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MathsMediumClass 12

Family of Planes

3D Geometry

16

JEE Qs

8%

Hard

60

min

Always clearly define the conditions given to accurately determine the value of the parameter λ, which is the key to solving family of planes problems.

🧮 Key Formulas

Equation of a plane passing through the intersection of planes P1: A1x + B1y + C1z + D1 = 0 and P2: A2x + B2y + C2z + D2 = 0 is P1 + λP2 = 0 => (A1x + B1y + C1z + D1) + λ(A2x + B2y + C2z + D2) = 0
Vector form: For planes r⋅n1 = d1 and r⋅n2 = d2, the family of planes is r⋅(n1 + λn2) = d1 + λd2

✅ Key Points for JEE

  • 1The equation P1 + λP2 = 0 represents any plane passing through the line of intersection of plane P1=0 and plane P2=0.
  • 2The parameter λ is a real number whose value is determined by an additional condition provided in the problem (e.g., passing through a specific point, parallel or perpendicular to another plane, at a certain distance from the origin).
  • 3The normal vector of the family of planes (P1 + λP2 = 0) is (A1 + λA2)i + (B1 + λB2)j + (C1 + λC2)k in Cartesian form, or n1 + λn2 in vector form. This is crucial for applying perpendicularity or parallelism conditions.
  • 4The equation P1 + λP2 = 0 represents all planes passing through the line of intersection except for P2 itself. If the required plane is P2, one must deduce it from the conditions or use an alternative method.

⚠️ Common Mistakes

  • Incorrectly determining the value of λ by misapplying conditions for parallelism, perpendicularity, or passing through a point.
  • Making algebraic errors when expanding the family of planes equation or solving for λ.
  • Forgetting to use the vector form for problems involving dot products or cross products with normal vectors, or mixing Cartesian and vector components incorrectly.

📝 Practice Questions

See all

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

2025·MCQMedium

Q16.Let a straight line L pass through the point P(2, −1, 3) and be perpendicular to the lines x−12 = y+11 = z−3−2 and x−3 1 = y−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

2025·MCQMedium

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

2025·MCQHard

Q25.Let L1 : x−13 = y−1−1 = z+10 and L2 : x−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, −1) on L2 , then the value of 26α( PB)2 is _________

2025·NumericalMedium

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

2025·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 −2^k and 2^i + ^j −^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)

2025·Multi conceptHard

NCERT Chapters

  • Class 12 Mathematics Ch 11: Three Dimensional Geometry