Question

In a three dimensional co-ordinate system $P,Q$ and $R$ are images of a point $A(a,b,c)$ in the $xy$, the $yz$ and $zx$ 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. $2(a^2+b^2+c^2)$

Answer 1

The correct option is A $0$

Given:A point $A(a,b,c)$ has its image in $xy,yz$ and $zx$ coordinate system.

Image of $A(a,b,c)$ in $xy-$plane:

Since we need its image in $xy-$ plane, only the sign of $z$ coordinate changes.

$\therefore$Image of $A(a,b,c)$ in $xy-$plane is $P(a,b,-c)$

Image of $A(a,b,c)$ in $yz-$plane:

Since we need its image in $yz-$ plane, only the sign of $x$ coordinate changes.

$\therefore$Image of $A(a,b,c)$ in $yz-$plane is $Q(-a,b,c)$

Image of $A(a,b,c)$ in $zx-$plane:

Since we need its image in $zx-$ plane, only the sign of $y$ coordinate changes.

$\therefore$Image of $A(a,b,c)$ in $zx-$plane is $R(a,-b,c)$

Given:$G$ is the Centroid of $△PQR$

$\therefore$centroid$G=(\frac{a-a+a}{3},\frac{b+b-b}{3},\frac{-c+c+c}{3})$

$\therefore G=(\frac{a}{3},\frac{b}{3},\frac{c}{3})$

Now, We have $A(a,b,c),G=(\frac{a}{3},\frac{b}{3},\frac{c}{3})$ and $O(0,0,0)$ is the origin

Since $AO=(a,b,c),OG=(\frac{a}{3},\frac{b}{3},\frac{c}{3})$ and $AG=(\frac{2a}{3},\frac{2b}{3},\frac{2c}{3})$

their direction ratios are of $AO=a:b:c,OG=a:b:c$ and $AG=a:b:c$ which are same.

Hence they lie on the same line.

$\therefore$Area of $△AOG=0$