# Find the Shortest Distance Between the Lines r=(4i-j)+λ(i+2j-3k) and r=(i-j+2k)+μ(i+4j-5k) - Mathematics and Statistics

Sum

Find the shortest distance between the lines

bar r = (4 hat i - hat j) + lambda(hat i + 2 hat j - 3 hat k)

and

bar r = (hat i - hat j + 2 hat k) + mu(hat i + 4 hat j -5 hat k)

where λ and μ are parameters

#### Solution

Equation of lines are.,

bar r = (4 bar"i"-bar "j") + lambda(bar"i" +2bar"j" -3bar"k") &

bar r = (hat i - hat j + 2 hat k) + mu(hat i + 4 hat j -5 hat k)

∴ above lines passes through

bar"a"_1 = (4 bar"i" - bar"j") "and" bar"a"_2 = (bar "i" - bar"j" + 2bar"k")

and parallel to

bar"b"_1 = bar"i" + 2bar"j" - 3bar"k"  &  bar"b"_2 = bar"i" - 4bar"j" -  5bar"k"

Shortest distance =|((bar"a"_2 - bar"a"_1).(bar"b"_1 xx bar"b"_2))/|(bar"b"_1 xx bar"b"_2)||

=> bar"a"_2 - bar"a"_1 = -3bar"j" + 2bar"k"

bar"b"_1 xx bar"b"_2 = |(bar"i",bar"j" , bar"k"),(1,2,-3),(1,4,-5)| = 2bar"i" +2bar"j" + 2bar"k"

therefore |bar"b"_1 xx bar"b"_2| = 2sqrt3

Shortest distance = |((-3bar"i"+2bar"k").(2bar"i" + 2bar"j" + 2bar"k"))/(2sqrt3)|

= |(-6+4)/(2sqrt3)|

= |-2/(2sqrt3)|

"d" = 1/sqrt3 "units"

Concept: Shortest Distance Between Two Lines
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2015-2016 (March)

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