Question

If the centroid of the tetrahedron OABC, cohere A, B, C are given by \[\left( \alpha ,5,6 \right),\left( 1,\beta ,4 \right),\left( 3,2,\delta \right)\] respectively be (1,-1,2) then the value of \[{{\alpha }^{2}}+{{\beta }^{2}}+{{\delta }^{2}}\] equals,
A. \[{{\alpha }^{2}}+{{\beta }^{2}}\]
B. \[{{\delta }^{2}}+{{\beta }^{2}}\]
C. \[{{\alpha }^{2}}+{{\delta }^{2}}\]
D. none of these

Answer 1

Hint:Tetrahedron is a triangular period. Substitute the given vertices in the formula to find the centroid of the tetrahedron. Substitute (1,-1,2) and get the value of \[\alpha ,\beta ,\delta \].

Complete step-by-step answer:
We have been given a tetrahedron OABC. A tetrahedron is also known as a triangular pyramid. It is a polyhedron composed of four triangles faces, six straight edges and four vertex corners.

Let the 4 vertices of the tetrahedron be,

\[O(0,0,0),A\left( \alpha ,5,6 \right),B\left( 1,\beta ,4 \right),C\left( 3,2,\delta \right)\].
Thus the centroid of the tetrahedron can be found using the formula,
Centroid \[=\left(
\dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}}{4},\dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}+{{y}_{4}}}{
4},\dfrac{{{z}_{1}}+{{z}_{2}}+{{z}_{3}}+{{z}_{4}}}{4} \right)\]
Here, \[\begin{align}
  & \left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)=(0,0,0) \\
 & \left( {{x}_{2}},{{y}_{2}},{{z}_{2}} \right)=\left( \alpha ,5,6 \right) \\
 & \left( {{x}_{3}},{{y}_{3}},{{z}_{3}} \right)=\left( 1,\beta ,4 \right) \\
 & \left( {{x}_{4}},{{y}_{4}},{{z}_{4}} \right)=\left( 3,2,\delta \right) \\
\end{align}\]
Thus, substitute these values in the formula of centroid.
Centroid \[=\left( \dfrac{0+\alpha +1+3}{4},\dfrac{{{0}_{1}}+5+\beta
+2}{4},\dfrac{0+6+4+\delta }{4} \right)\]\[=\left( \dfrac{4+\alpha }{4},\dfrac{7+\beta
}{4},\dfrac{10+\delta }{4} \right)\]
We have been given the values of the centroid as (1,-1,2).
\[\left( \dfrac{4+\alpha }{4},\dfrac{7+\beta }{4},\dfrac{10+\delta }{4} \right)=\left( 1,-1,2
\right)\]

Thus we can say that,
\[\dfrac{4+\alpha }{4}=1,\dfrac{7+\beta }{4}=-1,\dfrac{10+\delta }{4}=2\]
Cross multiply and get the values of \[\alpha ,\beta ,\delta \].
\[\begin{align}
  & 4+\alpha =4 \\
 & \therefore \alpha =4-4=0 \\
 & 7+\beta =-4 \\
 & \beta =-4-7 \\
 & \therefore \beta =-11 \\
 & 10+\delta =8 \\
 & \delta =8-10=-2 \\
\end{align}\]
Thus we got\[\alpha =0,\beta =-11,\delta =-2\].
Now we need to find the value of \[{{\alpha }^{2}}+{{\beta }^{2}}+{{\delta }^{2}}\]. If we put
\[\alpha =0\], we get,
\[\begin{align}
  & {{\alpha }^{2}}+{{\beta }^{2}}+{{\delta }^{2}}=0+{{\beta }^{2}}+{{\delta }^{2}} \\
 & \therefore {{\alpha }^{2}}+{{\beta }^{2}}+{{\delta }^{2}}={{\beta }^{2}}+{{\delta }^{2}} \\
\end{align}\]
Thus, option B is the correct answer.

Note:
Tetrahedron is the simplest of the entire ordinary convex polygon and the only one that has fewer than 5 faces. The tetrahedron is the 3D case of Euclideon simplex. So it can also be called 3-simplex.