The coefficients of  x, y  and  z  in the plane equation are the direction numbers  A, B  and  C  respectively of the line which is perpendicular to the plane so we can see that:  A = 1,  B = 0  and
C = 3. And the required parametric line equation that passes through the point  P₀  is:

This parametric line equations can be presented in another way as:

$$ {{x - x_0 \over A} = {y - y_0 \over B} = {z - z_0 \over C}}.$$

After substituting given values, we get the equation of the line.

$$ {{x + 2 \over 1} = {y \over 0} = {z - 1 \over 3}}.$$

Notice that the division of  y  by zero stands for the value of   y = y₀ = 0   also notice that the free term value in the plane equation  −4  is not used in the solution because it sets the location of the plane but not the attitude of it.

The attitude numbers of the given line is     A = 5 + 6 = 11 ,      B = 4 − 1 = 3       and       C = 3 − 7 = −4.

The equation of the plane is A(x − x0) + B(y − y0) + C(z − z0) = 0

11(x − x0) + 3(y − y0) − 4(z − z0) = 0

11(x − 5) + 3(y − 2) − 4(z + 3) = 0

11x − 55 + 3y − 6 − 4z − 12 = 0

Now we have to find the value of  D  that we can find from the fact that the point  (5,2,−1)  is located on the plane, therefore it should satisfy the equation of the plane:

11 · 5 + 3 · 2 − 4 · (−1) + D = 0

and D = −57   so the equation of the plane is:     11x + 3y − 4z −73 = 0

Because the given and the required planes are parallel their x, y and z coefficient denoted by  A, B and  C  respectively are the same so we have  A = 2,  B = 1  and C = −2 The new plane should pass through the point P₀. And the equation of the required plane is:

A (x − x₀) + B (y − y₀) + C (z − z₀) = 0

Substituting values of the point and the direction numbers we get the equation:

2 (x − 2) + 1 (y − 1) − 2(z + 1) = 0

After arranging terms, we get the final plane equation: 2x + y − 2z − 7 = 0

If the plane is given by Ax + By + Cz + D = 0

The parametric line equations connecting point   P₀ (x₀ , y₀ , z₀) are

If the intersection point of the line and the plane is:   P₁ (x₁ , y₁ , z₁)

The square distance from point P₀ to point P₁ is: d² = (x₀ − x₁)² + (y₀ − y₁)² + (z₀ − z₁)²(1)

Because point P₁ is on the line and also on the plane the line parametric equations are:

Point x₁ is located on the plane hence it satisfies the plane equation:

Ax₁ + By₁ + Cz₁ + D = 0

After substitute eq (2) into the plane equation we get:

A(x₀ + At₁) + B(y₀ + Bt₁) + C(z₀ + Ct₁) + D = 0(3)

Now substitute equation  (2)  into equation  (1)  we get

Substituting the value of t₁ from equation (4) into equation (3) we get:

$$ Ax_0 + {A^2d \over \sqrt{A^2+B^2+C^2}} + By_0 + {B^2d \over \sqrt{A^2+B^2+C^2}} + Cz_0 + {C^2d \over \sqrt{A^2+B^2+C^2}}+D=0 $$

After arranging terms and some algebraic steps we get the final value for d:

$$ {d \over \sqrt{A^2+B^2+C^2}} (A^2+B^2+C^2) +Ax_0+By_0+Cz_0+D=0$$

$$ d= {-(Ax_0+By_0+Cz_0+D)\sqrt{A^2+B^2+C^2} \over A^2+B^2+C^2}$$

$$ d={|Ax_0 + By_0 + Cz_0 + D| \over \sqrt{A^2+B^2+C^2}}$$

And after substituting values we get:

$$ d={|2‧2 + 1‧1 + 2‧1 + 5| \over \sqrt{4+1+4}} = {12\over \sqrt{9}} = {4}$$

Notice that two arbitrary lines should be located on the same plane otherwise no plane is existed, in another way if both lines intersect each other, they always are located on the same plane. The exception case is two parallel lines (see next example 9).

We can easily write the lines in vector form with the direction numbers, that is.

The cross product of this two lines vectors will give us the plane direction numbers

Because point  (1 , − 2 , 2)  is located on the line  L₁  it is also located on the plane.

And the equation of the plane is:   A (x − x₀) + B (y − y₀) + C (z − z₀) = 0

− 2(x − 1) + 2 (y + 2) + 1(z − 2) = 0

2x − 2y − z − 4 = 0

Notice That the two lines are parallel because their direction numbers   (2 , 3 , −1)) are the same.

We can see that the points   (0 , 1 , −2)   and   (2 , −1 , 0)   are both located on the lines and hence are located on the plane.

The direction numbers  a , b  and  c  of the line   L₃   containing both points are:

We can write the lines in vector form with the direction numbers, that is.

The cross product of this two lines vectors will give us the perpendicular line direction numbers which are the plane direction numbers

The direction numbers of the plane are     A = 2   B = −3   C = −5  

Because line L₁ is located on the plane the point (1 , − 2 , 2) is also located on the plane.

And the equation of the plane is:   A (x − x₀) + B (y − y₀) + C (z − z₀) = 0

2(x − 0) − 3(y − 1) − 5(z + 2) = 0

2x − 3y − 5z − 7 = 0

Find the equation of the plane that contains the point  P₁ (3 , -1 , 2)  and the line.

$$L_1: {{x-2 \over 2} = {y+1 \over 3} = {z \over -2}}$$

We choose a point on the line according to the given line   P₂(2 , −1 , 0)   this point is located on the required plane.

The direction numbers  a , b  and  c  of the line  L₂  connecting   P₁   to   P₂   are:

This line can be written in vector form as:     L₂ = i + 2k.

Given line L₁ in vector form will be         L₁ = 2i + 3j − 2k.

The cross product of this two lines  L1  and  L2  will give us the perpendicular line direction numbers which are the plane direction numbers

The direction numbers of the plane are     A = 2   B = −2   and   C = −1  

And the equation of the plane is:   A (x − x₀) + B (y − y₀) + C (z − z₀) = 0

2(x − 3) − 2(y + 1) − 1(z −2) = 0

2x − 2y − z − 6 = 0

Determine if the plane   2x − 3y + z − 2 = 0   is parallel to the line.

$$L_1: {{x-2 \over 1} = {y+2 \over 1} = {z+1 \over 1}}$$

The direction numbers  a, b  and  c  of line   L2   are:           a = 2         b = −3         c = 1. The vector presentation of line L2 is:     L2 = 2i − 3j + k The vector presentation of line L1 is:     L1 = i + j + k A, B and C are the direction numbers of line L1

From the vector dot product we know that the angle between two vectors is:

cosθ = a‧A + b‧B + c‧C

In our case cosθ = 1‧2 + 1(−3) + 1 = 0 :

the value of cos is zero,  cosθ = 0  this happen when angle θ = 90 degree. Because line  L₂  is perpendicular to the plane a line that is also perpendicular to line L₂ must be parallel to the plane.

Find the equation of a plane passing through the point   (1 , 3 , −2)

and is perpendicular to the line     L₁:     x = 2 + 4t        y = 5        z = 1 −3t

The direction numbers  A, B  and  C  of a line   that is perpendicular to a plane, determines the attitude numbers of that plane, therefore L1 is perpendicular to the plane and:               A = 4         B = 0         C = −3. Now we define another point on the plane as (x , y , z) connect this point to the point P1 to get line L2. The direction numbers of L2 are: a = (x − x1)         b = (y − y1)         c = (z − z1)

Line L₂ is perpendicular to line L₁ therefore their dot product must be equal to 0.

A (x − x₀) + B (y − y₀) + C (z − z₀) = 0

L₁‧L₂ = 4 (x − 1) − 0 (y − 3) − 3 (z + 2) = 0

4x − 4 − 3z − 6 = 0 :

And the final plane equation is:         4x − 3z − 10 = 0