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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    = 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   . 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    p . If 1 is the position vector of F, then  for some t1R, and      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  = − 3ˆ+ 2ˆ

We know that the perpendicular distance from the given point with position vector  to the planer 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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