Question 206220: Hi all, I was hoping someone could help me answer the followingl; I need to show that the lines:
[(x-2)/5] = [(y+3)/-3] = [(x-5)/2] and [(x-3)/4] = [(y-2)/6] = [(z-10)/-1] are orthagonal.
Im not sure how to do this so some help with steps and explanations would most appreciated.
Thanks, -nick
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! I need to show that the lines:
[(x-2)/5] = [(y+3)/-3] = [(x-5)/2] and [(x-3)/4] = [(y-2)/6] = [(z-10)/-1] are orthagonal.
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(x-2)/5 = (y+3)/-3 = (x-5)/2 (I think the last term is z)
x = 5t+2, y = -3t-3, z = 2t+5
Direction vector v1 = (5,-3,2)
ABS(v1) = sqrt(25 + 9 + 4) = sqrt(38)
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(x-3)/4 = (y-2)/6 = (z-10)/-1
Direction vector v2 = (4,6,-1)
ABS(v2) = sqrt(16+36+1) = sqrt(53)
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v1 dot v2 = (5*4 + -3*6 + 2*-1) = 0
Since the lengths (ABS) are not zero, the cosine of the angle between is zero.
So the angle between the vectors (and the 2 lines) is 90 degrees.
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