Compare two planes

Decide how two planes meet

Compare the normals before solving simultaneous equations. Parallel normals require a constant check to distinguish coincident and disjoint planes. Nonparallel normals give an intersection line: find a common point and a direction perpendicular to both normals.

The source considers two distinct planes. If their nonzero normals are parallel, the planes are parallel; otherwise they intersect in a straight line. In symbols, parallel normals satisfy $\mathbf n_1=k\mathbf n_2$ for a nonzero real k. For nonparallel normals no such k exists.

Distinguish coincident planes

If distinctness is not given, parallel normals can also describe the same plane. For $\mathbf r\cdot\mathbf n_1=d_1$ and $\mathbf r\cdot\mathbf n_2=d_2$, when $\mathbf n_1=k\mathbf n_2$ the planes coincide exactly when $d_1=kd_2$. Check both the normal and constant; a normal comparison alone does not prove disjointness.

The source’s two-plane relationship activity is retained as an external pointer from its QR code; external activity content has not been reviewed here.

Separate parallel planes by projection

Their separation is the perpendicular distance between them. Choose one point in each plane and project the joining displacement onto a unit common normal. Alternatively, first write both equations with the same normal $\mathbf n$ and then use

$$h=\frac{|d_1-d_2|}{|\mathbf n|}.$$

Matching the normals requires rescaling the constants too. Coincident planes have zero separation.

Find the line common to nonparallel planes

For planes with normals $\mathbf a=(a_1,a_2,a_3)$ and $\mathbf b=(b_1,b_2,b_3)$, the source gives two methods.

Method 1: simultaneous equations using the calculator. Expand to

$$\begin{aligned}a_1x+a_2y+a_3z&=d_1,\\b_1x+b_2y+b_3z&=d_2.\end{aligned}$$

Use the source calculator application PlySmlt2 to express two variables in terms of the third, then read a parametric equation. Choose a free variable that the equations permit; not every coordinate must be a valid free choice in every problem.

Method 2: a common point and cross product. If P is known to belong to both planes, their common line is perpendicular to both normals and therefore has direction $\mathbf m=\mathbf a\times\mathbf b$:

$$\mathbf r=\mathbf p+\lambda\mathbf m,\quad\lambda\in\mathbb R.$$

The source remark requires a point common to both planes, not merely a point in one. Verify that point in both equations. Nonparallel normals ensure $\mathbf m\ne\mathbf0$; parallel normals would give a zero cross product and cannot supply a line direction. The source’s intersection-line activity is retained from its second QR code, without a claim of external-content review.

Derive the acute angle between two planes

Use their normals, not arbitrary in-plane directions. The source refers to the earlier vector-angle formula in section 12.2:

$$\cos\theta=\frac{\mathbf n_1\cdot\mathbf n_2}{|\mathbf n_1||\mathbf n_2|}.$$

When $\theta$ is acute, the plane angle is $\alpha=\theta$. When $\theta$ is obtuse, the smaller plane angle is $\alpha=\pi-\theta$. The second case has two printed methods: take that supplementary angle directly, or use

$$\begin{aligned}\cos(\pi-\alpha)&=\cos\pi\cos\alpha+\sin\pi\sin\alpha\\&=-\cos\alpha.\end{aligned}$$

Hence the obtuse case gives $\alpha=\cos^{-1}[-(\mathbf n_1\cdot\mathbf n_2)/(|\mathbf n_1||\mathbf n_2|)]$. Both orientations combine as

$$\alpha=\cos^{-1}\!\left(\frac{|\mathbf n_1\cdot\mathbf n_2|}{|\mathbf n_1||\mathbf n_2|}\right).$$

The cross-sectional diagrams use equal and supplementary angles to relate the plane angle to the normal angle. Reversing either normal leaves $\alpha$ unchanged. Although the source says “acute”, the formula also covers perpendicular planes at $\pi/2$ and parallel/coincident planes at 0; both normals must remain nonzero. This is a cosine formula for two planes, whereas a line–plane angle uses the sine formula derived earlier.

Use a common line to meet a third plane

Find the common line of $\Pi_1:x+y=2$ and $\Pi_2:y+z=3$. Verify a point and direction for your answer in both planes. Then decide whether that line meets $\Pi_3:x+z=5$, and find the point if it does.

After your attempt: compare the reasoning

Choose $y=t$. Then $x=2-t$ and $z=3-t$, so the common line is

$\mathbf r=(2,0,3)+t(-1,1,-1).$

The base point satisfies both plane equations. The direction is perpendicular to both normals $(1,1,0)$ and $(0,1,1)$; their cross product is $(1,-1,1)$, the negative of our equally valid direction. Substitution into the third plane gives $5-2t=5$, so $t=0$ and the intersection is $(2,0,3)$. The sequential method first finds a line from two planes, then intersects that line with the third plane.

Review the source mastery objectives twice

The source provides separate blank self-assessment columns headed “At the end of lecture” and “At the end of tutorial”.

Revisit every objective at both checkpoints; a completed lesson is not by itself evidence of mastery.

Reflection of a point is explicitly included in the original objectives: find its perpendicular foot F and use the image position $2\mathbf f-\mathbf a$.

The source describes the plane angle by its normals; choose the smaller angle as derived above.