Key Concepts in 3D Analytic Geometry

Posted on May 3, 2025 in Mathematics

This document outlines fundamental procedures for solving common problems in three-dimensional analytic geometry.

1. Plane Perpendicular to Line Through Point

To find the equation of a plane perpendicular to a given line and passing through a given point:

Take the direction vector of the line. This vector is the normal vector (A, B, C) of the plane.
Substitute these values into the plane equation form: Ax + By + Cz + D = 0.
Substitute the coordinates (x, y, z) of the given point into the equation to solve for D.
The resulting equation is the equation of the plane.

2. Symmetric Point with Respect to a Line

To find the symmetric point P’ of a point P with respect to a line r:

Calculate the projection M of point P onto line r. This is done by finding the plane (let’s call it Pi) that contains P and is perpendicular to r.
To find the equation of Pi: Take the direction vector of line r; this is the normal vector (A, B, C) of plane Pi. Substitute these into Ax + By + Cz + D = 0. Substitute the coordinates of point P for x, y, z to solve for D.
Find the intersection point M of plane Pi and line r. This is done by substituting the parametric coordinates of line r (x(λ), y(λ), z(λ)) into the equation of plane Pi and solving for the parameter λ.
Substitute the value of λ back into the parametric equations of line r to obtain the coordinates of point M.
Point M is the midpoint of the segment PP’. Use the midpoint formula: M = ((P_x + P’_x)/2, (P_y + P’_y)/2, (P_z + P’_z)/2). Solve for the coordinates of P’. P’ = 2M – P.

3. Line Through Point and Its Symmetric Point

To find the equation of the line passing through a point P and its symmetric point P’ (with respect to a line r):

Calculate the vector PP’. This vector is the direction vector of the desired line.
Use point P (or P’) and the direction vector PP’ to write the equation of the line (e.g., in parametric or symmetric form). Note that this line is perpendicular to the original line r.

4. Plane Containing Line, Perpendicular to Plane

To find the equation of a plane containing a given line r and perpendicular to a given plane Pi₁:

Take the normal vector n₁ of the given plane Pi₁.
Take the direction vector v_r of the given line r.
The normal vector n of the desired plane is perpendicular to both n₁ and v_r. Calculate n as the cross product: n = v_r × n₁.
Take any point P_r on the given line r.
Use the normal vector n = (A, B, C) and the point P_r(x₀, y₀, z₀) to write the plane equation: A(x – x₀) + B(y – y₀) + C(z – z₀) = 0, or Ax + By + Cz + D = 0 where D = -(Ax₀ + By₀ + Cz₀).

5. Common Perpendicular Line to Two Lines

To find the equation of the common perpendicular line to two skew lines r₁ and r₂:

First, verify that the lines are skew (not parallel and not intersecting).
Calculate the direction vector w of the common perpendicular line. This vector is perpendicular to the direction vectors of both lines, so w = v₁ × v₂, where v₁ and v₂ are the direction vectors of r₁ and r₂, respectively.
Find a plane Pi that contains the first line r₁ and is parallel to the vector w. This plane is defined by a point P₁ on r₁ and two direction vectors: v₁ and w. The normal vector of Pi is v₁ × w. Use P₁ to find the plane equation.
Find the intersection point Q of the second line r₂ with the plane Pi. Substitute the parametric coordinates of r₂ into the equation of Pi and solve for the parameter. Substitute the parameter value back into the parametric equations of r₂ to get point Q.
The common perpendicular line passes through point Q and has direction vector w. Write the equation of this line using Q and w.

6. Plane Through Point, Perpendicular to Line

To find the equation of the plane passing through point A(1,0,0) and perpendicular to the line passing through points B(0,1,0) and C(0,0,1):

Calculate the direction vector of the line BC: BC = C – B = (0-0, 0-1, 1-0) = (0, -1, 1).
This vector BC is the normal vector (A, B, C) of the desired plane. So, the plane equation is 0x – 1y + 1z + D = 0, or -y + z + D = 0.
Substitute the coordinates of point A(1,0,0) into the equation to find D: -0 + 0 + D = 0, so D = 0.
The equation of the plane is -y + z = 0, or y – z = 0.

7. Plane Through Point, Perpendicular to Plane, Parallel to Line

To find the equation of a plane passing through point A(1,0,-1), perpendicular to the plane x – y + 2z + 1 = 0, and parallel to a given line r:

Identify the normal vector n_p of the given plane: n_p = (1, -1, 2).
Identify the direction vector v_r of the given line r. (Assume the line r is given, e.g., in parametric form. If not, convert it).
The normal vector n of the desired plane is perpendicular to both n_p and v_r. Calculate n as the cross product: n = n_p × v_r.
Use the normal vector n = (A, B, C) and the point A(1,0,-1) to write the plane equation: A(x – 1) + B(y – 0) + C(z – (-1)) = 0, or A(x – 1) + By + C(z + 1) = 0.
Expand and simplify to get the form Ax + By + Cz + D = 0.

8. Volume of a Tetrahedron (Pyramid)

To calculate the volume of a tetrahedron with vertices A, B, C, and D:

Form three vectors originating from one vertex, e.g., AB, AC, and AD.
Calculate the scalar triple product of these three vectors: AB ⋅ (AC × AD). This is equal to the determinant of the matrix formed by the coordinates of the three vectors.
The volume V of the tetrahedron is one-sixth of the absolute value of the scalar triple product: V = (1/6) |AB ⋅ (AC × AD)|.
The unit of volume is cubic units (e.g., m³).

9. Distance from Point to Plane Containing Line and Point

To find the distance from point P(1, -1, 2) to the plane containing line r (passing through point A(2, 1, 3) with direction vector v_r) and point P:

Note: The original description implies the plane contains point P, in which case the distance from P to the plane is 0. Assuming the intent was to find the distance from P to the plane containing line r (through A with direction v_r) and another point, say Q. Let’s follow the example’s structure and assume the plane contains line r (through A(2,1,3) with direction v_r) and an external point P(1,-1,2).

Identify a point on line r, A(2,1,3), and its direction vector v_r.
Identify the external point P(1,-1,2).
Form the vector AP = P – A = (1-2, -1-1, 2-3) = (-1, -2, -1).
The plane contains line r (defined by A and v_r) and point P. The direction vectors for this plane are v_r and AP.
The normal vector n of the plane is the cross product: n = v_r × AP.
Use the normal vector n = (A, B, C) and point A(2,1,3) (or P(1,-1,2)) to find the equation of the plane Ax + By + Cz + D = 0. Substitute the point’s coordinates to solve for D.
Once the plane equation Ax + By + Cz + D = 0 is found, use the distance formula from point P(x_p, y_p, z_p) to the plane: d(P, Pi) = |Ax_p + By_p + Cz_p + D| / √(A² + B² + C²).
Substitute the coordinates of P(1,-1,2) and the values A, B, C, D from the plane equation into the formula to compute the distance.

10. Condition for Line and Plane Parallelism

To determine if a given line r and a given plane Pi are parallel:

Identify the direction vector v_r of the line r. (Ensure the line equation is in a form that shows the direction vector, e.g., parametric form).
Identify the normal vector n_p of the plane Pi. (Ensure the plane equation is in the form Ax + By + Cz + D = 0, where n_p = (A, B, C)).
A line is parallel to a plane if and only if the line’s direction vector is perpendicular to the plane’s normal vector.
This condition is met when their dot product is zero: v_r ⋅ n_p = 0.
Example: If v_r = (a, 2, -3) and n_p = (a, a, 2), their dot product is (a)(a) + (2)(a) + (-3)(2) = a² + 2a – 6. For the line and plane to be parallel, a² + 2a – 6 = 0. Solve this quadratic equation for ‘a’.
If the dot product is zero, the line is either parallel to the plane or contained within the plane. To check if it’s contained, take any point on the line and substitute its coordinates into the plane equation. If the equation is satisfied, the line lies in the plane. If not, the line is strictly parallel to the plane.