    Question

# If equations of two non intersecting lines are - x−x1l1=y−y1m1=z−z1n1 & x−x2l2=y−y2m2=z−z2n2 then the shortest distance between them will be -

A
1Σ(l1m2l2m1)2∣ ∣x1y1z1l1m1n1l2m2n2∣ ∣
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B
1Σ(l1m2l2m1)2∣ ∣x2x1y2y1z2z1l1m1n1l2m2n2∣ ∣
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C
1Σ(l1m2l2m1)2∣ ∣x2y2z2l1m1n1l2m2n2∣ ∣
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D
None of these
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Solution

## The correct option is B 1√Σ(l1m2−l2m1)2∣∣ ∣∣x2−x1y2−y1z2−z1l1m1n1l2m2n2∣∣ ∣∣Let’s rewrite the first line’s equation - x−x1l1=y−y1m1=z−z1n1=r1 (say) Any point on first line( let’s call it “P” ) can be written as (x1+l1.r1,y1+m1.r1,z1+n1.r1) Similarly equation of second line can be rewritten as - x−x2l2=y−y2m2=z−z2n2=r2 (say) And any point on second line( let’s call it “Q” )can be written as (x2+l2.r2,y2+m2.r2,z2+n2.r2) Direction ratios of PQ will be - (x1+l1.r1−(x2+l2.r2),y1+m1.r1−(y2+m2.r2),z1+n1.r1−(z2+n2.r2)) This line will be the shortest line only if it’s perpendicular to both the given lines. PQ is perpendicular to the first line then - (x1+l1.r1−(x2+l2.r2)).(l1)+(y1+m1.r1−(y2+m2.r2))(m1)(z1+n1.r1−(z2+n2.r2)).(n1)=0 Similarly, PQ will be perpendicular to the second line - (x1+l1.r1−(x2+l2.r2)).(l2)+(y1+m1.r1−(y2+m2.r2))(m2)(z1+n1.r1−(z2+n2.r2)).(n2)=0 On solving these equations we get - 1√Σ(l1m2−l2m1)2∣∣ ∣∣x2−x1y2−y1z2−z1l1m1n1l2m2n2∣∣ ∣∣  Suggest Corrections  0      Similar questions  Explore more