A= (1, -2, 3), B = (2, 1, 3), C = (4, 2, 1) and G = (-1, 3, 5) is the centroid of the tetrahedron ABCD. If p= Dy and q = Dz then 13p-1.1q =

Question

A= (1, -2, 3), B = (2, 1, 3), C = (4, 2, 1) and G = (-1, 3, 5) is the centroid of the tetrahedron ABCD. If p= Dy and q = Dz then 13p-1.1q = (A) 0 (B) 1 (C) -1 (D) 2

Tardigrade Admin · Accepted Answer

((1+2+4+x/4),(-2 +1 +2 +y /4),(3+3+1+z/4)) = (-1 ,3,5)

Q. $A = (1, - 2, 3), B = (2, 1, 3), C = (4, 2, 1)$ and $G = (- 1, 3, 5)$ is the centroid of the tetrahedron ABCD. If $p = D_{y}$ and $q = D_{z}$ then $13 p - 1.1 q =$

0

1

-1

2

$(\frac{1 + 2 + 4 + x}{4}, \frac{- 2 + 1 + 2 + y}{4}, \frac{3 + 3 + 1 + z}{4})$
$= (- 1, 3, 5)$

Q. A=(1,−2,3),B=(2,1,3),C=(4,2,1) and G=(−1,3,5) is the centroid of the tetrahedron ABCD. If p=Dy​ and q=Dz​ then 13p−1.1q=

0

1

-1

2

(41+2+4+x​,4−2+1+2+y​,43+3+1+z​) =(−1,3,5)

Q. $A = (1, - 2, 3), B = (2, 1, 3), C = (4, 2, 1)$ and $G = (- 1, 3, 5)$ is the centroid of the tetrahedron ABCD. If $p = D_{y}$ and $q = D_{z}$ then $13 p - 1.1 q =$

$(\frac{1 + 2 + 4 + x}{4}, \frac{- 2 + 1 + 2 + y}{4}, \frac{3 + 3 + 1 + z}{4})$
$= (- 1, 3, 5)$