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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, 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 it vector form is `vecr = (-hatj - hatk) + lambda(4hati + 6hatj + 2hatk)`
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 `vecr = (3hati + 9hatj + 4hatk) + mu(-7hati - 5hatj)`
Now here `veca_1 = -hatj - hatk, vecb_1 - 4hati + 6hatj + 2hatk`
`veca_2 = 3hati + 9hatj + 4hatk, vecb_2 = -7hati - 5hatj`
Shortest distance between AB and CD
S.D. = `|((veca_2 - veca_1)*(vecb_1 xx vecb_2))/|vecb_1 xx vecb_2||`
`veca_2 - veca_1 = (3hati + 9hatj + 4hatk) - (-hatj - hatk) = 3hati + 10hatj + 5hatk`
`vecb_1 xx vecb_2 = |(hati, hatj, hatk),(4, 6, 2),(-7, -5, 0)|`
= `hati(0 + 10) - hatj(0 + 14) + hatk(-20 + 42)`
= `10hati - 14hatj + 22hatk`
`|vecb_1 xx vecb_2| = sqrt((10)^2 + (-14)^2 + (22)^2)`
= `sqrt(100 + 196 + 484)`
= `sqrt(780)`
∴ S.D. = `((3hati + 10hatj + 5hatk)*(10hati - 14hatj + 22hatk))/sqrt(780)`
= `(30 - 140 + 100)/sqrt(780)`
= 0
Thus, the two lines intersect each other.
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