Equation of the plane through the points (2, 1, -1) and (-1, 3, 2) and perpendicular to the plane x – 2y + 4z = 0 is given by

The four points (0, 4, 3) , (-1, -5, -3) , (-2, -2, 1) and (1, 1, -1) lie in the plane

The equation of a plane and condition of two planes being parallel or perpendicular.
A) Every equation of first degree in x, y, z, i. e., Ax + By + Cz + D = 0 represents a plane. The coefficients of x, y, z are the direction ratios of the normal to the plane.
B) Angle between two planes is equal to the angle between the normals to the planes.
\ cos q =
Planes are perpendicular if å A₁ A₂ = 0 and parallel if
C) Planes parallel to co – ordinate planes are x = l, y = l or z = l.
Planes perpendicular to co – ordinate planes x = 0, y = 0, z = 0 are
by + cz + d = 0 (x missing), ax + cz + d = 0
(y missing), ax + by + d = 0 (z missing) Find the equation of the plane through the intersection of the planes x + 2y + 3z – 4 = 0 and 2x + y – z + 5 = 0 and perpendicular to the plane 5x + 3y + 6z + 8 = 0 is

The direction cosines of the line joining the points (4, 3, - 5) and (-2, 1, -8) are

For such questions, we should know formula to find direction cosines.

The locus of x² + y² + z² = 0 is

For such questions, we have to be careful about points satisfying the equation.

Let f (x) be a quadratic expression which is positive for all real x. If g(x) = f (x) + f ¢ (x) + f ¢¢ (x), then for any real x

=

For such questions, we should know different the condition of limit to exist.