Anonymous

# 3d vectors?

Are the lines AB and CD parallel, intersect, or skewed.

A(3,2,4) B(-3,-7,-8) C(0,1,3) D(-2,5,9)

Relevance

Line AB = A + sAB

Line CD = C + tCD

Calculate the directional vectors.

u = AB = <B - A> = <-3-3, -7-2, -8-4> = <-6, -9, -12>

v = CD = <D - C> = <-2-0, 5-1, 9-3> = <-2, 4, 6>

u and v are not non-zero multiples of each other so the lines are not parallel.

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Check to see if there is a point of intersection.

Line AB = A + sAB = A + su

Line CD = C + tCD = C + tv

Any non-zero multiple of the directional vectors of the lines are also directional vectors. Divide u by -3 and divide v by -2.

u = <2, 3, 4>

v = <1, -2, -3>

Line AB = <3, 2, 4> + s<2, 3, 4>

Line CD = <0, 1, 3> + t<1, -2, -3>

x = 3 + 2s = 0 + t

y = 2 + 3s = 1 - 2t

z = 4 + 4s = 3 - 3t

-2x + z = -2 = 3 - 5t

5t = 5

t = 1

Plug back into the equation for x and solve for s.

x = 3 + 2s = 0 + t

3 + 2s = 0 + 1

2s = -2

s = -1

Now plug both values into the equation for y to see if the solution is consistent. If it is the lines intersect.

y = 2 + 3s = 1 - 2t

y = 2 + 3*(-1) = 1 - 2*(1)

2 - 3 = 1 - 2

-1 = -1

The equations are consistent. The lines intersect.