Definition, Theorem, Proof, Solved Example Problems, Solution - Distance of a point from a plane | 12th Mathematics : UNIT 6 : Applications of Vector Algebra
Chapter: 12th Mathematics : UNIT 6 : Applications of Vector Algebra
Distance of a point from a plane
Distance
of a point from a plane
(a) Vector form of equation
Theorem 6.20
The perpendicular distance from a point with position vector 
to
the plane 
â‹… 
= p
is given by
Proof
Let A be the point whose position vector is .
Let F be the foot of the perpendicular from the point A
to the plane â‹…= p . The line joining F and A is parallel to the normal vector and hence its equation is = + t .
But F is the point of intersection of the line = +
t and the given plane â‹… = p . If 1 is the position vector of F, then 
for
some t1 ∈ R, and ⋅ = p Eliminating 1 we get
Therefore, the length of the perpendicular from the point A to the
given plane is
The position vector of the foot F of the perpendicular AF is given
by
(b) Cartesian form of equation
In Caretesian form if A( x1 , y1
, z1 ) is the given point with position vector and
ax + by + cz = p is the Cartesian equation of the
given plane, then = x1ˆi +
y1ˆ j + z1ˆk and n =
aˆi + bˆj + cˆk.
Therefore, using these vectors in 
we get the
perpendicular distance from a point to the plane in Cartesian form as
Remark
The perpendicular distance from the origin to the plane ax +
by + cz + d = 0 is given by
Example 6.49
Find the distance of a point (2, 5, −3) from the plane ⋅ (6ˆi − 3ˆj + 2ˆk ) = 5 .
Solution
Comparing the given equation of the plane with ⋅= p we have = 6ˆi − 3ˆj + 2ˆk
We know that the perpendicular distance from the given point
with position vector to the planer â‹…= p is given by 
.Therefore, substituting 
in the
formula, we get
Example 6.50
Find the distance of the point (5, −5, −10) from the point of
intersection of a straight line passing through the points A(4,1, 2) and
B (7, 5, 4) with the
plane x − y + z = 5 .
Solution
The Cartesian equation of the straight line joining A and
B is
Therefore, an arbitrary point on the straight line is of the form
(3t + 4, 4t +1, 2t + 2) . To find the point of
intersection of the straight line and the plane, we substitute x = 3t
+ 4, y = 4t +1, z = 2t + 2 in x − y + z =
5 , and we get t = 0 . Therefore,the point of intersection of the
straight line is (2, −1, 2) .
Now, the distance between the two points (2, −1, 2) and (5, −5,
−10) is
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