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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 (b) Cartesian form of equation

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‹… . 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   â‹…   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  â‹… (6ˆi − 3ˆj + 2ˆk ) = 5 .

Solution

Comparing the given equation of the plane with â‹…p we have  = 6ˆ− 3ˆ+ 2ˆ

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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12th Mathematics : UNIT 6 : Applications of Vector Algebra : Distance of a point from a plane | Definition, Theorem, Proof, Solved Example Problems, Solution

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12th Mathematics : UNIT 6 : Applications of Vector Algebra


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