Question

In a three-dimensional coordinate system, P, Q and R are images of a point A$(a,b,c)$ in the $x-y,y-z$ and $z-x$ planes, respectively. If G is the centroid of triangle PQR, then area of triangle AOG is (O is the origin)-

• A. $0$
• B. $a^2+b^2+c^2$
• C. $\frac{2}{3}(a^2+b^2+c^2)$
• D. None of these.

Answer 1

Answer: (A)

$0$

Given $P,Q,R$ are images of point $A(a,b,c)$ in $XY,YZ,ZX$ planes

$G$ is centroid of triangle $PQR$

Formula: The image of point $(x,y,z)$ in XY plane is $(x,y,-z)$ and $YZ$ plane is $(-x,y,z)$, $ZX$ plane is $(x,-y,z)$.

In this case point is $A(a,b,c)$

Image of point $A$ in $XY$ plane is $P(a,b,-c)$

$YZ$ plane is $Q(-a,b,c)$

$ZX$ plane is $R(a,-b,c)$

Formula: centroid of triangle is $\left(\frac{sumofcoordinates}{3}\right)$

Centroid of triangle is $(\frac{a-a+a}{3},\frac{b-b+b}{3},\frac{-c+c+c}{3})(\frac{a}{3},\frac{b}{3},\frac{c}{3})$

Formula: Area of triangle is $\frac{1}{2}|AO\times GO|$

Vector $AO$ is $(a,b,c)$

Vector $GO$ is $(\frac{a}{3},\frac{b}{3},\frac{c}{3})$

By formula

$\frac{1}{2}|AO\times GO|$$=\frac{1}{2}\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ a& b& c\\ \frac{a}{3}& \frac{b}{3}& \frac{c}{3}\end{array}\right|=\frac{1}{2}\left|\hat{i}\right(0)+\hat{j}(0)+\hat{k}(0\left)\right|=0$

Area of triangle $AOG$ is zero