Distance of a point from a plane (a) Vector form of equation (b) Cartesian form of equation

**Distance
of a point from a plane**

The perpendicular distance from a point with position vector * *to
the plane * *â‹… * *= *p
*is given by

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 t_{1} âˆˆ **R,** and * *â‹… * *= *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

In Caretesian form if *A*( *x*_{1} , *y*_{1}
, *z*_{1} ) is the given point with position vector * *and
*ax *+ *by *+ *cz *= *p *is the Cartesian equation of the
given plane, then * *= *x*_{1}Ë†*i *+
*y*_{1}Ë† *j *+ *z*_{1}Ë†*k *and *n *=
*a*Ë†*i *+ *b*Ë†*j *+ *c*Ë†*k*.
Therefore, using these vectors in we get the
perpendicular distance from a point to the plane in Cartesian form as

The perpendicular distance from the origin to the plane *ax *+
*by *+ *cz *+ *d *= 0 is given by

Find the distance of a point (2, 5, âˆ’3) from the plane * *â‹… (6Ë†*i *âˆ’ 3Ë†*j *+ 2Ë†*k *) = 5 .

Comparing the given equation of the plane with â‹…= *p* we have = 6Ë†*i *âˆ’ 3Ë†*j *+ 2Ë†*k *

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

Find the distance of the point (5, âˆ’5, âˆ’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 *âˆ’ *y *+ *z *= 5 .

The Cartesian equation of the straight line joining *A *and
*B *is

Therefore, an arbitrary point on the straight line is of the form
(3*t *+ 4, 4*t *+1, 2*t *+ 2) . To find the point of
intersection of the straight line and the plane, we substitute *x *= 3*t
*+ 4, *y *= 4*t *+1, *z *= 2*t *+ 2 in *x *âˆ’ *y *+ *z *=
5 , and we get *t *= 0 . Therefore,the point of intersection of the
straight line is (2, âˆ’1, 2) .

Now, the distance between the two points (2, âˆ’1, 2) and (5, âˆ’5,
âˆ’10) is

Tags : Definition, Theorem, Proof, Solved Example Problems, Solution , 12th Mathematics : UNIT 6 : Applications of Vector Algebra

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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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