Exercises — Equation of a plane — normal form, intercept form, general form

WHAT we do: just read the coefficients of $x,y,z$. WHY it works: the general form $ax+by+cz+d=0$ is literally $\vec{n}\cdot \vec{r}+d=0$, so the coefficients are the components of $\vec{n}$ — see the pencil in the figure above. (a) $\vec{n}=(4,-3,12)$. (b) $|\vec{n}|=\sqrt{4^2+(-3{)}^2+12^2}=\sqrt{16+9+144}=\sqrt{169}=13$.

The test for normal form $lx+my+nz=p$: the coefficients must be direction cosines, i.e. $l^2+m^2+n^2=1$, and $p\ge 0$. (i) $1^2+2^2+2^2=9\ne 1$ — not normal form (it's general form). (ii) ${\left(\frac{1}{3}\right)}^2+{\left(\frac{2}{3}\right)}^2+{\left(\frac{2}{3}\right)}^2=\frac{1}{9}+\frac{4}{9}+\frac{4}{9}=1$ ✓ and $p=3\ge 0$ ✓ — this is normal form. (iii) RHS $=1$ with each variable divided by a number — that's intercept form, not normal.

Intercept form $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ reads intercepts off the denominators. Intercepts: $a=5$, $b=-2$, $c=4$. Meaning the plane pierces the axes at $(5,0,0)$, $(0,-2,0)$, $(0,0,4)$.

Step 1 — move constant across. $2x-2y+z=-6$. Step 2 — compute $|\vec{n}|$. $|\vec{n}|=\sqrt{2^2+(-2{)}^2+1^2}=\sqrt{4+4+1}=3$. Step 3 — divide, choosing the sign so RHS $\ge 0$. Dividing by $+3$ gives RHS $=-6/3=-2<0$ — not allowed, since $p$ is a distance. So divide by $-3$: $\frac{-2}{3}x+\frac{2}{3}y-\frac{1}{3}z=2.$ Result: $\hat{n}=\left(-\frac{2}{3},\frac{2}{3},-\frac{1}{3}\right)$, $p=2$. Check unit length: $\frac{4}{9}+\frac{4}{9}+\frac{1}{9}=1$ ✓. This $p$ is exactly the origin-to-plane distance from Distance of a point from a plane with the point being $O$.

WHY divide by 12: intercept form needs RHS $=1$. $\frac{3x}{12}+\frac{4y}{12}-\frac{6z}{12}=1\Rightarrow \frac{x}{4}+\frac{y}{3}+\frac{z}{-2}=1.$ Intercepts: $x$: $4$, $y$: $3$, $z$: $-2$.

Point–normal form: $\vec{n}\cdot (\vec{r}-{\vec{r}}_0)=0$, i.e. $1(x-2)+4(y+1)-2(z-3)=0.$ Expand: $x-2+4y+4-2z+6=0\Rightarrow x+4y-2z+8=0.$ Check: plug $(2,-1,3)$: $2-4-6+8=0$ ✓.

WHY cross product: the normal must be perpendicular to two directions lying in the plane; the cross product $\overrightarrow{AB}\times \overrightarrow{AC}$ is perpendicular to both at once. $\overrightarrow{AB}=B-A=(1,-1,1)$, $\overrightarrow{AC}=C-A=(-1,1,1)$. $\vec{n}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -1& 1\\ -1& 1& 1\end{array}\right|=\hat{i}(-1\cdot 1-1\cdot 1)-\hat{j}(1\cdot 1-1\cdot (-1))+\hat{k}(1\cdot 1-(-1)(-1)).$ $=\hat{i}(-1-1)-\hat{j}(1+1)+\hat{k}(1-1)=(-2,-2,0).$ Use point $A(1,1,0)$: $-2(x-1)-2(y-1)+0=0\Rightarrow -2x-2y+4=0\Rightarrow x+y=2.$ Check: $B$: $2+0=2$ ✓, $C$: $0+2=2$ ✓. The zero $z$-normal means the plane is parallel to the $z$-axis — sensible, since $A,B,C$ climb in $z$ but the plane leans only in $x,y$.

WHY this formula: the signed distance is the projection of $\overrightarrow{QP}$ (from any plane-point $Q$ to $P$) onto the unit normal — see Distance of a point from a plane. It collapses to $\text{dist}=\frac{|a{x}_0+b{y}_0+c{z}_0+d|}{\sqrt{a^2+b^2+c^2}}.$ Numerator: $|2\cdot 3-2+2\cdot 1+3|=|6-2+2+3|=9$. Denominator: $\sqrt{4+1+4}=3$. Distance $=9/3=3$.

WHY the normals: the angle between planes equals the angle between their normals (see Angle between two planes), found via the dot product. ${\vec{n}}_1=(1,1,1)$, ${\vec{n}}_2=(1,-1,1)$. $\cos \theta =\frac{|{\vec{n}}_1\cdot {\vec{n}}_2|}{|{\vec{n}}_1||{\vec{n}}_2|}=\frac{|1-1+1|}{\sqrt{3}\cdot \sqrt{3}}=\frac{1}{3}.$ $\theta =\arccos \frac{1}{3}\approx 70.53^{\circ }$. (We take absolute value in the numerator to force the acute angle.)

Idea: if the plane is parallel to two directions, its normal is perpendicular to both — take $\vec{n}=\vec{u}\times \vec{v}$. $\vec{n}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 1\\ 2& -1& 3\end{array}\right|=\hat{i}(2\cdot 3-1\cdot (-1))-\hat{j}(1\cdot 3-1\cdot 2)+\hat{k}(1\cdot (-1)-2\cdot 2).$ $=\hat{i}(6+1)-\hat{j}(3-2)+\hat{k}(-1-4)=(7,-1,-5).$ Through $(1,0,2)$: $7(x-1)-1(y)-5(z-2)=0\Rightarrow 7x-y-5z+3=0.$ Check $(1,0,2)$: $7-0-10+3=0$ ✓.

WHY they're parallel: identical normals $(2,-1,2)$. Distance between parallel planes = difference of their $p$-values along the same unit normal. Write both as $\dots =d$: constants $5$ and $-4$, so $|d_1-d_2|=|5-(-4)|=9$. Divide by $|\vec{n}|=\sqrt{4+1+4}=3$: $\text{distance}=\frac{|5-(-4)|}{3}=\frac{9}{3}=3.$

Two in-plane directions: the line's direction $\vec{u}=(1,1,0)$, and the vector from the line's base $(0,1,2)$ to the point $(2,0,1)$, namely $\vec{w}=(2,-1,-1)$. Normal $\vec{n}=\vec{u}\times \vec{w}$: $\vec{n}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& 0\\ 2& -1& -1\end{array}\right|=\hat{i}(1\cdot (-1)-0\cdot (-1))-\hat{j}(1\cdot (-1)-0\cdot 2)+\hat{k}(1\cdot (-1)-1\cdot 2).$ $=\hat{i}(-1)-\hat{j}(-1)+\hat{k}(-3)=(-1,1,-3).$ Through base $(0,1,2)$: $-1(x)+1(y-1)-3(z-2)=0\Rightarrow -x+y-3z+5=0,$ i.e. $x-y+3z-5=0.$ Check point $(2,0,1)$: $2-0+3-5=0$ ✓. Check line base $(0,1,2)$: $0-1+6-5=0$ ✓.

Geometry: the reflection $O^{\prime }$ lies along the normal direction from $O$, at twice the foot-of-perpendicular distance. Move from $O$ by $2p$ along $\hat{n}$. Unit normal: $\hat{n}=\frac{1}{3}(2,-1,2)$ (since $|\vec{n}|=3$). Distance $p=9/3=3$ (RHS already positive). Foot of perpendicular $N=O+p\hat{n}=3\cdot \frac{1}{3}(2,-1,2)=(2,-1,2)$. Reflection $O^{\prime }=O+2p\hat{n}=2N=(4,-2,4)$. Check: midpoint of $O$ and $O^{\prime }$ is $(2,-1,2)$, which satisfies $2(2)-(-1)+2(2)=4+1+4=9$ ✓.

Idea (pencil of planes): any plane through the intersection line can be written $(x+y+z-1)+\lambda (2x+3y-z-4)=0.$ Impose the point $(1,1,1)$: $(1+1+1-1)+\lambda (2+3-1-4)=0\Rightarrow 2+\lambda (0)=0.$ This gives $2=0$ — impossible! So no finite $\lambda$ works: the required plane is the second one alone, the $\lambda \to \infty$ member $2x+3y-z-4=0$. Check $(1,1,1)$: $2+3-1-4=0$ ✓. Lesson: the pencil $P_1+\lambda P_2$ misses exactly $P_2$ itself; when the point already lies on $P_2$, that is your answer. Equation: $2x+3y-z=4$.

Intercept form with equal intercepts: $\frac{x}{a}+\frac{y}{a}+\frac{z}{a}=1\Rightarrow x+y+z=a.$ Normal $\vec{n}=(1,1,1)$, $|\vec{n}|=\sqrt{3}$. Distance from origin: $p=\frac{|a|}{\sqrt{3}}=\sqrt{3}\Rightarrow |a|=3.$ Positive intercepts ⇒ $a=3$. Plane: $x+y+z=3$. Check: distance $=|-3|/\sqrt{3}=3/\sqrt{3}=\sqrt{3}$ ✓ (writing it as $x+y+z-3=0$, origin gives $|-3|/\sqrt{3}$).