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MathsHardClass 12

Shortest Distance + Skew Lines

3D Geometry

65

JEE Qs

25%

Hard

75

min

Master the vector derivation for shortest distance; it clarifies the underlying geometry and helps solve complex problems involving common perpendiculars and relative positions of lines.

🧮 Key Formulas

Shortest distance (d) between two skew lines r1 = a1 + Ī»b1 and r2 = a2 + μb2 is: d = |(a2 - a1) ā‹… (b1 x b2)| / |b1 x b2|
Shortest distance (d) between two skew lines (x-x1)/l1 = (y-y1)/m1 = (z-z1)/n1 and (x-x2)/l2 = (y-y2)/m2 = (z-z2)/n2 is: d = |det([[x2-x1, y2-y1, z2-z1], [l1, m1, n1], [l2, m2, n2]])| / sqrt(((m1n2-m2n1)^2 + (n1l2-n2l1)^2 + (l1m2-l2m1)^2))
Shortest distance (d) between two parallel lines r1 = a1 + λb and r2 = a2 + μb is: d = |(a2 - a1) x b| / |b|

āœ… Key Points for JEE

  • 1Skew lines are lines that are neither parallel nor intersecting; they exist only in 3D space.
  • 2The shortest distance between two skew lines is the length of the unique common perpendicular segment connecting them.
  • 3The vector representing the shortest distance is perpendicular to the direction vectors of both lines (b1 and b2), hence it is parallel to b1 x b2.
  • 4If the scalar triple product (a2 - a1) ā‹… (b1 x b2) = 0, the lines are coplanar (either intersecting or parallel). If b1 x b2 is also 0, the lines are parallel. If b1 x b2 is not 0, the lines are intersecting, and the shortest distance is 0.

āš ļø Common Mistakes

  • āœ•Mistaking parallel or intersecting lines for skew lines, or vice-versa, and applying the wrong formula.
  • āœ•Errors in calculating the cross product (b1 x b2) or the dot product, leading to incorrect numerator or denominator values.
  • āœ•Sign errors or arithmetic mistakes when evaluating the determinant in the Cartesian formula, which is critical.
  • āœ•Forgetting to take the absolute value of the scalar result, as distance must be non-negative.

šŸ“ Practice Questions

See all

Q4. Let P be the foot of the perpendicular from the point (1, 2, 2) on the line L : xāˆ’11 = y+1āˆ’1 = zāˆ’22 . Let the line →r = (āˆ’^i + ^j āˆ’2^k) + Ī»(^i āˆ’^j + ^k), Ī» ∈R, intersect the line L at Q . Then 2(PQ)2 is equal to : (1) 25 (2) 19 (3) 29 (4) 27

2025Ā·MCQMedium

Q16.Let a straight line L pass through the point P(2, āˆ’1, 3) and be perpendicular to the lines xāˆ’12 = y+11 = zāˆ’3āˆ’2 and xāˆ’3 1 = yāˆ’23 = z+24 . If the line L intersects the yz -plane at the point Q , then the distance between the points P and Q is : (1) √10 (2) 2√3 (3) 2 (4) 3

2025Ā·MCQMedium

Q8. Let L1 : xāˆ’12 = yāˆ’23 = zāˆ’34 and L2 : xāˆ’23 = yāˆ’44 = zāˆ’55 be two lines. Then which of the following points lies on the line of the shortest distance between L1 and L2 ? (1) ( 143 , āˆ’3, 223 ) (2) (āˆ’53 , āˆ’7, 1) (3) (2, 3, 13 ) (4) ( 83 , āˆ’1, 13 )

2025Ā·MCQHard

Q25.Let L1 : xāˆ’13 = yāˆ’1āˆ’1 = z+10 and L2 : xāˆ’22 = 0y = z+4α , α ∈R, be two lines, which intersect at the point B. If P is the foot of perpendicular from the point A(1, 1, āˆ’1) on L2 , then the value of 26α( PB)2 is _________

2025Ā·NumericalMedium

Q14.The perpendicular distance, of the line xāˆ’1 2 = āˆ’1 = z+32 from the point P(2, āˆ’10, 1), is : (1) 6 (2) 5√2 (3) 4√3 (4) 3√5

2025Ā·MCQMedium

Q3. Let the position vectors of the vertices A, B and C of a tetrahedron ABCD be ^i + 2^j + ^k,^i + 3^j āˆ’2^k and 2^i + ^j āˆ’^k respectively. The altitude from the vertex D to the opposite face ABC meets the median line segment through A of the triangle ABC at the point E . If the length of AD is √110 and the volume of the 3 tetrahedron is √805 , then the position vector of E is 6√2 (1) 12 1 (7^i + 4^j + 3^k) (2) 12 (^i + 4^j + 7^k) (3) 1 6 (12^i + 12^j + ^k) (4) 16 (7^i + 12^j + ^k)

2025Ā·Multi conceptHard

NCERT Chapters

  • Class 12 Maths Ch 11: Three Dimensional Geometry