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Question
Prove that the line through A(0, – 1, – 1) and B(4, 5, 1) intersects the line through C(3, 9, 4) and D(– 4, 4, 4).
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Solution
Given points are A(0, – 1, – 1) and B(4, 5, 1) C(3, 9, 4) and D(– 4, 4, 4)
Cartesian form of equation AB is
`(x - 0)/(4 - 0) = (y + 1)/(5 + 1) = (z + 1)/(1 + 1)`
⇒ `x/4 = (y + 1)/6 = (z + 1)/2`
And its vector form is `vec"r" = (-hat"j" - hat"k") + lambda(4hat"i" + 6hat"j" + 2hat"k")`
Similarly, equation of CD is
`(x - 3)/(-4 - 3) = (y - 9)/(4 - 9) = (z - 4)/(4 - 4)`
⇒ `(x - 3)/(-7) = (y - 9)/(-5) = (z - 4)/0`
And its vector form is `vec"r" = (3hat"i" + 9hat"j" + 4hat"k") + mu(-7hat"i" - 5hat"j")`
Now, here `vec"a"_1 = -hat"j" - hat"k", vec"b"_1 = 4hat"i" + 6hat"j" + 2hat"k"`
`vec"a"_2 = 3hat"i" + 9hat"j" + 4hat"k", vec"b"_2 = -7hat"i" - 5hat"j"`
Shortest distance between AB and CD
S.D. = `|((vec"a"_2 - vec"a"_1)*(vec"b"_1 xx vec"b"_2))/|vec"b"_1 xx vec"b"_2||`
`vec"a"_2 - vec"a"_1 = (3hat"i" + 9hat"j" + 4hat"k") - (-hat"j" - hat"k")`
= `3hat"i" + 10hat"j" + 5hat"k"`
`vec"a" xx vec"b"_2 = |(hat"i", hat"j", hat"k"),(4, 6, 2),(-7, -5, 0)|`
= `hat"i"(0 + 10) - hat"j"(0 + 14) + hat"k"(-20 + 42)`
= `10hat"i" - 14hat"j" + 22hat"k"`
`|vec"b"_1 xx vec"b"_2| = sqrt((10)^2 + (-14)62 + (22)^2)`
= `sqrt(100 + 196 + 484)`
= `sqrt(780)`
∴ S.D. = `((3hat"i" + 10hat"j" + 5hat"k")*(10hat"i" - 14hat"j" + 22hat"k"))/sqrt(780)`
= `(30 - 140 + 110)/sqrt(780)`
= 0
Hence, the two lines intersect each other.
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