbyntza · Accepted Answer

first of all write the equation for the plane...3(x+5)+2(y+5)+1(z-5)=03x+2y+z+20=0The shortest distance from the plane to the point will be a line orthogonal to the plane...thus the direction of the line will bethe same as the normal direction of the plane...Since the line passes through the point (6, 1, 2) then its equation must be...&amp;lt;6,1,2&amp;gt;+t&amp;lt;3,2,1&amp;gt;=&amp;lt;6+3t,1+2t, 2+t&amp;gt;all we need to know now is what value of the parameter t satisfies the equation of the plane then plug it back into the equation for our line to find the point...3x+2y+z+20=03(6+t)+2(1+2t)+(2+t)+20=0so we find that when t = -3 the line intersects the plane, thus from &amp;lt;6+3t, 1+2t, 2+t&amp;gt; we know that the point is.(-3,-5,-1)

Nash Frank · Accepted Answer

The normal is normal to the plane.The point normal form of a plane is normal vector dot product local vector minus distance equals null. The distance is to the origin. Every point in the plane has this distance. (-5,-5,5) (dotproduct) (3,2,1)=-15-10+5=-20. the point normal form of the given plane is 3x+2y+z+20=0.The point nearest to the point is on the line (6,1,2)+s(3,2,1)and in the plane. So:(3,2,1) (dotproduct)((6,1,2)+s(3,2,1))+20=18+2+2+s(9+4+1)+20=14s+42=0, so s=-3.The point is (6,1,2)-3(3,2,1)=(-3,-5,-1).

Let P be the plane containing the point(-5,-5,5) and with normal vector <3,2,1>. Find thepoint of P closest to the point (6,1,2).

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