วันจันทร์ที่ 27 เมษายน พ.ศ. 2569

The line of intersection of two planes using vectors (Part 2)

There’s a clean, fully vector-based formula for a point on the intersection line, with no elimination at all.

Setup

Two planes:

\mathbf{n}_1 \cdot \mathbf{r} = d_1,\quad \mathbf{n}_2 \cdot \mathbf{r} = d_2

Let:

  • n1, n2 = normal vectors
  • d = n1 × n2 = direction of the line

Key idea

The point ( r0 ) lies:

  • in both planes
  • perpendicular to the direction ( d )

So we construct it directly using cross products.

Formula for the point

\mathbf{r}_0 = \frac{ d_1 (\mathbf{n}_2 \times \mathbf{d}) + d_2 (\mathbf{d} \times \mathbf{n}_1) }{ |\mathbf{d}|^2 }

Why this works (intuition)

  • ( n2×d ) lies in plane 2 and points toward satisfying plane 1
  • ( d×n1 ) lies in plane 1 and points toward satisfying plane 2
  • The combination balances both constraints exactly
  • Dividing by |d|2 normalizes the result

Example

Planes:

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

Step 1: normals

	n1 = (1, 1, 1),
	n2 = (1, -1, 1)
    
( d1 = 1, d2 = 3 )

Step 2: direction

	d = n1 × n2 = (2, 0, -2)

Step 3: compute pieces

	n2 × d = (1, -1, 1) × (2, 0, -2) = (2, 4, 2)

	d × n1 = (2, 0, -2) × (1, 1, 1) = (2, -4, 2)

Step 4: plug into formula

\mathbf{r}_0 = \frac{ 1(2,4,2) + 3(2,-4,2) }{ (2)^2 + 0^2 + (-2)^2 }

	r0 = (1, -1, 1)

Final line

Direction: (1, 0, -1)

	r(t) = (1, -1, 1) + t (1, 0, -1)

Takeaway

  • No solving equations
  • No choosing variables
  • Pure vector operations: cross products + scaling
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