Plane Through Three Points, and Its Intersection Line With Another Plane: 04-BS-1, May 2015, Q6

Question 6: 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 $A=(2,1,-2)$, $B=(1,2,0)$, $C=(1,0,-1)$ defining plane $P$; a second plane $x+y-2z=3$.

Find. (a) An equation for $P$. (b) The line $P\cap\{x+y-2z=3\}$.

Plane $P$ (through $A,B,C$) meets the plane $x+y-2z=3$ along a line, direction $\mathbf n_P\times\mathbf n_2$.

Approach. (a) Form two edge vectors from the three points and cross them to get the normal $\mathbf n_P$. (b) The intersection line's direction is $\mathbf n_P\times\mathbf n_2$ (orthogonal to both plane normals); find one point common to both planes to anchor it.

(a) Edge vectors and normal. $\overrightarrow{AB}=B-A=(-1,1,2)$, $\overrightarrow{AC}=C-A=(-1,-1,1)$. $$\mathbf n_P=\overrightarrow{AB}\times\overrightarrow{AC}=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\-1&1&2\\-1&-1&1\end{vmatrix}=\big(1(1)-2(-1)\big)\mathbf i-\big((-1)(1)-2(-1)\big)\mathbf j+\big((-1)(-1)-1(-1)\big)\mathbf k=(3,-1,2).$$
(a) Plane equation. Through $A=(2,1,-2)$ with normal $(3,-1,2)$: $$3(x-2)-1(y-1)+2(z+2)=0\ \Rightarrow\ \boxed{3x-y+2z=1}$$ (check: $B$: $3(1)-2+0=1$ ✓; $C$: $3(1)-0-2=1$ ✓).
(b) Direction of the intersection line. With $\mathbf n_P=(3,-1,2)$ and $\mathbf n_2=(1,1,-2)$ (normal of $x+y-2z=3$), $$\mathbf n_P\times\mathbf n_2=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\3&-1&2\\1&1&-2\end{vmatrix}=\big((-1)(-2)-2(1)\big)\mathbf i-\big(3(-2)-2(1)\big)\mathbf j+\big(3(1)-(-1)(1)\big)\mathbf k=(0,8,4).$$ This simplifies to the direction $(0,2,1)$.
(b) A point common to both planes. Setting $z=0$: $3x-y=1$ and $x+y=3$. Adding, $4x=4\Rightarrow x=1$, then $y=2$. The point $(1,2,0)$ (which happens to be $B$ itself) satisfies both plane equations. So $$\boxed{(x,y,z)=(1,2,0)+t(0,2,1)}.$$

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

Concept & Theory

Topic. Plane through three points via the cross product, and the intersection line of two planes.

Key relations. Given three non-collinear points, $\mathbf n=\overrightarrow{AB}\times\overrightarrow{AC}$ is normal to the plane through them. Two non-parallel planes with normals $\mathbf n_1,\mathbf n_2$ intersect in a line with direction $\mathbf n_1\times\mathbf n_2$ (orthogonal to both normals, hence lying in both planes); any single point satisfying both plane equations anchors it.

Why this works. A line lying in both planes must be perpendicular to both normals simultaneously, and the cross product is exactly the (unique up to scale) vector orthogonal to two given vectors — this is why the same cross-product operation recurs for both the plane-normal step and the intersection-direction step.

Common pitfall. Forgetting to verify the found point against both plane equations (it is easy to satisfy only the system used to solve for $x,y$ while silently assuming $z=0$ is consistent) — always back-substitute into both original equations.

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

Question 6 of 8: Plane Through Three Points, and Its Intersection Line With Another Plane

Question 6: Plane Through Three Points, and Its Intersection Line With Another Plane (a) 10, (b) 10 marks