Question

In a three-dimensional coordinate system P, Q and R are images of a point A (a, b, c) in the xy, yz and zx planes respectively. If G is the centroid of triangle PQR then area of triangle AOG is (O is the origin)
\[{
  A.0 \\
  B.{a^2} + {b^2} + {c^2} \\
  C.\dfrac{2}{3}({a^2} + {b^2} + {c^2}) \\
  D.2({a^2} + {b^2} + {c^2}) \\
} \]

Answer 1

Hint: At first, read the question carefully & note down the given data in the question & also what has to be found as an answer. Then, we have to find P and we will find this by finding the image in \[xy\]-plane. Then \[z\]coordinate will be replaced by \[( - z)\]coordinate then we have p \[(a,b,c)\]. In this way we can find the coordinate of point Q and R.

Complete step-by-step answer:
First, we have to find the coordinate of P. Since P is in $xy$-plane. Therefore, $z$ can be replaced by $( - z)$Therefore, Coordinate of P is $(a,b, - c)$
Similarly, for the Q coordinate
$y$ will be replaced by $( - y)$
Therefore, coordinate of Q is $(a, - b,c)$\[\]
Similarly, for the R coordinate
$x$ will be replaced by $( - x)$
Therefore, coordinate of R is $( - a,b,c)$
To find the centroid, we know the formula i.e., $(\dfrac{{{x_1} + {x_2} + {x_3}}}{3},\dfrac{{{y_1} + {y_2} + {y_3}}}{3},\dfrac{{{z_1} + {z_2} + {z_3}}}{3})$
Therefore, coordinate of centroid $(G)$, where $G = $Centroid
${
   = (\dfrac{{a - a + a}}{3},\dfrac{{b - b + b}}{3},\dfrac{{c + c - c}}{3}) \\
   = (\dfrac{a}{3},\dfrac{b}{3},\dfrac{c}{3}) \\
} $
We know that coordinate of origin in Three Dimension is $(0,0,0)$
Let $O = (0,0,0)$[Where $O$ is origin]
Therefore, we have $A(a,b,c),O(0,0,0),G(\dfrac{a}{3},\dfrac{b}{3},\dfrac{c}{3})$
To find the direction cosine, we can use the above expression.
Therefore, direction cosine of $GO = (\dfrac{a}{3},\dfrac{b}{3},\dfrac{c}{3})$
Therefore, direction cosine of $AO = (a,b,c)$ and direction cosine of \[AG = (\dfrac{{2a}}{3},\dfrac{{2b}}{3},\dfrac{{2c}}{3})\]
Now, From the above direction cosine, it has concluded that all the points lie on the same straight line.
Therefore, it is collinear.

Hence, the area of the triangle \[AOG\] is \[0\].

Note: The area of any collinear points is zero. Concept of straight line & its formulations should be known. It will be absolutely zero because it does not have any area. By using this method, we can easily reach out to the final answer.