Family of Planes Passing through the Intersection of Two Planes
A variable pl...
Question
A variable plane which remains at a constant distance 3p from the origin cuts the coordinate axes at A,B,C. Show that the locus of the centroid of triangle ABC is 1x2+1y2+1z2=1p2
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Solution
Let the equation of the plane be xa+yb+zc=1 ...(i) where a,b,c are variables. This plane meets X,Y and Z axes at points A(a,0,0),B(0,b,0) and C(0,0,c) respectively. Let (α,β,γ) be the coordinates of the centroid of ΔABC. Then, α=a+0+03=a3,β=0+b+03=b3, γ=0+0+c3=c3 ...(ii) The plane represented by equation (i) is at a distance 3p from the origin. ∴3p = Length of perpendicular from (0,0,0) to the plane (i) ⇒3p=∣∣0a+0b+0c−1∣∣√(1a)2+(1b)2+(1c)2 ⇒3p=1√1a2+1b2+1c2 ⇒19p2=1a2+1b2+1c2 ...(iii) From (ii), we have a=3α,b=3β and c=3γ Substituting the values of a,b,c, in (iii), we obtian ⇒19p2=19α2+1β2+19γ2 ⇒1p2=1α2+1β2+1γ2 So, the locus of centroid of triangle is 1p2=1x2+1y2+1z2