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Distance of a Point From a Plane is considered one the most difficult concept.
24 Questions around this concept.
The distance of the point (1, 3, −7) from the plane passing through the point (1, −1, −1), having normal perpendicular
to both the lines and
is:
If the points (1, 1, ) and (-3, 0, 1) are equidistant from the plane, 3x+4y-12z+13=0, then
satisfies the equation :
Distance between two parallel planes is :
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The distance of the point (1, −2, 4) from the plane passing through the point (1, 2, 2) and perpendicular to the planes x − y + 2z = 3 and 2x −2y + z + 12=0, is :
The perpendicular distance (D) from a point having position vector to the plane
is given by
Consider a point P with position vector and a plane π1 whose equation is
Let R be a point in the plane such that is orthogonal to the plane π1.
since, line PR passes through P(a) and is parallel to the vector which is normal to the plane π1. So, vector equation of line PR is
Point R is the intersection of Eq. (i) and the given plane π1.
On putting the value of λ in Eq. (i), we obtain the position vector of R given by
Cartesian Form
Let P(x1, y1, z1) be the given point with position vector and ax + by + cz + d = 0 be the Cartesian equation of the given plane. Then
Hence, from Vector form of the perpendicular from P to the plane is
Distance Between The Parallel Planes
The distance between the two parallel planes ax + by + cz + d1 = 0 and ax + by + cz + d2 = 0 is given by
Let P(x1, y1, z1) be any point in the plane ax + by + cz + d1 = 0.
Then the distance of the point P from plane ax + by + cz + d2 = 0 is
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