Question 947678: Which of the following is the point of intersection (if it exists) of the line in R3 with vector
equation (x, y, z) = (2a, 0, −c) +t(a, −b, −2c), t ∈ R and the line in R3 with vector equation
(x, y, z) = (0, 2b, c) + s(2a, −2b, 0), s ∈ R (where a, b and c are non-zero real numbers)?
really need help with this cant figure it out
cheers
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! we are given two lines such that
equation A is (x,y,z) = 2a+0b-c+ta-tb-2tc
equation B is (x,y,z) = 0a+2b+c+2sa-2sb+0sc
If the two lines intersect, then the point P where they intersect must lie on both lines. Let's call its position vector p. The point P can only exist if there are values of s and t such that
p = 2a+0b-c+ta-tb-2tc = 0a+2b+c+2sa-2sb+0sc
For this equation to have a solution, the components in the a, b and c directions must each seperately be equal. This means that
1) 2a+ta = 0a+2sa gives 2+t = 2s
2) 0b-tb = 2b-2sb gives -t = 2-2s
3) -c-2tc = c+0sc gives -1-2t = 1
equation 3 gives t = -1 and substituting for t in equation 2 gives 1 = 2-2s and s = 1/2
THE LINES ONLY MEET if these values of s and t also fit equation (1)
2-1 = 2*(1/2) gives 1 = 1
now substituting for t=-1 in equation A gives
p = 2a+0b-c-a+b+2c gives a+b+c
now substituting for s=1/2 in equation B gives
p = 0a+2b+c+a-b = a+b+c
we see that the position vector p is the same for both lines, thus the lines intersect at p = a+b+c
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