find the point that trisect the line segment from A(3,6,8) to B(6,0,-1)?

Question

vectors

Answer 1

Point O (0,0,0)
OA = 3i + 6j + 8k
OB = 6i + 0j - k
AB = AO + OB = OB - OA
=(6i + 0j - k) - (3i + 6j + 8k)
=3i - 6j -9k

trisect means 1/3 of the line
let the point P be the point you are trying to find
AP= 1/3 * AB
=3/3i - 6/3j -9/3k
= i - 2j - 3k

OP=OA+AP
OP= (3i+6j+8k) + (i-2j-3k)
=4i+4j+5k

The point is (4,4,5)

Answer 2

If i understand "trisect" correctly, there are actually 2 points that trisect a line: first one (call it P) is at the 1/3 mark and another (call it Q) is at the 2/3 mark. So, P=A+1/3(B-A) and Q=A+2/3(B-A). Or P = (3,6,8)+1/3(6-3,0-6,-1-8) = (3,6,8) + 1/3(3,-6,-9) = (3,6,8)+(1,-2,-3) = (4,4,5). Similarly, Q = (3,6,8) + 2/3(3,-6,-9) = (3,6,8)+(2,-4,-6) = (5,2,2)

Answer 3

Let p be position vector of point that trisects line.
p ,a and b are VECTORS
p = a + (1/3).(b - a)
p = (2/3).a + (1/3).b
p = (2/3) [ 3 6 8 ] + (1/3) [6 0 - 1]
p = [ 2 4 16/3] + [2 0 - 1/3]
p = [ 4 4 5 ]-----vector
Point is P(4, 4, 5)

Answer 4

There are two such points, (4,4,5) and (5,2,2).