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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).
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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