vectors A(5,-1,3), B(-2,2,4)&C(0,3,-4) are extended outward from origin of a cartesian coordinate sys.?

Question

well i must find the unit vector of point A to a point equidistant from B&C on the line BC. i'm really in trouble.

Answer 1

Point, P, equidistant from B and C on the line BC is the midpoint of BC.
=> P = (1/2) [ -2 + 0, 2 + 3, 4 - 4 ] = (-1, 5/2, 0)

AP vector
= position vector of P - position vector of A
= (-1, 5/2, 0) - ( 5, -1, 3)
= (-6, 7/2, -3)

Magnitude of AP vector
= √[ (-6)^2 + (7/2)^2 + (-3)^2 ]
= √[ 45 + 49/4 ]
= 1/ [(1/2)√(229)] *(-6, 7/2, -3)

Answer 2

Calculate the midpoint M of line segment BC. It is on line BC.

M = [(-2+0)/2, (2+3)/2, (4-4)/2] = (-1, 5/2, 0)

Calculate the directional vector v = AM.

v = AM = = <-1-5, 5/2+1, 0-3> = <-6, 7/2, -3>

Any positive multiple of v is also a directional vector. Multiply by 2.

v = <-12, 7, -6>

Calculate the magnitude of v.

|| v || = √[(-12)² + 7² + (-6)²] = √(144 + 49 + 36) = √229

The unit vector in the direction from A to M is:

v / || v || = <-12/√229, 7/√229, -6/√229>