Plane Through Three Points, and Its Intersection Line With Another Plane: 04-BS-1, December 2013, Q4

Question 4: Plane Through Three Points, and Its Intersection Line With Another Plane (a) 10, (b) 10 marks

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Three points defining plane $P$: $A=(2,1,-2)$, $B=(1,2,0)$, $C=(1,0,-1)$; a second plane $x+y-2z=3$.

Find. (a) the equation of $P$. (b) the line where $P$ meets the second plane.

Approach. (a) Form two in-plane vectors from the three points, cross them to get a normal, then write the point-normal plane equation. (b) The intersection line's direction is the cross product of the two planes' normals; find one point on the line by solving the two plane equations simultaneously.

(a) In-plane vectors. $\vec{AB}=B-A=(-1,1,2)$, $\vec{AC}=C-A=(-1,-1,1)$.
(a) Normal vector. $$\vec n=\vec{AB}\times\vec{AC}=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\-1&1&2\\-1&-1&1\end{vmatrix}=(1(1)-2(-1))\mathbf i-((-1)(1)-2(-1))\mathbf j+((-1)(-1)-1(-1))\mathbf k=(3,-1,2).$$
(a) Plane equation. Using point $A=(2,1,-2)$: $3(x-2)-1(y-1)+2(z+2)=0\Rightarrow 3x-y+2z=1$. (Check with $B$: $3(1)-2+0=1$ ✓; with $C$: $3(1)-0-2=1$ ✓.) $$\Rightarrow\quad \boxed{3x-y+2z=1}$$
(b) Direction of the intersection line. The line lies in both planes, so its direction is perpendicular to both normals: $\vec d=\vec n\times(1,1,-2)$. $$\vec d=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\3&-1&2\\1&1&-2\end{vmatrix}=((-1)(-2)-2(1))\mathbf i-(3(-2)-2(1))\mathbf j+(3(1)-(-1)(1))\mathbf k=(0,8,4)\ \parallel\ (0,2,1).$$
(b) A point on the line. Solve $3x-y+2z=1$ and $x+y-2z=3$ simultaneously. Adding the two equations eliminates $z$: $4x=4\Rightarrow x=1$. From the second equation, $y=3-x+2z=2+2z$; taking $z=0$ gives $y=2$, so $(1,2,0)$ is on the line (this is point $B$ itself, a convenient check).

$$\boxed{(x,y,z)=(1,2,0)+t(0,2,1),\quad t\in\mathbb R}$$

Quantity	Result
Plane $P$	$3x-y+2z=1$
Intersection line	$(x,y,z)=(1,2,0)+t(0,2,1)$

Concept & Theory

Topic. Planes and lines in $\mathbb R^3$ via the cross product.

Key relations. A plane through point $A$ with normal $\vec n$ is $\vec n\cdot(\mathbf x-A)=0$; the normal to the plane through three points is $\vec{AB}\times\vec{AC}$; the line of intersection of two planes with normals $\vec n_1,\vec n_2$ has direction $\vec n_1\times\vec n_2$ (perpendicular to both normals, hence lying in both planes).

Why this works. A line lying in both planes must be perpendicular to both planes' normal vectors — the cross product is exactly the vector perpendicular to two given vectors, so it is the unique (up to scale) direction of the intersection. Solving the two plane equations for a specific point (e.g. setting $z=0$) then anchors that direction into a full parametric line.

Common pitfall. Sign errors expanding the $3\times3$ cross-product determinant (the middle cofactor carries a minus sign); also forgetting to double-check the plane equation against all three original points, not just the one used to fix the constant.

Source. Stewart, Calculus: Early Transcendentals, Ch. 12 (§12.5 Equations of Lines and Planes in Space).

Question 4 of 8: Plane Through Three Points, and Its Intersection Line With Another Plane

Question 4: Plane Through Three Points, and Its Intersection Line With Another Plane (a) 10, (b) 10 marks