Swatchai's Web Adventure: The line of intersection of two planes using vectors (Part 1)

วันอาทิตย์ที่ 26 เมษายน พ.ศ. 2569

The line of intersection of two planes using vectors (Part 1)

To find the line of intersection of two planes using vectors, you combine two key ideas:

The direction of the line comes from the cross product of the planes’ normals
A point on the line is found by solving the two plane equations together

Step 1: Write the planes

Suppose the planes are:

$\text{Plane 1: } a_1 x + b_1 y + c_1 z = d_1$

$\text{Plane 2: } a_2 x + b_2 y + c_2 z = d_2$

Their normal vectors are:

$\mathbf{n}_1 = (a_1, b_1, c_1), \quad \mathbf{n}_2 = (a_2, b_2, c_2)$

Step 2: Direction of the intersection line

The direction vector d is perpendicular to both normals:

$\mathbf{d} = \mathbf{n}_1 \times \mathbf{n}_2$

If this cross product is zero → planes are parallel (no intersection or infinite overlap).

Step 3: Find a point on the line

To get a specific point:

Solve the two plane equations simultaneously
Usually set one variable (like (z = 0)) and solve for (x, y)

Step 4: Write the vector equation of the line

Once you have:

A point $\mathbf{r}_0 = (x_0, y_0, z_0)$
Direction $\mathbf{d}$

The line is:

$\mathbf{r}(t) = \mathbf{r}_0 + t \mathbf{d}$

Example

Find the intersection of:

    x + y + z = 1
    x - y + z = 3

1. Normals

$\mathbf{n}_1 = (1,1,1), \quad \mathbf{n}_2 = (1,-1,1)$

2. Direction

$\mathbf{d} = \mathbf{n}_1 \times \mathbf{n}_2 = (2, 0, -2)$

(Simplify → (1,0,-1))

3. Find a point

Set (z = 0):

    x + y = 1
    x - y = 3

Add:

$2x = 4 \Rightarrow x = 2$

Then:

	y = -1

So point = (2, -1, 0)

4. Final line

$\mathbf{r}(t) = (2,-1,0) + t(1,0,-1)$

Final takeaway

Cross product → gives direction
Solve equations → gives point
Combine → gives line

Swatchai's Web Adventure