MathsMediumClass 12

Equation of Plane — Normal form, intercept form

3D Geometry

JEE Qs

Hard

min

Master the conversion between general and normal forms of the plane, paying close attention to the sign of the constant term to ensure 'p' (distance from origin) is always positive.

🧮 Key Formulas

Normal Form (Vector): r . n_hat = p

Normal Form (Cartesian): lx + my + nz = p (where l, m, n are direction cosines of normal, p is perpendicular distance from origin, p > 0)

Intercept Form (Cartesian): x/a + y/b + z/c = 1 (where a, b, c are x, y, z intercepts respectively)

Conversion from General Form (Ax + By + Cz + D = 0) to Normal Form: Divide by sqrt(A^2 + B^2 + C^2) after ensuring constant term is positive on the RHS. If D is positive, divide by -sqrt(A^2 + B^2 + C^2) to make the RHS positive.

✅ Key Points for JEE

1The parameter 'p' in the normal form always represents the perpendicular distance of the plane from the origin, hence 'p' must always be a positive value.
2The coefficients (l, m, n) in the cartesian normal form are the direction cosines of the normal vector to the plane. If given a general equation Ax + By + Cz + D = 0, the vector (A, B, C) is a normal vector, and its direction cosines are A/sqrt(A^2+B^2+C^2), etc.
3The intercept form is particularly useful when the plane's intersections with the coordinate axes are directly given or are required as part of the problem.
4Converting between different forms of plane equations (especially general to normal) is a crucial skill for solving problems involving distance from the origin or properties related to the normal vector.

⚠️ Common Mistakes

✕Failing to ensure that the constant term 'p' in the normal form (r . n_hat = p or lx + my + nz = p) is always positive, often leading to sign errors when converting from the general equation.
✕Confusing the components of a normal vector (A, B, C from Ax+By+Cz+D=0) with the direction cosines (l, m, n), which require normalization (division by magnitude).
✕Incorrectly identifying intercepts when the plane equation is not directly in intercept form, or when an axis is parallel to the plane (implying infinite intercept).

📝 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 : 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 Maths Ch 11: Three Dimensional Geometry