Question

The vertices of a figure are given by A(-8,4,-2), B(-6,3,5), and C(-10,5,-9). What type of figure is ABC? Justify your answer (Hint: Calculate the magnitudes of the sides of the figure)

Answer 1

Have you considered calculating the three magnitudes?

Answer 2

yup done

Answer 3

Given A(-8,4,-2),B(-6,3,5),C(-10,5,-9)
(AB) ⃗=((-6)-(-8)),(3-4),(5)-(-2)=(2,-1,7)
|(AB) ⃗ |= √(〖(2)〗^2+〖(-1)〗^2+〖(7)〗^2 )=√54=∛6
(AC) ⃗=(-10)-(-8),(5-4),(-9)-(-2)=(-2,1,-7)
|(AC) ⃗ |=√((〖-2)〗^2+(1)^2+(〖-7)〗^2 )=√54= ∛6
(BC) ⃗=(-10)-(-6),(5-3),( -9-5)=(-4,2,-14)
|(BC) ⃗ |=√(〖(-4)〗^2+〖(2)〗^2+〖(-14)〗^2 )=√216=6√6
|(BC) ⃗ |^2=|(AB) ⃗ |^2+|(AC) ⃗ |^2
(6√6)^2=(∛6)^2+〖(∛6)〗^2
216=54+54
216=108

is what i got

Answer 4

How did you get a cube root?

Please do it over until you get, \(3\sqrt{6},\;3\sqrt{6},\;and\;6\sqrt{6}\)