Let
be the vector normal to the plane and
be the position vector of the point through which the plane passes. Then find the equation of plane.
Asked by Topperlearning User
| 4th Jun, 2014,
01:23: PM
Expert Answer:
Let P be a arbitrary point on the plane with position vector .


Thus, equation of the plane with
as a normal vector and
as a position vector of the point through which the plane passes is



Answered by
| 4th Jun, 2014,
03:23: PM
Concept Videos
Q 48
- If
is the normal from the origin to the plane, and
is the unit vector along
. P(x, y, z) be any point on the plane and
is perpendicular to
. Find the equation of plane in the normal form.
- Find the equation of the plane, which is at a distance of 5 unit from the origin and has
as a normal vector.
- Find the equation of the plane whose distance from the origin is 8 units and the direction ratios of the normal are 6, – 3, – 2.
- Find the equation of the plane is
. Find the perpendicular distance of the plane from the origin.
- Write the normal and Cartesian form of the plane
.
- Find the equation of the plane passing through the points (–1, 4, – 3), (3, 2, – 5) and (– 3, 8, – 5).
- Find the equation of the plane passing through the point (3, – 3, 1) and perpendicular to the line joining the points (3, 4, – 1) and (2, – 1, 5).
- Find the co-ordinates of the foot of the perpendicular drawn from the origin to the plane 2x – 3y + 6z + 7 = 0.
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