Exercises — Angle between two planes

The normal is just the list of coefficients of $x,y,z$ (see Equation of a plane).

• First plane: ${\vec{n}}_1=⟨4,-1,2⟩$.
• Second plane: ${\vec{n}}_2=⟨1,3,-5⟩$. The constant on the right ($7$, or the $+1$) plays no role in orientation — it only slides the sheet, it does not tilt it.

Rewrite as $1x+0y+0z=0$ and $0x+0y+1z=0$, so ${\vec{n}}_1=⟨1,0,0⟩$, ${\vec{n}}_2=⟨0,0,1⟩$.

• Dot: $1\cdot 0+0\cdot 0+0\cdot 1=0$.
• A zero dot product means the arrows are perpendicular, so $\cos \theta =0$ and $\theta =90^{\circ }$. What it looks like: two walls of a room meeting along the $y$-axis in a clean right angle.

• ${\vec{n}}_1=⟨2,-2,1⟩$, ${\vec{n}}_2=⟨1,2,2⟩$.
• Dot: $2(1)+(-2)(2)+1(2)=2-4+2=0$.
• Zero again $\Rightarrow \theta =90^{\circ }$. These planes are perpendicular even though neither is an axis plane.

• ${\vec{n}}_1=⟨2,-1,2⟩$, ${\vec{n}}_2=⟨1,2,2⟩$.
• Dot (WHAT: measure directional agreement): $2(1)+(-1)(2)+2(2)=2-2+4=4$.
• Magnitudes (WHY: normalise out arrow length): $\Vert {\vec{n}}_1\Vert =\sqrt{4+1+4}=3$, $\Vert {\vec{n}}_2\Vert =\sqrt{1+4+4}=3$.
• $\cos \theta =\frac{|4|}{3\cdot 3}=\frac{4}{9}\Rightarrow \theta ={\cos }^{-1}\frac{4}{9}\approx 63.6^{\circ }.$

• ${\vec{n}}_1=⟨1,1,1⟩$, ${\vec{n}}_2=⟨-1,-1,1⟩$.
• Dot: $(1)(-1)+(1)(-1)+(1)(1)=-1-1+1=-1$. Negative!
• $\Vert {\vec{n}}_1\Vert =\Vert {\vec{n}}_2\Vert =\sqrt{3}$.
• The modulus rescues us: $\cos \theta =\frac{|-1|}{\sqrt{3}\sqrt{3}}=\frac{1}{3}\Rightarrow \theta ={\cos }^{-1}\frac{1}{3}\approx 70.5^{\circ }.$ What it looks like (see Figure 2 below): the coral arrow ${\vec{n}}_1$ and the lavender arrow ${\vec{n}}_2$ separate by the small green $70.5^{\circ }$ arc; if we forgot the bars and flipped ${\vec{n}}_2$ to $-{\vec{n}}_2$ (the dotted grey arrow — same plane!), we'd read off the large yellow $109.5^{\circ }$ arc instead. The modulus always keeps the small one.

First, what is $\vec{r}$? It is the position vector $\vec{r}=⟨x,y,z⟩$ — the arrow from the origin to a general point $(x,y,z)$ on the plane. So the statement $\vec{r}\cdot \vec{n}=d$ literally says "for every point on the plane, its position dotted with the fixed vector $\vec{n}$ gives the same number $d$." That fixed $\vec{n}$ is precisely the normal (see Normal vector and dot product), so in this form the normal is read off directly.

• ${\vec{n}}_1=⟨2,2,-3⟩$, ${\vec{n}}_2=⟨3,-3,1⟩$.
• Dot: $2(3)+2(-3)+(-3)(1)=6-6-3=-3$.
• $\Vert {\vec{n}}_1\Vert =\sqrt{4+4+9}=\sqrt{17}$, $\Vert {\vec{n}}_2\Vert =\sqrt{9+9+1}=\sqrt{19}$.
• $\cos \theta =\frac{|-3|}{\sqrt{17}\sqrt{19}}=\frac{3}{\sqrt{323}}\approx 0.1669\Rightarrow \theta \approx 80.4^{\circ }.$

WHY a zero dot product means perpendicular. The Scalar (dot) product obeys ${\vec{n}}_1\cdot {\vec{n}}_2=\Vert {\vec{n}}_1\Vert \Vert {\vec{n}}_2\Vert \cos \phi$, where $\phi$ is the angle between the two arrows. The lengths $\Vert {\vec{n}}_1\Vert ,\Vert {\vec{n}}_2\Vert$ are never zero for real normals, so the whole product is zero exactly when $\cos \phi =0$, i.e. $\phi =90^{\circ }$: the normal arrows are at right angles. And two planes are perpendicular precisely when their normals are perpendicular (tilt the sheets to a right angle and the arrows sticking out of them turn to a right angle too). So: $\text{planes}\perp ⟺\text{normals}\perp ⟺{\vec{n}}_1\cdot {\vec{n}}_2=0.$ Now apply it:

• ${\vec{n}}_1=⟨2,-\lambda ,3⟩$, ${\vec{n}}_2=⟨1,2,-1⟩$.
• Dot: $2(1)+(-\lambda )(2)+3(-1)=2-2\lambda -3=-1-2\lambda$.
• Set to zero: $-1-2\lambda =0\Rightarrow \lambda =-\frac{1}{2}.$

WHY scalar-multiple normals mean parallel planes. Recall the normal points straight out of its sheet, so the normal's direction fixes the sheet's tilt completely (its length is irrelevant — a longer arrow in the same direction sticks out of the same-tilted sheet). If ${\vec{n}}_2=k{\vec{n}}_1$ for some number $k$, the two arrows point along the same line, hence the two sheets have the same tilt — they are parallel (or the very same plane). And parallel arrows meet at $0^{\circ }$, so the angle between the planes is $0^{\circ }$. Now check:

• ${\vec{n}}_1=⟨2,-1,2⟩$, ${\vec{n}}_2=⟨-4,2,-4⟩$.
• Check ratios: $\frac{-4}{2}=\frac{2}{-1}=\frac{-4}{2}=-2$. All equal, so ${\vec{n}}_2=-2{\vec{n}}_1$: parallel.
• Angle between parallel planes is $\theta =0^{\circ }$ (they never open into a wedge). Sanity via formula: $\cos \theta =\frac{|-8-2-8|}{3\cdot 6}=\frac{18}{18}=1\Rightarrow \theta =0^{\circ }.$ ✔

• Dot: $1(1)+1(0)+0(1)=1$.
• $\Vert {\vec{n}}_1\Vert =\sqrt{2}$, $\Vert {\vec{n}}_2\Vert =\sqrt{2}$, product $=2$.
• $\cos \theta =\frac{|1|}{2}=\frac{1}{2}\Rightarrow \theta =60^{\circ }.$ Geometry: both normals have length $\sqrt{2}$ and share only the $x$-component. The overlap ($1$) is exactly half of $\Vert {\vec{n}}_1\Vert \Vert {\vec{n}}_2\Vert =2$, and "half" is the signature cosine of $60^{\circ }$.

The point is a decoy for the angle question — angle needs only the normals.

• ${\vec{n}}_1=⟨3,-1,2⟩$, ${\vec{n}}_2=⟨1,2,1⟩$.
• Dot: $3(1)+(-1)(2)+2(1)=3-2+2=3$.
• $\Vert {\vec{n}}_1\Vert =\sqrt{9+1+4}=\sqrt{14}$, $\Vert {\vec{n}}_2\Vert =\sqrt{1+4+1}=\sqrt{6}$.
• $\cos \theta =\frac{|3|}{\sqrt{14}\sqrt{6}}=\frac{3}{\sqrt{84}}=\frac{3}{2\sqrt{21}}\approx 0.3273\Rightarrow \theta \approx 70.9^{\circ }.$ (The point $(1,2,-1)$ would only matter if we were asked for the plane's equation or a distance.)

The plane $z=0$ has normal ${\vec{n}}_2=⟨0,0,1⟩$. We want $\theta =45^{\circ }$, so $\cos 45^{\circ }=\frac{1}{\sqrt{2}}$.

• Our normal ${\vec{n}}_1=⟨1,0,c⟩$, $\Vert {\vec{n}}_1\Vert =\sqrt{1+c^2}$, $\Vert {\vec{n}}_2\Vert =1$.
• Dot: $1(0)+0(0)+c(1)=c$.
• Equation: $\frac{|c|}{\sqrt{1+c^2}}=\frac{1}{\sqrt{2}}$.
• WHY we may square both sides (WHAT/WHY): WHAT — we square to clear the square root and the modulus. WHY it is safe — both sides are non-negative here ($\frac{|c|}{\sqrt{1+c^2}}\ge 0$ and $\frac{1}{\sqrt{2}}>0$), and squaring is a reversible step whenever both sides are $\ge 0$ (it creates no false solutions and loses none). Squaring also replaces $|c{|}^2$ by $c^2$, erasing the modulus cleanly: $\frac{c^2}{1+c^2}=\frac{1}{2}\Rightarrow 2c^2=1+c^2\Rightarrow c^2=1\Rightarrow c=1(\text{taking }c>0).$
• Normal $⟨1,0,1⟩$; through origin gives $\boxed{x+z=0}$.

• ${\vec{n}}_1=⟨1,1,0⟩$, ${\vec{n}}_2=⟨0,1,1⟩$.
• Dot: $1(0)+1(1)+0(1)=1$; $\Vert {\vec{n}}_1\Vert =\Vert {\vec{n}}_2\Vert =\sqrt{2}$.
• $\cos \theta =\frac{|1|}{2}=\frac{1}{2}\Rightarrow \theta =60^{\circ }.$
• Line of intersection: solve $x=-y$ and $z=-y$; let $y=t$, so $(x,y,z)=(-t,t,-t)=t⟨-1,1,-1⟩$. It runs along $⟨-1,1,-1⟩$ through the origin.

We are given directions inside the plane, not the normal. WHY the cross product? For the angle formula we must feed in the plane's normal, and the cross product $\vec{u}\times \vec{v}$ is by construction perpendicular to both $\vec{u}$ and $\vec{v}$ — so it is perpendicular to the whole plane they span, i.e. it is a normal. That is exactly the direction we lack, so the cross product is the right tool (no other combination of $\vec{u},\vec{v}$ is guaranteed perpendicular to both). How to compute it — the determinant formula: $\vec{u}\times \vec{v}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 0& 1\\ 0& 1& 1\end{array}\right|=\hat{i}(0\cdot 1-1\cdot 1)-\hat{j}(1\cdot 1-1\cdot 0)+\hat{k}(1\cdot 1-0\cdot 0).$

• So ${\vec{n}}_1=⟨-1,-1,1⟩$.
• ${\vec{n}}_2=⟨0,0,1⟩$ for $z=0$.
• Dot: $(-1)(0)+(-1)(0)+(1)(1)=1$; $\Vert {\vec{n}}_1\Vert =\sqrt{3}$, $\Vert {\vec{n}}_2\Vert =1$.
• $\cos \theta =\frac{|1|}{\sqrt{3}}=\frac{1}{\sqrt{3}}\Rightarrow \theta ={\cos }^{-1}\frac{1}{\sqrt{3}}\approx 54.7^{\circ }.$

• ${\vec{n}}_1=⟨1,k,1⟩$, ${\vec{n}}_2=⟨k,-1,k⟩$.
• Dot: $1(k)+k(-1)+1(k)=k-k+k=k$.
• $\Vert {\vec{n}}_1\Vert =\sqrt{2+k^2}$, $\Vert {\vec{n}}_2\Vert =\sqrt{2k^2+1}$.
• Set $\frac{|k|}{\sqrt{2+k^2}\sqrt{2k^2+1}}=\cos 60^{\circ }=\frac{1}{2}$.
• WHY we may square both sides: both sides here are non-negative — the left side is a modulus over positive square roots ($\ge 0$), the right side is $\frac{1}{2}>0$. When both sides of an equation are $\ge 0$, squaring is a reversible step (it neither creates nor loses solutions), and it conveniently kills the $|\cdot |$ since $|k{|}^2=k^2$. Squaring: $\frac{k^2}{(2+k^2)(2k^2+1)}=\frac{1}{4}\Rightarrow 4k^2=(2+k^2)(2k^2+1)=2k^4+5k^2+2.$
• Rearranging: $2k^4+k^2+2=0$. Since $k^2\ge 0$, every term is $\ge 0$ and the constant $2$ keeps the left side $\ge 2>0$ — so there is no real $k$. (Because our squaring was reversible, this "no solution" is genuine, not an artefact.) Interpretation: these two normals can never be tilted only $60^{\circ }$ apart — a genuinely valuable "no solution" answer.

Both have $\vec{n}=⟨2,-1,2⟩$; they are parallel (same tilt, different shift).

• Dot: $4+1+4=9$; $\Vert \vec{n}\Vert =3$ each, product $9$.
• $\cos \theta =\frac{|9|}{9}=1\Rightarrow \theta =0^{\circ }.$ Why: identical normals means identical tilt, so there is no wedge to open — the angle is genuinely $0^{\circ }$, not "undefined". This is the limiting case that closes off the parallel scenario.

For a line and a plane we use sine, because the line vs the normal is the complement (see Angle between a line and a plane): $\sin \theta =\frac{|\vec{b}\cdot \vec{n}|}{\Vert \vec{b}\Vert \Vert \vec{n}\Vert }.$ Step 1 — identify the vectors. The plane $x-y+z=0$ has normal $\vec{n}=⟨1,-1,1⟩$; the line's direction is $\vec{b}=⟨1,1,1⟩$. Step 2 — dot product (numerator). $\vec{b}\cdot \vec{n}=1(1)+1(-1)+1(1)=1-1+1=1.$ Step 3 — magnitudes (denominator). $\Vert \vec{b}\Vert =\sqrt{1+1+1}=\sqrt{3}$, $\Vert \vec{n}\Vert =\sqrt{1+1+1}=\sqrt{3}$, so the product is $3$. Step 4 — assemble and invert. $\sin \theta =\frac{|1|}{3}=\frac{1}{3}\Rightarrow \theta ={\sin }^{-1}\frac{1}{3}\approx 19.5^{\circ }.$ Contrast: for two planes we would have used the same fraction with cosine, giving ${\cos }^{-1}\frac{1}{3}\approx 70.5^{\circ }$. Note $19.5^{\circ }+70.5^{\circ }=90^{\circ }$: the line–plane angle and the "normal-to-normal" angle are complementary — that is the whole reason $\cos \to \sin$. "Cosine for planes, Sine for the line."

L1 Recognition ::: read off $⟨a,b,c⟩$; ignore the constant $d$. L2 Application ::: full formula with modulus and magnitudes. L3 Analysis ::: solve for unknowns; perpendicular ⇔ dot $=0$, parallel ⇔ ratios match. L4 Synthesis ::: build the plane/normal from partial data; ignore decoy points for angles. L5 Mastery ::: cross product for in-plane data, no-solution cases, degenerate $0^{\circ }$, and $\cos$ vs $\sin$.