How to Show That a Point Lies on a Line Vector

To demonstrate that a specific point lies on a given line vector, you must show that the point's coordinates satisfy the line's vector equation by yielding a consistent scalar parameter across all components. This method involves substituting the point's coordinates into the line's equation and solving for the scalar parameter (often denoted as lambda, λ). If a unique value for λ can be found that satisfies all coordinate equations simultaneously, then the point is on the line.

Understanding Line Vector Equations

A line in 2D or 3D space can be represented by a vector equation. This equation typically takes the form:

r = a + λd

Where:

r is the position vector of any point on the line (e.g., (x, y, z)).
a is the position vector of a known point on the line (e.g., (x₀, y₀, z₀)).
λ (lambda) is a scalar parameter that can be any real number. As λ varies, r traces out all points on the line.
d is the direction vector of the line (e.g., (dx, dy, dz)), indicating the line's orientation in space.

The Core Principle

A point P with position vector p lies on the line if and only if its position vector p can be expressed in the form a + λd for a single, unique value of λ. In other words, if you substitute the coordinates of point P into the line's vector equation, you should be able to solve for λ from each coordinate (x, y, z) equation, and all these λ values must be identical.

Step-by-Step Method to Prove a Point Lies on a Line

Follow these steps to determine if a point lies on a vector line:

1. Identify the Line's Vector Equation

Ensure the line's equation is in the standard vector form:
(x, y, z) = (x₀, y₀, z₀) + λ(dx, dy, dz)

2. Substitute the Point's Coordinates

Replace the general position vector (x, y, z) with the specific coordinates of the point P(x_p, y_p, z_p) you are testing.
This will create a system of three scalar equations (one for each coordinate):

x_p = x₀ + λdx
y_p = y₀ + λdy
z_p = z₀ + λdz

3. Solve for the Scalar Parameter (λ)

From each of the three scalar equations, solve for λ independently.
For example, from the x-component equation, you would find λ = (x_p - x₀) / dx. If the line's x-component equation is x = 4 + λ(1) and you substitute a point's x-coordinate of 1 into this equation, you would get 1 = 4 + λ. Solving for λ yields λ = 1 - 4 = -3. This process must be repeated for the y and z components.

4. Check for Consistency

This is the most critical step:

If the calculated λ values from all three coordinate equations are identical, then the point lies on the line. The consistent λ value indicates that the point is indeed part of the line traced by varying λ.
If even one of the λ values is different from the others, the point does not lie on the line. This means there is no single scalar parameter that can transform the known point a along the direction d to reach point P.

Practical Example

Let's illustrate with an example.

Problem Statement

Does the point A(1, 5, 7) lie on the line given by the vector equation r = (4, -1, 3) + λ(-1, 2, 1)?

Solution

Line Equation: (x, y, z) = (4, -1, 3) + λ(-1, 2, 1)
Substitute Point A(1, 5, 7):
- 1 = 4 + λ(-1) => 1 = 4 - λ
- 5 = -1 + λ(2) => 5 = -1 + 2λ
- 7 = 3 + λ(1) => 7 = 3 + λ
Solve for λ in each equation:
- From x: 1 = 4 - λ => λ = 4 - 1 => λ = 3
- From y: 5 = -1 + 2λ => 6 = 2λ => λ = 3
- From z: 7 = 3 + λ => λ = 7 - 3 => λ = 4
Check for Consistency:
The calculated λ values are 3, 3, 4. Since the value from the z-component (4) is different from the values from the x and y components (3), the λ values are not consistent.

Conclusion

Therefore, the point A(1, 5, 7) does not lie on the line r = (4, -1, 3) + λ(-1, 2, 1).

Another Example (Point lies on the line)

Does the point B(3, 1, 4) lie on the line r = (4, -1, 3) + λ(-1, 2, 1)?

Substitute Point B(3, 1, 4):
- 3 = 4 - λ
- 1 = -1 + 2λ
- 4 = 3 + λ
Solve for λ:
- From x: λ = 4 - 3 => λ = 1
- From y: 2λ = 1 + 1 => 2λ = 2 => λ = 1
- From z: λ = 4 - 3 => λ = 1
Check for Consistency:
The calculated λ values are 1, 1, 1. All values are identical.

Conclusion

Therefore, the point B(3, 1, 4) does lie on the line r = (4, -1, 3) + λ(-1, 2, 1).

Why This Method Works

The vector equation r = a + λd is essentially a recipe for generating every single point on the line. The parameter λ acts as a scaling factor for the direction vector d. If a point P truly lies on the line, there must be one specific λ value that, when applied to the direction vector and added to the starting point a, perfectly lands on P. If different λ values are required for different coordinates, it implies the point P is not reachable by moving along the direction d from a using a single scaling factor.

Common Pitfalls and Tips

Double-check calculations: Even a minor arithmetic error can lead to inconsistent λ values and an incorrect conclusion.
Ensure consistency across all components: It's not enough for two out of three λ values to match; all must be identical.
Understand the components: Clearly distinguish between the position vector of a point on the line (a), the direction vector of the line (d), and the coordinates of the point being tested (p).

Here's a quick summary table:

Input	Process	Output (λ values)	Conclusion
Point coordinates & Line equation	Substitute point coordinates into x, y, z components of the line equation. Solve for λ from each component separately.	`λx, λy, λz`	If `λx = λy = λz`, point is on the line.
			If `λx ≠ λy` or `λy ≠ λz`, point is not on the line.

For further reading on vector lines and their properties, you can refer to resources like Khan Academy on Vector Equations of Lines.