Q.Find the parametric vector, non-parametric vector and Cartesian form of the equations of the plane passing through the three non-collinear points (3,6,−2),(−1,−2,6), and (6,4,−2).
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Equation of a Plane
Equation of a Plane
A plane is fixed by a point on it and a direction perpendicular to it (its normal n). Every standard form below is really that one idea written differently.
Point + normal form
If the plane passes through a with normal n, then for any point r on it, r−a lies in the plane, so it is perpendicular to n:
(r−a)⋅n=0⟺r⋅n=a⋅n.
In Cartesian form with n=(A,B,C): A(x−x1)+B(y−y1)+C(z−z1)=0, i.e. Ax+By+Cz=d. The coefficients of x,y,z are the normal's direction ratios.
Normal (perpendicular) form
If n^ is the unit normal and the plane is at distance p from the origin: r⋅n^=p, i.e. lx+my+nz=p with l2+m2+n2=1.
Intercept form …
Normal =(b−a)×(c−a), where a,b,c are the three given points. …
Take one point as base, form the two edge vectors to the other two points, cross them for the normal, then write all three requested forms of the plane's equation.
Step 1. Base point and edge vectors. A(3,6,−2),B(−1,−2,6),C(6,4,−2): b−a=(−4,−8,8), c−a=(3,−2,0).
Step 2. Parametric vector equation.
r=(3i^+6j^−2k^)+s(−4i^−8j^+8k^)+t(3i^−2j^).
Step 3. Normal =(b−a)×(c−a).
i^−43j^−8−2k^80=i^(0+16)−j^(0−24)+k^(8+24)=16i^+24j^+32k^.
Simplify (divide by 8): normal ∝(2,3,4). …
Two edge vectors from a common point, cross for the normal, …
- Sign slip on the middle cofactor of the normal cross product …
Showing the 12 most recent of 28 on this concept.
- CBSE 2026Set ANNUAL1 markMCQQ.The equation of the plane having intercepts 2, −3, 1 on X, Y and Z axis respectively is ................. .(a) 2x − 3y + z = 0(b) 3x − 2y + 6z = 6(c) 3x + 2y + z = 3(d) x + 2y + 3z = 2
›Reveal solutionSolution
Use the intercept form of a plane: x/a+y/b+z/c=1.
With a=2,b=−3,c=1:
2x+−3y+1z=1
…
- CBSE 2025Set E1 markMCQQ.Equation of a plane parallel to the plane 9x−8y+7z=10 is(a) 9x−8y−7z=5(b) 9x−8y+7z=5(c) 9x+8y+7z=5(d) 9x−y+7z=5
›Reveal solutionSolution
Parallel planes have identical coefficients of x,y,z (same normal), only the constant differs.
The plane 9x−8y+7z=10 has normal direction (9,−8,7). Any parallel plane must have the same normal, i.e. the same coefficients of x,y,z, differi …
- CBSE 2024Set D1 markMCQQ.The equation of the xy-plane is(a) x=0(b) y=0(c) z=0(d) none of these
›Reveal solutionSolution
Points on the xy-plane have z=0.
The xy-plane consists of all points (x,y,0); the defining condition is that the z-coordinate vanishes. Hence its equation is z=0. (Simila …
- CBSE 2024Set D1 markMCQQ.The equation of the plane parallel to the plane 3x−5y+4z=11 is(a) 3x−5y+4z=21(b) 3x+5y+4z=25(c) 3x+5y+4z=35(d) none of these
›Reveal solutionSolution
Parallel planes share the same normal, so the coefficients of x,y,z must be identical: 3x−5y+4z=21.
Two planes are parallel iff their normal vectors are proportional. The given plane 3x−5y+4z=11 has normal (3,−5,4). A parallel plane must therefore have the form
3x−5y+4z=k …
- CBSE 2024Set ANNUAL1 markMCQQ.The Cartesian equation of the plane r⃗ · (î + ĵ − k̂) = 2 is :(a) x − y − z = 2(b) x + y − z = 2(c) x + y + z = 2(d) x + y − z = −2
›Reveal solutionSolution
Writing r⃗ = xî + yĵ + zk̂ and dotting with (î + ĵ − k̂) directly gives the Cartesian form x + y − z = 2.
The vector equation of a plane is r⋅n=d, where r=x^+y^+zk^ is the position vector of a general point and n is the plane's normal.
Here n=^+^−k^ and d=2.
…
- CBSE 2023Set ANNUAL1 markMCQQ.The Cartesian equation of the plane r⋅(i^+j^−k^)=2 is-(a) x+y−z=0(b) x+y−z=2(c) x+y−z=1(d) x+y+z+2=0
›Reveal solutionSolution
Substitute r=xi^+yj^+zk^ into the vector equation of the plane and take the dot product.
Given r⋅(i^+j^−k^)=2, with r=xi^+yj^+zk^:
…
- CBSE 2023Set E1 markMCQQ.Direction ratios of the normal to the plane x+2y−3z+15=0 are(a) 1,2,3(b) 1,−2,3(c) 1,2,−3(d) 1,2,15
›Reveal solutionSolution
The normal to ax+by+cz+d=0 has direction ratios a,b,c, i.e. 1,2,−3.
For a plane written as ax+by+cz+d=0, the normal vector is ai+bj+ck, so its direction ratios are the coe …
- CBSE 2023Set E1 markMCQQ.Equation of a plane parallel to the plane x−8y−9z=12 is(a) x+8y+9z=12(b) x−8y−9z=2023(c) 8x−y−9z=12(d) x−9y−8z=12
›Reveal solutionSolution
A plane parallel to x−8y−9z=12 keeps the coefficients 1,−8,−9; only the constant changes, e.g. x−8y−9z=2023.
Two planes are parallel iff their normal vectors are proportional, i.e. the coefficients of x,y,z are the same (up to a common factor). Only the constant term may differ.
…
- CBSE 2023Set ANNUAL1 markMCQQ.The equation of the plane with intercepts of 2, 3 and 4 on the x,y and z-axes respectively is:(a) 4x+6y+3z=12(b) 6x+4y+3z=12(c) 3x+4y+6z=12(d) 5x+4y+3z=0
›Reveal solutionSolution
Use the intercept form of a plane, ax+by+cz=1, then clear denominators.
With intercepts a=2, b=3, c=4:
2x+3y+4z=1
…
- CBSE 2023Set ANNUAL1 markQ.Find the intercepts cut off by the plane 2x+y−z=5 on co-ordinate axes.
›Reveal solutionSolution
Rewrite the plane equation in intercept form ax+by+cz=1 by dividing through so the RHS becomes 1.
2x+y−z=5
Divide both sides by 5:
5/2x+5y+−5z=1
…
- CBSE 2022Set HE2191 markQ.Write true or false: Equation of a plane in normal form is lx+my+nz=d.
›Reveal solutionSolution
This is exactly the standard normal (or perpendicular) form of the equation of a plane.
The equation of a plane in normal form is lx+my+nz=d, where (l,m,n) are the direction cosines of the normal to the plane from the origin, and d (≥0) is th …
- CBSE 2022Set HE2191 markQ.Write true or false: The planes 2x−y+4z=5 and 5x−2.5y+10z=6 are parallel.
›Reveal solutionSolution
Two planes are parallel if their normal vectors are proportional; check the ratio of coefficients.
Plane 1: 2x−y+4z=5, normal (2,−1,4).
Plane 2: 5x−2.5y+10z=6, normal (5,−2.5,10).
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