Question

In the tetrahedron ABCD, \[A=\left( 1,2,-3 \right)\]and \[G=\left( -3,4,5 \right)\]is the centroid of tetrahedron. If P is centroid of triangle BCD, then AP= . . . . . . . . . . . . . . . . . . .
A. \[\dfrac{8\sqrt{21}}{3}\]
B. \[\dfrac{4\sqrt{21}}{3}\]
C. \[4\sqrt{21}\]
D. \[\dfrac{\sqrt{21}}{3}\]

Answer 1

Hint: If G is the centroid of tetrahedron ABCD and P is the centroid of triangle BCD then G which is centroid of tetrahedron ABCD divides the line AP in the ratio 3:1 it is theorem proved theoretically in coordinate geometry which you can refer it.

Complete step by step answer:
Given, \[A=\left( 1,2,-3 \right)\]and \[G=\left( -3,4,5 \right)\]

Therefore, distance between two points AG is given by
\[AG=\sqrt{{{\left( -3-1 \right)}^{2}}+{{\left( 4-2 \right)}^{2}}+{{\left( 5-\left( -3 \right) \right)}^{2}}}\]
\[AG=\sqrt{16+4+64}\]
\[AG=\sqrt{84}=2\sqrt{21}\]. . . . . . . . . . . . . . . . . . . . (1)
Given that P is the centroid of triangle BCD
So, G divides AP in the ratio 3:1
Let AG=3x . . . . . . . . . . . . . . . . . . . .(2)
 then GP becomes x. . . . . . . . . . . . . . . .(3)
We know the value of AG as we calculated above using distance formula,
So now evaluate the value of x using AG value
\[3x=2\sqrt{21}\]
\[x=\dfrac{2\sqrt{21}}{3}\]. . . . . . . . . . . . . . . . . . . . . . . . (4)
Now \[AP=AG+GP\]
\[AP=3x+x\]
\[AP=4x=4\left( \dfrac{2\sqrt{21}}{3} \right)\]
\[AP=\dfrac{8\sqrt{21}}{3}\]

So, the correct answer is “Option A”.

Note: The point of concurrence of medians of triangle is called centroid. It is denoted by the letter G. the centroid of the triangle also divides the medians of the triangle in the ratio 2:1. The centroid also divides the line segment joining orthocentre and circumcentre in the ratio 2:1 internally