Class 12 Maths Three Dimensional Geometry Plane

Plane

A plane is determined uniquely if any one of the following is known:

(i) The normal to the plane and its distance from the origin is given, i.e., equation ofa plane in normal form.

(ii) It passes through a point and is perpendicular to a given direction.

(iii) It passes through three given non collinear points.

Equation of a plane in normal form: Vector

Consider a plane whose distance from the origin is‘d’.

Equation of a plane in normal form: Cartesian form

Let P(x, y, z) be any point on the plane.

Note:- Equation (3) shows that if   r⃗ = xî+ yĵ+ zk̂ =d is the vector equation of the plane, then ax+by+cz=d is the Cartesian equation of the plane, where a,b and c are the direction ratios of the normal to the plane.

Problem:

Determine the direction cosines of the normal to the plane and the distance from origin:

x+y+z=1  ,  2.2x+3y-z=5

x + y + z = 1 ... (1)

The direction ratios of normal are 1, 1, and 1.

Therefore √ ((1)2+ (1)2+ (1)2)

Dividing both sides of equation (1) by √3, we obtain

(1/√3) x+ (1/√3) y+ (1/√3) z= (1/√3) … (2)

This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines ofnormal to the plane and d is the distance of normal from the origin.

Therefore, the direction cosines of the normal are (1/√3),(1/√3) and (1/√3) the distance ofnormal from the origin is (1/√3) units.

2x + 3y − z = 5 ... (1)

The direction ratios of normal are 2, 3, and −1.

Therefore, √ (2)2+ (3)2+ (-1)2 =√ (14)

Dividing both sides of equation (1) by √ (14), we obtain

(2/√ (14)) x+ (3/√ (14)) y– (1/√ (14)) z= (5/√ (14))

This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines ofnormal to the plane and d is the distance of normal from the origin.

Therefore, the direction cosines of the normal to the plane are (2/√ (14)),(3/√ (14)) and (-1/√ (14)) and the distance of normal from the origin is (5/√ (14)) units.

5y + 8 = 0

=> 0 x − 5y + 0z = 8................. (1)

The direction ratios of normal are (0, −5, and 0).

Therefore √ (0+ (-5)2+0) =5.

Dividing both sides of equation (1) by 5, we obtain

-y= (8/5)

This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines ofnormal to the plane and d is the distance of normal from the origin.

Therefore, the direction cosines of the normal to the plane are (0, −1, and 0) and thedistance of normal from the origin is (8/5) units.

Problem:-

Find the distance of the plane 2x-3y+4z-6=0 from the origin.

Since the direction ratios of the normal to the plane are 2,-3, 4; the direction cosines of it are

(2)/ (√ (2)2+ (-3)2+ (4)2),

(-3)/(√ (2)2+ (-3)2+ (4)2),

(4)/(√ (2)2+ (-3)2+ (4)2),

i.e. (2)/√ (29),

(-3)/√ (29),

(4)/√ (29)

Hence, dividing the equation (2x-3y+4z-6)=0 i.e. 2x-3y+4z=6 throughout by √ (29), we get

((2)/√ (29))x + ((-3)/√ (29))y + ((4)/√ (29)) z = (6)/√ (29)

This is the form lx+my+nz=d, where d is the distance of the plane from the origin.So, the distance of the plane from the origin is (6)/√ (29)

Problem:-

Find the Cartesian equation of the following planes:

It is given that equation of the plane is

r-> . (î+ ĵ-2k̂)=2   (1)

For any arbitrary point P(x, y, z) on the plane, position vector r⃗is given by,

r-> = xî+ yĵ+ zk̂

Substituting the value of r⃗ in equation (1), we obtain

(xî+ yĵ+ zk̂)(î+ ĵ-2k̂) =2

=>x+y-z=2.

This is the Cartesian equation of the plane.

r->. (2î+3 ĵ-4k̂)=1 (1)

For any arbitrary point P (x, y, z) on the plane, position vector r-> is given by,

r-> = xî+ yĵ+ zk̂

Substituting the value of r-> in equation (1), we obtain

(xî+ yĵ+ zk̂)(2î+3 ĵ-4k̂) = 1

=>2x+3y-4z=2.

This is the Cartesian equation of the plane.

It is given that equation of the plane is

r->. [(s-2t)î +(3-t)ĵ +(2s+t)k̂]   (1)

For any arbitrary point P (x, y, z) on the plane, position vector r-> is given by,

r-> = xî+ yĵ+ zk̂

Substituting the value of r⃗ in equation (1), we obtain

(xî+ yĵ+ zk̂) [(s-2t)î +(3-t)ĵ +(2s+t)k̂]

⇒ (s − 2t) x + (3 − t) y + (2s + t) z = 15

This is the Cartesian equation of the plane.

Problem:-

Find the coordinates of the foot of the perpendicular drawn from the origin to the plane 2x-3y+4z-6=0?

Let the coordinates of the foot of the perpendicular P from the origin to the plane is (x1, y1, z1).

Then, the direction ratios of the line OP are (x1, y1, z1).

Writing the equation of the plane in the normal form, we have

(2/√29) x-(3/√29) y+ (4/√29) z= (6/√29) where (2/√29), (3/√29) y, (4/√29) are the direction cosines of the OP.

Since d.c.’s and direction ratios of a line are proportional, we have

x1/ (2/√29)) = (y1/ -3/√29)) = (z1/ (4/√29)) = k

i.e. x1=(2k)/(√29), y1 =((-3k)/(√29), z1 =(4k)/(√29)

Substituting these in the equation of the plane, we get k = (6/√29)

Hence, the foot of the perpendicular is (12/29), (- 18/29), (24/29).

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