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MathsMediumClass 12

Equation of Line — Symmetric, parametric form

3D Geometry

16

JEE Qs

8%

Hard

75

min

Master the parametric form to efficiently represent any point on a line, simplifying problems involving intersections, distances, and conditions on points.

🧮 Key Formulas

Symmetric form of a line passing through (x1, y1, z1) with direction ratios a, b, c: (x - x1)/a = (y - y1)/b = (z - z1)/c
Parametric form of a line: x = x1 + λa, y = y1 + λb, z = z1 + λc (where λ is the parameter)
Vector form of a line: r_vec = a_vec + λb_vec (where a_vec is the position vector of a point on the line and b_vec is the direction vector of the line)
Symmetric form of a line passing through two points (x1, y1, z1) and (x2, y2, z2): (x - x1)/(x2 - x1) = (y - y1)/(y2 - y1) = (z - z1)/(z2 - z1)

āœ… Key Points for JEE

  • 1The parametric form (x1+Ī»a, y1+Ī»b, z1+Ī»c) allows representing any arbitrary point on the line, which is critical for solving problems involving distances, intersections, or specific conditions.
  • 2The parameter Ī» in the parametric form represents a scaled distance along the line from the initial point (x1, y1, z1); if a, b, c are direction cosines, then |Ī»| is the actual distance.
  • 3To convert between vector form r_vec = a_vec + Ī»b_vec and Cartesian symmetric/parametric forms: a_vec gives the fixed point's coordinates (x1, y1, z1), and b_vec gives the direction ratios (a, b, c).
  • 4If any direction ratio (a, b, or c) is zero, the symmetric form cannot have that term in the denominator. Instead, that numerator term is set to zero (e.g., if a=0, then x - x1 = 0 and (y - y1)/b = (z - z1)/c).
  • 5Direction ratios (a, b, c) define the direction of the line, while direction cosines (l, m, n) are normalized direction ratios where l^2 + m^2 + n^2 = 1. Either can be used, but direction cosines simplify distance calculations.

āš ļø Common Mistakes

  • āœ•Incorrectly writing the symmetric form when one or more direction ratios are zero (e.g., dividing by zero instead of equating the numerator to zero).
  • āœ•Confusing the coordinates of the fixed point (x1, y1, z1) with the direction ratios (a, b, c) while applying the formulas.
  • āœ•Failing to understand that Ī» is a general parameter, not necessarily the distance from the fixed point unless the direction vector is a unit vector (i.e., a, b, c are direction cosines).
  • āœ•Algebraic errors when substituting the parametric point coordinates into equations of other geometric entities (e.g., planes, spheres) to find intersection points.

šŸ“ Practice Questions

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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