Let two vertices of a triangle ABC be (2,4,6) and (0,−2,−5), and its centroid be (2,1,−1). If the image of the third vertex in the plane x+2y+4z=11 is (α,β,γ), then αβ+βγ+γα is equal to :
Explanation
Given that A(2,4,6),B(0,−2,−5) and centroid G(2,1,−1)
Let C(x1,y1,z1)
Since G is centroid of ΔABC therefore
(2,1,−1)≡(32+0+x1,34−2+y1,36−5+z1)
⇒32+x1=2,32+y1=1,31+z1=−1
⇒x1=4,y1=1,z1=−4⇒C(4,1,−4)
Also given that the image of C(4,1,−4) in the plane x+2y+4z=11 is (α,β,γ)
⇒1α−4=2β−1=4γ+4=12+22+42−2(4+2×1+4(−4)−11)
=21−2(−21)=2
⇒α=6,β=5,γ=4
and αβ+βγ+γα=6×5+5×4+4×6=30+20+24=74
Explanation
Given that A(2,4,6),B(0,−2,−5) and centroid G(2,1,−1)
Let C(x1,y1,z1)
Since G is centroid of ΔABC therefore
(2,1,−1)≡(32+0+x1,34−2+y1,36−5+z1)
⇒32+x1=2,32+y1=1,31+z1=−1
⇒x1=4,y1=1,z1=−4⇒C(4,1,−4)
Also given that the image of C(4,1,−4) in the plane x+2y+4z=11 is (α,β,γ)
⇒1α−4=2β−1=4γ+4=12+22+42−2(4+2×1+4(−4)−11)
=21−2(−21)=2
⇒α=6,β=5,γ=4
and αβ+βγ+γα=6×5+5×4+4×6=30+20+24=74
