Consider a triangular pyramid ABCD the position vectors of whone agular points are `A(3,0,1),B(-1,4,1),C(5,3, 2) and D(0,-5,4)` Let G be the point of intersection of the medians of the triangle BCT. The length of the vector `bar(AG)` is

Options

A. `sqrt17`
B. `sqrt51//3`
C. `3//sqrt6`
D. `sqrt59//4`

Answer

b

Explanation

Point G is `( 4/3, 1/3,8/3) `. Therefore,
`|vec(AG)|^(2)= (5/3)^(2)+ 1/9+(5/2)^(2)= 51/9`
`or |vec(AG)|= sqrt51/3`
`vec(AB)= -4hati + 4hatj + 0hatk`
` vec(AC)= 2hati + 2hatj + 2hatk`
`vec(AB)xxvec(AC)= -8-8|{:(hati,hatj,hatk),(1,-1,0),(1,1,1):}|`
`8 (hati + hatj - 2hatk)`
Area of `triangleABC = 1/2|vec(AB)xx vec(AC)|=4 sqrt6`
`vec(AD)= -3hati - 5hatj + 3hatk`
The length of the perpendicular from the vertex D on the opposite face
`=|" projection of " vec(AD) " on "vec(AB)xx vec(AC)|`
`|((-3hati-5hatj+ 3hatk)(hati+hatj-2hatk))/sqrt6|`
`|(-3-5-6)/sqrt6|=14/sqrt6`