Question

In a three-dimensional coordinate system P, Q and R are images of 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. a2+b2+c2
C. \[\dfrac{2}{3}\](a2+b2+c2)
D. None of these.

Answer 1

Hint: In this question, the first thing that you should do is to write down the formula for finding the area of the triangle. Area of the triangle is $\dfrac{1}{2}$ $ \times $ base $ \times $height; here AO vector is base and GO vector is height. Centroid of the triangle is (sum of coordinates of x divided by 3, sum of coordinates of y divided by 3, sum of coordinates of z divided by 3). Finally put the value in the formula to get the answer.

Complete step-by-step answer:
In this question it is given that,

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

G is the centroid of triangle PQR.
 The image of point (x, y, z) in XY plane is (x, y,-z),
 YZ plane is (-x, y, z) and the 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)
Centroid of triangle is (sum of coordinates of x divided by 3, sum of coordinates of y divided by 3, sum of coordinates of z divided by 3)
Centroid of triangle is$\left( {\dfrac{{a - a + a}}{3},\dfrac{{b + b - b}}{3},\dfrac{{ - c + c + c}}{3}} \right)$
                                         = $\left( {\dfrac{a}{3},\dfrac{b}{3},\dfrac{c}{3}} \right)$
Area of triangle = $\dfrac{1}{2}$ $ \times $ base $ \times $height
                              = $\dfrac{1}{2}$|AO $ \times $GO|
 Vector AO = (a, b, c)
And vector GO = $\left( {\dfrac{a}{3},\dfrac{b}{3},\dfrac{c}{3}} \right)$
 By formula
              $\dfrac{1}{2}$|AO $ \times $GO| = $\dfrac{1}{2}$$\left| {\left( {\begin{array}{*{20}{c}}
  {\hat{i} }&{\hat{j} }&{\hat{k}} \\
  a&b&c \\
  {\dfrac{a}{3}}&{\dfrac{b}{3}}&{\dfrac{c}{3}}
\end{array}} \right)} \right|$
                                          =$\dfrac{1}{2}$$\left| {\hat{i} \left( 0 \right) +\hat{j} \left( 0 \right) + \hat{k} \left( 0 \right)} \right|$
                                          = 0
             Area of triangle = 0
Thus, the area of triangle AOG is zero, so option A is correct.

Note: In XY plane the coordinate of x and y is positive and coordinate of z is negative similarly in YZ plane the coordinate of z and y is positive and coordinate of x is negative, in ZX plane the coordinate of x and z is positive and coordinate of y is negative. In 3D and vector$\hat{i}.\hat{i}$=1,$\hat{j}.\hat{j}$=1 and $\hat{k}.\hat{k}$=1; where $\hat{i}$ is unit vector along x-axis, $\hat{j}$ is unit vector along y-axis and $\hat{k}$ is unit vector along z-axis.