● 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 of a 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.

Now we shall find vector and Cartesian equations of the planes.

`=>` Consider a plane whose perpendicular distance from the origin is `d (d ≠ 0).` Fig.

`=>` If `bar (ON)` is the normal from the origin to the plane, and `hat n` is the unit normal vector along `bar (ON)` . Then `bar (ON) = d hat n`.

`=>` Let P be any point on the plane. Therefore,

`bar (NP)` is perpendicular to `bar (ON),` Therefore, `bar (ON) * bar (ON) = 0` ..........(1)

`=>` Let `vec r` be the position vector of the point P, then `bar (NP) = vec r -d hat n` (as `bar (ON) + bar (NP) = bar (OP)`)

Therefore, (1) becomes

` ( vec r - d hat n) * d hat n = 0`


or `( vec r - d hat n) * hat n =0` `( d ≠ 0)`

or `vec r * hat n - d hat n * hat n = 0`

i.e., `color{red}{vec r * hat n = d}` (as `hat n * hat n =1`) ..........(2)

This is the vector form of the equation of the plane.


`color{brown} "Cartesian form of the equation of the plane "`

`=>` Equation (2) gives the vector equation of a plane, where `hat n` is the unit vector normal to the plane. Let P(x, y, z) be any point on the plane. Then

`bar (OP) = vec r = x hat i + y hat j + z hat k`

`=>` Let `l, m, n` be the direction cosines of `hat n` . Then

`hat n = l hat i + m hat j + n hat k`

Therefore, (2) gives

`( x hat i + y hat j + z hat k) * (l hat i + m hat j + n hat k) =d`

i.e., `color{red}{lx+ my +nz = d}` .............(3)

`=>` This is the cartesian equation of the plane in the normal form.

`=>` In the space, there can be many planes that are perpendicular to the given vector, but through a given point `P(x_1, y_1, z_1)`, only one such plane exists (see Fig).

`=>` Let A plane pass through a point A with position vector`vec a` and perpendicular to the vector`bar N`

Let `vec r` be the position vector of any point P(x, y, z) in the plane. (Fig 11.13).

`=>` Then the point P lies in the plane if and only if

`bar (AP) ` is perpendicular to `bar (N)` , i.e., `bar (AP) *bar N =0`. But

`bar (AP) = vec r - vec a` . Therefore `(vec r - vec a) * vec N = 0` .......(1)

This is the vector equation of the plane.

`color{brown} "Cartesian form"`

`=>` Let the given point A be `(x_1, y_1, z_1)`, P be (x, y, z) and direction ratios of `bar N` are A, B and C. Then,

`vec a = x_1 hat i + y_1 hat j + z_1 hat k , vec r = x hat i + y hat j + z hat k` and `bar(N) = A hat i + B hat j + C hat k`

Now `(vec r - vec a) * vec N= 0`

So ` [ (x-x_1) hat i + (y-y_1) hat j + (z-z_1) hat k]* ( A hat i +B hat j + C hat k) =0`

i.e., `color{red}{A (x- x_1) +B(y- y_1) +C(z-z_1) = 0}`

`=>` Let `R, S` and `T` be three non collinear points on the plane with position vectors `vec a ,vec b` and `vec c` respectively (Fig).

`=>` The vectors `bar (RS) ` and `vec (RT)` are in the given plane. Therefore, the vector `bar (RS) xx vec (RT)` is perpendicular to the plane containing points R, S and T. Let `vec r` be the position vector of any point P in the plane. Therefore, the equation of the plane passing through R and perpendicular to the vector `bar (RS) xx bar (RT) ` is

`(vec r - vec a) * (bar (RS) xx bar (RT) ) = 0`

or `color{red}{( vec r - vec a) xx [ (vec b - vec a) xx (vec c - vec a) ] = 0}` ........(1)

`=>` This is the equation of the plane in vector form passing through three noncollinear points.





`=>` These planes will resemble the pages of a book where the line containing the points R, S and T are members in the binding of the book.

`color{brown} "Cartesian form"`

`=>` Let `(x_1, y_1, z_1), (x_2, y_2, z_2)` and `(x_3, y_3, z_3)` be the coordinates of the points R, S and T respectively. Let `(x, y, z)` be the coordinates of any point P on the plane with position vector `vec r` . Then

`bar (RP) = (x-x_1)hat i + (y-y_1) hat j + (z-z_1) hat k`

`bar (RS) = (x_2 - x_1) hat i + (y_2 - y_1) hat j + (z_2 - z_1) hat k`

`bar (RT) = (x_3 -x_1) hat i + (y_3- y_1) hat j + (z_3 -z_1) hat k`

`=>` Substituting these values in equation (1) of the vector form and expressing it in the form of a determinant, we have




`=>` which is the equation of the plane in Cartesian form passing through three non collinear points `(x_1, y_1, z_1), (x_2, y_2, z_2)` and `(x_3, y_3, z_3)`.

`=>` In this section, we shall deduce the equation of a plane in terms of the intercepts made by the plane on the coordinate axes. Let the equation of the plane be

`Ax + By + Cz + D = 0 \ \ (D ≠ 0)` ... (1)

`=>` Let the plane make intercepts a, b, c on `x, y` and `z` axes, respectively (Fig).

`=>` Hence, the plane meets x, y and z-axes at `(a, 0, 0), (0, b, 0), (0, 0, c),` respectively.

Therefore ` Aa + D = 0` or `A = (-D)/a`

`Bb +D =0` or `B = (-D)/b`

`Cc +D =0` or `C= (-D)/c`

`=>` Substituting these values in the equation (1) of the plane and simplifying, we get

`color{red}{x/a + y/b + z/c = 1}` .......(1)

which is the required equation of the plane in the intercept form.

`=>` Let `π_1` and `π_2` be two planes with equations `vec r * hat n_1 = d_1` and `vec r * hat n_2 = d_2` respectively. The position vector of any point on the line of intersection must satisfy both the equations (Fig).

`=>` If `vec t` is the position vector of a point on the line, then

`vec t * hat n_1 = d_1` and `vec t * hat n_2 = d_2`

Therefore, for all real values of λ, we have

` vec t * ( hat n_1 + lambda hat n_2) = d_1 + lambda d_2`

`=>` Since `vec t` is arbitrary, it satisfies for any point on the line.

`=>` Hence, the equation `vec r * ( hat n_1 + lambda hat n_2) = d_1 + lambda d_2` represents a plane `pi_31 ` which is such that if any vector `vec r` satisfies both the equations `π_1` and `π_2`, it also satisfies the equation `π_3` i.e., any plane passing through the intersection of the planes

`vec r * vec n_1 = d_1` and `vec r * vec n_2 = d_2`

has the equation `color{red}{vec r * (vec n_1 + lambda vec n_2) = d_1 +lambda d_2}` ............(1)

`color{brown} "Cartesian form"`

In Cartesian system, let

`vec n_1 = A_1 hat i + B_2 hat j + C_1 hat k`

`vec n_2 = A_2 hat i + B_2 hat j + C_2 hat k`

and `vec r = x hat i + y hat j + z hat k`

Then (1) becomes

`x (A_1 + λA_2) + y (B_1 + λB_2) + z (C_1 + λC_2) = d_1 + λd_2`

or`color{red}{ (A_1 x + B_1 y + C_1 z – d_1) + λ(A_2 x + B_2 y + C_2 z – d_2) = 0}` ... (2)

which is the required Cartesian form of the equation of the plane passing through the intersection of the given planes for each value of λ.

`=>` Let the given lines be

`vec r = vec a_1 + lambda vec b_1` .....(1)

and `vec r = vec a_2 + mu vec b_2` .....(2)

`=>` The line (1) passes through the point, say A, with position vector `vec a_1` and is parallel to`vec b_1`. The line (2) passes through the point, say B with position vector `vec a_2` and is parallel to `vec b_2` .

Thus , `bar (AB) = vec a_2 - vec a_1`

`=>` The given lines are coplanar if and only if `bar (AB)` is perpendicular to `vec b_1 xx vec b_2`.

i.e., `bar (AB) * (vec b_1 xx vec b_2) = 0` or `(vec a_2 - vec a_1)* ( vec b_1 xx vec b_2) =0`

`color{brown} "Cartesian form"`

`=>` Let `(x_1, y_1, z_1)` and `(x_2, y_2, z_2)` be the coordinates of the points A and B respectively.

`=>` Let `a_1, b_1, c_1` and `a_2, b_2, c_2` be the direction ratios of `vec b_1` and `vec b_2` , respectively. Then

`bar (AB) = (x_2 - x_1) hat i + (y_2 -y_1) hat j + (z_2 - z_1) hat k`

`vec b_1 = a_1 hat i + b_1 hat j + c_1 hat k` and `vec b_2 = a_2 hat i + b_2 hat j + c_2 hat k`

`=>` The given lines are coplanar if and only if `bar (AB) * (vec b_1 xx vec b_2) =0` . In the cartesian form, it can be expressed as

`color{red}{ | ( x_2 -x_1 , y_2- y_1 , z_2- z_1 ) , ( a_1 , b_2 ,c_1) , (a_2, b_2, c_2) | = 0}` ........ (4)

`=>` The angle between two planes is defined as the angle between their normals (Fig (a)).

`=>` If `θ` is an angle between the two planes, then so is `180 – θ` (Fig (b)). We shall take the acute angle as the angles between two planes.

`=>` If `vec n_1` and `vec n_2` are normals to the planes and θ be the angle between the planes

`vec r * vec n_1 = d_1` and `vec r * vec n_2 = d_2` .

`=>` Then `θ` is the angle between the normals to the planes drawn from some common point.

We have, `color{red}{cos theta = | (vec n_1 *vec n_2 )/( | vec n_1 | | vec n_2 |) |}`



`color{brown} "Cartesian form"`

`=>` Let `θ` be the angle between the planes,

`A_1 x + B_1 y + C_1z + D_1 = 0` and `A_2x + B_2 y + C_2 z + D_2 = 0`

The direction ratios of the normal to the planes are `A_1, B_1, C_1` and `A_2, B_2, C_2` respectively


Therefore,`color{red}{ cos θ = | ( A_1 A_2+ B_1 B_2 +C_1 C_2 )/( sqrt (A_(1)^2 + B_(1)^2 + C_(1)^2) sqrt (A_(2)^2 + B_(2)^2 +C_(2)^2) ) |}`

`"Vector form"`

`=>` Consider a point P with position vector `vec a` and a plane `π_1` whose equation is `vec r * hat n = d` (Fig).


`=>` Consider a plane `π_2` through P parallel to the plane `π_1`. The unit vector normal to `π_2` is `hat n` Hence, its equation is ` (vec r - vec a) * hat n = 0`

i.e., `vec r * hat n = vec a * hat n`

`=>` Thus, the distance ON′ of this plane from the origin is ` | vec a * hat n |` . Therefore, the distance PQ from the plane `π_1` is (Fig. (a))

i.e.,

`=>` which is the length of the perpendicular from a point to the given plane.

`\color{green} ✍️` We may establish the similar results for (Fig 11.19 (b)).




`color{brown} "Cartesian form"`-

`=>` Let `P(x_1, y_1, z_1) `be the given point with position vector `vec a` and

`Ax + By + Cz = D` be the Cartesian equation of the given plane. Then

`vec a = x_1 hat i + y_1 hat j + z_1 hat k`

`bar N = A hat i + B hat j + C hat k`

Hence, from Note 1, the perpendicular distance from P to the plane is

` color{blue}{d = | ( (x_1 hat i + y_1 hat j + z_1 hat k ) * ( A hat i + b hat j + C hat k) -D )/(sqrt (A^2 +B^2 +C^2 ) ) |}`

`=>` The angle between a line and a plane is the complement of the angle between the line and normal to the plane (Fig).

`"Vector form"` If the equation of the line is `vec r = vec a + lambda vec b` and the equation of the plane is

`vec r * vec n = d`. Then the angle θ between the line and the normal to the plane is

`color{red}{cos theta = | (vec b * vec n )/( | vec b| * | vec n| )|}`

and so the angle `φ` between the line and the plane is given by `90 – θ,` i.e.,
`sin (90 – θ) = cos θ`

i.e.,