Equation of Plane in Different Forms:
- General equation of a plane is ax + by + cz + d = 0
- Equation of the plane in Normal form is lx + my + nz = p where p is the length of the normal from the origin to the plane and (l, m, n) be the direction cosines of the normal.
- The equation to the plane passing through P(x1, y1, z1) and having direction ratios (a, b, c) for its normal is a(x – x1) + b(y – y1) + c (z – z1) = 0
- The equation of the plane passing through three non-collinear points (x1, y1, z1), (x2, y2, z2) and (x3, y3 , z3) is
- The equation of the plane whose intercepts are a, b, c on the x, y, z axes respectively is
x/a + y/b + z/c = 1 (a b c ≠ 0)
- Equation of YZ plane is x = 0, equation of plane parallel to YZ plane is x = d.
- Equation of ZX plane is y = 0, equation of plane parallel to ZX plane is y = d.
- Equation of XY plane is z = 0, equation of plane parallel to XY plane is z = d.
- Four points namely A (x1, y1, z1), B (x2, y2, z2), C (x3, y3, z3) and D (x4, y4, z4) will be coplanar if one point lies on the plane passing through other three points.
Example-1:
Find the equation to the plane passing through the point (2, -1, 3) which is the foot of the perpendicular drawn from the origin to the plane.
Solution:
The direction ratios of the normal to the plane are 2, -1, 3.
The equation of required plane is 2(x –2) –1 (y + 1) + 3 (z –3) = 0
⇒ 2x – y + 3z –14 = 0
Angle between the Planes
Angle between the planes is defined as angle between normals of the planes drawn from any point to the planes.
Angle between the planes a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 is
Note:
If a1a2 +b1b2 +c1c2 = 0, then the planes are perpendicular to each other.
If a1/a2 = b1/b2 = c1/c2 then the planes are parallel to each other.
Example-2:
Find angle between the planes 2x – y + z = 11 and x + y + 2z = 3.
Solution:
Example-3:
Find the equation of the plane passing through (2, 3, –4), (1, –1, 3) and parallel to x-axis.
Solution:
The equation of the plane passing through (2, 3, –4) is
a(x – 2) + b(y – 3) + c(z + 4) = 0 …… (1)
Since (1, –1, 3) lie on it, we have
a + 4b – 7c = 0 …… (2)
Since required plane is parallel to x-axis i.e. perpendicular to YZ plane i.e.
1.a + 0.b + 0.c = 0 Þ a = 0 Þ 4b – 7c = 0 => b/7 = c/4
∴ Equation of required plane is 7y + 4z = 5.
Perpendicular Distance:
The length of the perpendicular from the point P(x1, y1, z1) to the plane ax + by + cz + d = 0 is
|ax1+ by1 + cz1 + d / √(a2 + b2 + c2)|.
Family of Planes:
Equation of plane passing through the line of intersection of two planes u = 0 and v = 0 is
u + λv = 0
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