Two vectors in the plane → cross them for the normal: Three points A, B, C fix a plane (like a three-legged stool that never wobbles). The vectors AB and AC both lie flat in the plane, so their cross product AB × AC points straight out — that's the normal n.
Then use one point (say A) to finish, exactly like in 3.17.1: build ax + by + cz = d and find d by substituting A.
IB-style question — plane through three points
Find the Cartesian equation of the plane through A(1, 0, 2), B(3, 1, 2) and C(2, −1, 4).
Step by step
- Two vectors lying in the plane.
- Cross them to get the normal n = AB × AC.
- Build the equation: coefficients are the normal.
- Find d by substituting A(1, 0, 2).
- State the plane.
Final answer
2x − 4y − 3z = −4.
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A line gives one in-plane vector; a connecting vector gives the second: If a plane contains a line r = a + λd and a point P not on it, you already have two directions inside the plane: the line's direction d, and the vector AP from a point on the line to P. Cross them: n = d × AP.
Converting forms is just rearranging: scalar-product r·n = k expands to Cartesian ax + by + cz = k; parametric r = a + λu + μv uses two in-plane directions u, v, whose cross product u × v gives the normal.
IB-style question — plane through a point and a line
A plane contains the line r = (1, 0, 1) + λ(2, 1, −1) and the point P(0, 3, 2).
Find the Cartesian equation of the plane.
Step by step
- The line gives one in-plane direction d, and the point Q(1, 0, 1) on the line lets us build a second, QP.
- Cross them for the normal n = d × QP.
- Build the equation and find d by substituting Q(1, 0, 1).
- State the plane.
Final answer
4x − y + 7z = 11.
IB-style question — convert Cartesian to scalar-product form
Write the plane 3x − 2y + z = 8 in the scalar-product form r·n = k.
Step by step
- The coefficients are the normal; the constant stays as k.
- Write it out.
Final answer
r·(3, −2, 1) = 8.