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प्रश्न
Find the distance between the parallel lines `x/2 = y/(-1) = z/2` and `(x - 1)/2 = (y - 1)/(-1) = (z - 1)/2`
उत्तर
Line `x/2 = y/(-1) = z/2` passes through (0, 0, 0) and has direction ratios 2, –1, 2
∴ Vector equation of the line is
r = `(0hat"i" + 0hat"j" + 0hat"k") + lambda(2hat"i" - hat"j" + 2hat"k")`
i.e., r =`lambda(2hat"i" - hat"j" + 2hat"k")`
Line `(x - 1)/2 = (y - 1)/(-1) = (z - 1)/2` passes through (1, 1, 1) and has direction ratios 2, –1, 2.
∴ Vector equation of the line is
r = `(hat"i" + hat"j" + hat"k") + lambda(2hat"i" - hat"j" + 2hat"k")`
The distance between parallel lines `bar"r" = bar"a"_1 + lambdabar"b"` and `bar"r" = bar"a"_2 + lambdabar"b"` is `|(bar"a"_2 - bar"a"_1) xx hat"b"|`
Here, `bar"a"_1 = 0, bar"a"_2 = hat"i" + hat"j" + hat"k", bar"b" = 2hat"i"- hat"j" + 2hat"k"`
∴ `bar"b" = (bar"b")/|bar"b"|`
= `(2hat"i" - hat"j" + 2hat"k")/sqrt(2^2 + (-1)^2 + 2^2)`
= `2/3hat"i" - 1/3hat"j" + 2/3hat"k"`
∴ `(bar"a"_2 - bar"a"_1) xx hat"b" = |(hat"i", hat"j", hat"k"),(1, 1, 1),(2/3, (-1)/3, 2/3)|`
= `hat"i"(2/3 + 1/3) - hat"j"(2/3 - 2/3) + hat"k"((-1)/3 - 2/3)`
= `hat"i" - hat"k"`
∴ `|(bar"a"_2 - bar"a"_1) xx hat"b"| = sqrt((1)^2 + (-1)^2`
= `sqrt(2)`
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