Plane Through Three Points and Its Intersection with a Second Plane: 04-BS-1, May 2017, Q5

Question 5: Plane Through Three Points and Its Intersection with a Second 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 $P$, and a second plane $x+y-2z=3$.

Plane $P$ through the three given points meets the second plane along a line through $(1,2,0)$, direction $(0,2,1)$.

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

Approach. (a) Build two vectors from the three points and cross them for a normal. (b) Cross the two planes' normals for the line's direction, then find one shared point.

(a) Two in-plane vectors. With $A=(2,1,-2)$, $B=(1,2,0)$, $C=(1,0,-1)$: $\overrightarrow{AB}=(-1,1,2)$, $\overrightarrow{AC}=(-1,-1,1)$.
(a) Normal via cross product. $$\mathbf n_P=\overrightarrow{AB}\times\overrightarrow{AC}=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\-1&1&2\\-1&-1&1\end{vmatrix}=(1(1)-2(-1),\,-(-1(1)-2(-1)),\,(-1)(-1)-1(-1))=(3,-1,2).$$
(a) Write the plane. Through $A=(2,1,-2)$ with normal $(3,-1,2)$: $$3(x-2)-(y-1)+2(z+2)=0\ \Rightarrow\ \boxed{3x-y+2z=1}$$ (Check: all three points satisfy $3x-y+2z=1$ — $3(2)-1+2(-2)=1$, $3(1)-2+0=1$, $3(1)-0+2(-1)=1$ ✓.)
(b) Direction of the intersection line. With $\mathbf n_P=(3,-1,2)$ and $\mathbf n_2=(1,1,-2)$ (normal of $x+y-2z=3$), $$\mathbf d=\mathbf n_P\times\mathbf n_2=\big((-1)(-2)-2(1),\,-\!\big(3(-2)-2(1)\big),\,3(1)-(-1)(1)\big)=(0,8,4)\ \parallel\ (0,2,1).$$
(b) A point common to both planes. The point $(1,2,0)$ (one of the three given points) already lies on $P$ by construction; checking the second plane: $1+2-2(0)=3$ ✓, so it lies on both. Using it directly avoids solving a fresh system.
(b) Write the line. $$\boxed{(x,y,z)=(1,2,0)+t(0,2,1),\quad t\in\mathbb R}$$

Quantity	Result
(a) Normal $\mathbf n_P$	$(3,-1,2)$
(a) Plane $P$	$3x-y+2z=1$
(b) Direction $\mathbf d$	$(0,2,1)$
(b) Intersection line	$(x,y,z)=(1,2,0)+t(0,2,1)$

Concept & Theory

Topic. Plane through three points and the line of intersection of two planes.

Key relations. A plane through $A,B,C$ has normal $\overrightarrow{AB}\times\overrightarrow{AC}$. Two non-parallel planes with normals $\mathbf n_1,\mathbf n_2$ intersect in a line with direction $\mathbf n_1\times\mathbf n_2$ (perpendicular to both normals, hence lying in both planes).

Why this works. Part (b) reuses part (a)'s normal $(3,-1,2)$ directly in the cross product, and one of the three original points turns out to already satisfy the second plane's equation — recognizing this saves solving a $2\times2$ system for an anchor point from scratch.

Common pitfall. Sign errors in the $3\times3$ cross-product's middle ($\mathbf j$) component, which carries an extra minus sign; not checking whether a known point already lies on both planes before setting up a fresh system to find an anchor point.

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

Question 5 of 8: Plane Through Three Points and Its Intersection with a Second Plane

Question 5: Plane Through Three Points and Its Intersection with a Second Plane (a) 10, (b) 10 marks