 # I want vector equations of a plane and line please send me the formulas

4 years ago
Dear student

## 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 = 0

• 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.

CXmhaU -8: 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.

hb: 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.

CXmhaU -9: Find angle between the planes 2x – y + z = 11 and x + y + 2z = 3.

hb:  CXmhaU -10: Find the equation of the plane passing through (2, 3, –4), (1, –1, 3) and parallel to x-axis.

hb: 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.

Intersection of a Line and Plane:

If equation of a plane is ax + by + cz + d = 0, then direction cosines of normal to this plane are a, b, c. So angle between normal to the plane and a straight line having direction cosines l, m ,n is given by cos θ = al + bm + cn / √a2 + b2 + c2.

Then angle between the plane and the straight line is π/2 – θ.

• Plane and straight line will be parallel if al + bm + cn = 0
• Plane and straight line will be perpendicular if a/l = b/m = c/n.