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

__Answer:-__

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.

__Answer:-__

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:

__Answer:-__

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?

__Answer:-__

Let the coordinates of the foot of the perpendicular P from the origin to the plane is (x_{1}, y_{1}, z_{1}).

Then, the direction ratios of the line OP are (x_{1}, y_{1}, z_{1}).

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

x_{1}/ (2/√29)) = (y_{1}/ -3/√29)) = (z_{1}/ (4/√29)) = k

i.e. x_{1}=(2k)/(√29), y_{1} =((-3k)/(√29), z_{1} =(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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