Question

Find the Cartesian equation of the following planes: (a) (b) (c)

Answer 1

The vector form of the equation of plane is expressed as,

r → ⋅ n ^ =d(1)

Here,

r → is the position vector.

n ^ is the normal unit vector of the plane.

d is the distance of the plane from origin.

The Cartesian equation of plane in normal form is expressed as,

lx+my+nz=d (2)

Comparing equation (1) and (2), we get,

r → ⋅ n ^ =lx+my+nz(3)

The position vector is a general vector expressed as,

r → =x i ^ +y j ^ +z k ^ (4)

The normal vector is,

n ^ = n x i ^ + n y j ^ + n z k ^ (5)

Substitute equation (4) in equation (3).

( x i ^ +y j ^ +z k ^ )⋅( n x i ^ + n y j ^ + n z k ^ )=lx+my+nz(6)

Comparing both sides of the equation (6),

n x =l

n y =m

n z =n

Thus,

n ^ =l i ^ +m j ^ +n k ^ (7)

(a)

The given equation is,

r → ⋅( i ^ + j ^ − k ^ )=2(8)

Compare equation (8) with equation (1).

n ^ = i ^ + j ^ − k ^ (9)

Compare equation (7) and (9).

l=1

m=1

n=−1

Also,

d=2

Substitute the values of l,m,n and d in equation (2).

x+y−z=2

Hence, the Cartesian equation of the plane is x+y−z=2.

(b)

The given equation is,

r → ⋅( 2 i ^ +3 j ^ −4 k ^ )=1(10)

Compare equation (10) with equation (1).

n ^ =2 i ^ +3 j ^ −4 k ^ (11)

Compare equation (7) and (11).

l=2

m=3

n=−4

Also,

d=1

Substitute the values of l,m,n and d in equation (2).

2x+3y−4z=1

Hence, the Cartesian equation of the plane is 2x+3y−4z=1.

(c)

The given equation is,

r → ⋅( ( s−2t ) i ^ +( 3−t ) j ^ −( 2s+t ) k ^ )=15(12)

Compare equation (12) with equation (1).

n ^ =( s−2t ) i ^ +( 3−t ) j ^ −( 2s+t ) k ^ (13)

Compare equation (7) and (13).

l=s−2t

m=3−t

n=2s+t

Also,

d=15

Substitute the values of l,m,n and d in equation (2).

( s−2t )x+( 3−t )y−( 2s+t )z=15

Thus, the Cartesian equation of the plane is ( s−2t )x+( 3−t )y−( 2s+t )z=15.