Planes intersection calculator (2024)

Any three points not locating on a straight line determines a plane.

This definition of a plane is quite simple, but it is not convenient for calculation purposes. Instead, we use the fact that there is a unique plane which passes through a given point (x₀ , y₀ , z₀) and is perpendicular to a given line L which has direction numbers A, B and C.

Let the point (x₀ , y₀ , z₀) and the point (x , y , z) be located on a plane, and another point
(x₁ , y₁ , z₁) located on the line L which is perpendicular to the plane.

The direction numbers of line L₁ are:(x − x₀) , (y − y₀) , (z − z₀)

If line L has direction numbers A, B, C then these two lines L and L₁ will be perpendicular if and only if their direction numbers satisfy the relation (dot product of 2 vectors): $$ {A(x − x_0) + B(y − y_0) + C(z − z_0) = 0}$$

After arranging terms, we get the final plane equation: Ax + By + Cz − Ax₀ − By₀ − Cz₀ = 0

Notice that the point (x₁ , y₁ , z₁) is not a part of the plane equation that is because the plane is defined for all the parallel lines with the same direction numbers A B and C.

Then the equation of the plane gets the more recognizable form of:

Equation of a plane: Ax + By + Cz + D = 0

A, B, and C (called attitude numbers) are not all zero.

The equation of a plane parallel to the x-y axis: z + D = 0

The equation of a plane parallel to the x-z axis: y + D = 0

The equation of a plane parallel to the y-z axis: x + D = 0

The equation of a plane parallel to the x axis: y + z + D = 0

The equation of a plane parallel to the y axis: x + z + D = 0

The equation of a plane parallel to the z axis: x + y + D = 0

D − is the distance to the plane origin axis.

A₁x + B₁y + C₁z + D₁ = 0

A₂x + B₂y + C₂z + D₂ = 0

Equation of a plane passing through 3
points P₁ , P₂ , P₃.
Since the three points lies in the
plane, each of them satisfies the
plane equation:

And the plane equation will be: Ax + By + Cz + D = 0

These two vectors lie on the plane, so the cross product of this two vectors n₁ × n₂ gives a vector perpendicular to the plane, these values are the slopes of the plane equation.

The values of the cross product in the directions of x, y and z are:

Plane equation is:

The value of D is established by substituting a given point for example the point (x₁ , y₁ , z₁) in the plane equation.

Normal vectors to the planes are:

Planes are parallel if:
Planes are coincided if:

Those conditions can also be expressed as:

Find the intersection line equation between the two planes: 3x − y + 2z − 4 = 0 and 2x − y + 4z − 3 = 0

Any point along the intersection line between two planes, should satisfies both equations of the planes. But because we have three unknowns and only two equations, we can set the value t for one variable value, for example z = t then we get the equations:

These equations can be solved easily by Cramer's rule.

Hence the parametric equation of the two planes intersection line is:

These equations can be also represented by the equation:

Note that we could choose x = t instead of z = t and still get the same solution.

And the parametric equation of the two planes intersection line is:

And the parametric equation of the two planes intersection line is:

At first look it seems that we get a different line compare to the first solution, But if we set any value for t or &nbsp t = 0 and t = 1 &nbsp in the first solution we get the points &nbsp (1, -1, 0) and (3, 7, 1). Now we can set t = 1 and t = 3 to the second line, we get the points (1, -1, 0) and (3, 7, 1) which are the same points as in the first solution, so both lines are the same.

The intersection line can also be found by vector method.

The general vector direction of the perpendicular lines to the first and second planes are the coefficients x, y and z of the plane’s equations.

The cross product of these two lines will give the general direction of the intersection line between the two planes (the values provide the incline part marked with the prefix t of the intersection line).

Now we have to find a point that is located on the intersection line, this will be done by solving the planes equations, we will predefine the value of z = 0.

And the point is: (x, y, z) = (1, -1, 0), this points are the free values of the line parametric equation.

Finally, the planes intersection line equation is: x = 1 + 2t y = − 1 + 8t z = t

The type of solution depends on the parameter set to 0 (x = 0 or y = 0 or z = 0) and the solution method, by vector or by substitution.

Find the equation of a plane that passes through the points (1,0, 2), (-3, 5, 0) and (6, - 4, 2).

we choose the point (1, 0, 2) as the origin of the axes and will solve by vector method.

There are two vectors extending from the origin to the other two points:

The cross product of these two vectors gives the general direction of the perpendicular vector to the plane, this is also the direction coefficients of the plane.

Therefore the plane equation is: 8x + 10y + 9z + D = 0 (after multiplying all terms by -1)

Now D should be found, the origin point fulfills the plane equation so: 8*1 + 10*0 + 9*2 + D = 0

And the plane equation is: 8x + 10y + 9z − 26 = 0

The intersection line between two planes passes through the points (1,0,-2) and (1,-2,3) We also know that the point (2,4,-5)is located on the plane. Find the equation of the given plan and the equation of another plane with a tilted by 60 degrees to the given plane and has the same intersection line given for the first plane.

All three points are located on the given plane, so each of the points satisfies the equation of the plane.

We set arbitrary the values of D₁ = D₂ = D₃ = -1 then by Cramer's rule we have:

Now the equation of the plane is:
Multiplying by 18 we get the plane:	14x − 5y − 2z − 18 = 0

The angle between the two planes is given by vector dot product.

Because we have only one equation with 3 unknowns, we can set two values arbitrary for example
y₂ = 1 z₂ = 1 then we have:

Taking the square of both sides we get: 4x₂² − 4x₂ + 1 = 1.148(x₂² + 2)

2.852x₂² − 4x₂ − 1.296 = 0

So the plane equation are: &nbsp &nbsp 1.674x + y + z + D = 0 And 0.271x − y − z + D = 0

In order to find the value of D we substitute one of the points of the intersection line for example (1,0,-2) which is also located on the tilted plane to the plane equation 1.674x + y + z + D = 0.

The same way we handle the second solution for x₂ = − 0.271

We have two planes that is because they describe two planes tilted by 60 degrees either side of the given plane.

Find the equation of the plane that passes through the points (1,3,2), (-1,2,4) and (2, 1, 3).

Since all the points satisfies the plane equation we can substitute the values of x, y and z of each point into the plane equation Ax + By + Cz + D = 0 to get the following set of equations:

We have now three equations with four unknowns A, B, C and D theoretically there is no exact answer but we can solve the equations related to the unknown D.

Because the term D is a part of the solution of all the unknowns we can choose any value for D without changing the final answer, for example take D = −25 then the plane equation will be:

3x + 4y + 5z − 25 = 0

Find the equation of the line that passes through the point (-2,3,1) and is perpendicular to the plane
2x + 3y + z - 3 = 0 also find the intersection point and the distance of the point to the plane.

The attitude numbers of the line that are perpendicular to the plane are given by the coefficients A, B and C so the line attitudes are A = 2, B = 3 and C = 1.

After eliminating t we get the line form as fractions:

The distance of the point to the plane can be solved by vectors method.

Denote V as the plane vector V = iA + jB + kC where i, j and k are in the x, y and z directions, this vector is perpendicular to the plane.

The unit vector in the plane direction is:

The value of the vector P from a point (x₀ , y₀ , z₀) to the given point is:

We can choose the point (x₀ , y₀ , z₀) to be at the origin and therefore their values are equal to 0.

The distance from the point to the plane will be the projection of P on the unit vector direction this is the dot product of the vector P and the unit vector.

Another way to find the distance is by finding the plane and the line intersection point and then calculate distance between this point and the given point.

At the intersection point the values of x, y and z should be the same, so first we will find the value of t that satisfies both equations:

A(x1 + At) + B(y1 + Bt) + C(z2 + Ct) + D = 0

And the intersection point of the given line and the plane is (this line is perpendicular to the plane):

The distance between the given point and the plane is now the distance of the point to the intersection point and is given by the equation.

Find the equation of the line that passes through the point P₀ (-2, 0, 1) and is perpendicular to the plane x + 3z − 4 = 0.

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.

Find the equation of the plane that passes through the point (5,2,−1) and is perpendicular to a line that passes through the points (5,4,3) and (−6,1,7).

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

Find the equation of a plane that passes through the point P₀: (2, 1, −1) and is parallel to the plane M₁: 2x + y − 2z + 5 = 0. Also find the distance of point P₀ from plane M₁

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

d² = (At₁)² + (Bt₁)² + (Ct₁)² = t₁² (A² + B² + C²)

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:

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

Find the equation of the plane that contains the two lines.

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

Find the equation of the plane that contains the two parallel lines.

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