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

• 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