Exercises — Lines and planes in 3D — vector equations

(a) Set $t=0$: you are sitting at the start point $\mathbf{a}=(2,-1,4)$. Why: $t\mathbf{d}$ vanishes at $t=0$, leaving only the position part.

(b) The direction is the vector multiplied by $t$: $\mathbf{d}=(3,0,-2)$.

(c) We ask: is there one value of $t$ that hits every coordinate of $Q$? $5=2+3t\Rightarrow t=1,-1=-1+0\cdot t✓\text{ (any }t),2=4-2t\Rightarrow t=1.$ All coordinates agree on $t=1$, so yes, $Q$ is on the line (at $t=1$).

(a) The coefficients of $x,y,z$ are the normal (parent, "read it straight off"): $\mathbf{n}=(2,-1,2)$. The constant $7$ is not part of the normal — it only slides the plane along $\mathbf{n}$.

(b) Pick easy coordinates and solve for the last one. Let $x=0,y=0$: then $2z=7\Rightarrow z=3.5$. So $(0,0,3.5)$ works. Check: $2(0)-0+2(3.5)=7$ ✓.

Direction = displacement between the points (parent: "line through two points"): $\mathbf{d}=B-A=(4-1,2-2,5-(-1))=(3,0,6).$ Vector form: $\mathbf{r}=(1,2,-1)+t(3,0,6)$. Parametric: $x=1+3t,y=2,z=-1+6t$. Symmetric: solve each for $t$. But $y$ has direction-component $0$ — we cannot write $\frac{y-2}{0}$. So we state $y=2$ separately: $\frac{x-1}{3}=\frac{z+1}{6},y=2.$ Why the split: dividing by the zero component is illegal; that coordinate is simply frozen.

(a) Use $(\mathbf{r}-\mathbf{a})\cdot \mathbf{n}=0$. Write $\mathbf{r}=(x,y,z)$ and expand the dot product term by term: $(\mathbf{r}-\mathbf{a})\cdot \mathbf{n}=2(x-1)-2(y-1)+1(z-1)=0.$ Why this splits into the standard form: the dot product is linear — distribute it, and it separates into a "moving" part $\mathbf{r}\cdot \mathbf{n}=2x-2y+z$ and a "fixed" part $\mathbf{a}\cdot \mathbf{n}$. Rearranging $(\mathbf{r}-\mathbf{a})\cdot \mathbf{n}=\mathbf{r}\cdot \mathbf{n}-\mathbf{a}\cdot \mathbf{n}=0$ gives $\mathbf{r}\cdot \mathbf{n}=\mathbf{a}\cdot \mathbf{n}$. That right-hand constant is therefore forced to be $d=\mathbf{a}\cdot \mathbf{n}$ — it is just the fixed number the moving part must equal: $2x-2y+z=d,d=\mathbf{a}\cdot \mathbf{n}=2-2+1=1.$ So the plane is $2x-2y+z=1$.

(b) Point-to-plane distance $=\frac{|\mathbf{p}\cdot \mathbf{n}-d|}{|\mathbf{n}|}$. Here $\mathbf{p}=O$, so $\mathbf{p}\cdot \mathbf{n}=0$: $\text{dist}=\frac{|0-1|}{\sqrt{2^2+(-2{)}^2+1^2}}=\frac{1}{\sqrt{9}}=\frac{1}{3}.$

Step 1 — parallel? Directions are ${\mathbf{d}}_1=(1,1,0)$, ${\mathbf{d}}_2=(0,0,1)$. One is not a scalar multiple of the other (no $k$ gives $(1,1,0)=k(0,0,1)$), so not parallel.

Step 2 — do they meet? Set the two general points equal and see if some $(t,s)$ solves all three coordinates (this is a small system): $1+t=0,t=1,0=s.$ From the first equation $t=-1$; from the second $t=1$. Contradiction — no single $t$ works. Therefore the equations have no solution.

Conclusion: not parallel and never meeting $\Rightarrow$ the lines are skew (they pass at different heights, like an overpass over a road — see the figure).

Idea: a point of the line has coordinates $(1+t,1+2t,1+3t)$. It lies on the plane exactly when it satisfies the plane's equation — so substitute (see Intersection of lines and planes): $(1+t)+(1+2t)+(1+3t)=12\Rightarrow 3+6t=12\Rightarrow t=\frac{3}{2}.$ Put $t=\frac{3}{2}$ back into the line: $\left(1+\frac{3}{2},1+3,1+\frac{9}{2}\right)=\left(\frac{5}{2},4,\frac{11}{2}\right).$ Check: $\frac{5}{2}+4+\frac{11}{2}=\frac{5+8+11}{2}=\frac{24}{2}=12$ ✓.

(a) Build two spanning directions inside the plane, then take their cross product for the normal (parent: $\mathbf{n}=\mathbf{u}\times \mathbf{v}$): $\mathbf{u}=B-A=(1,1,0),\mathbf{v}=C-A=(0,1,1).$ $\mathbf{n}=\mathbf{u}\times \mathbf{v}=(1\cdot 1-0\cdot 1,0\cdot 0-1\cdot 1,1\cdot 1-1\cdot 0)=(1,-1,1).$ Plane: $(\mathbf{r}-\mathbf{a})\cdot \mathbf{n}=0$ with $\mathbf{a}=A=(1,0,0)$: $1(x-1)-1(y)+1(z)=0\Rightarrow x-y+z=1.$ Check each point: $A:1-0+0=1$ ✓, $B:2-1+0=1$ ✓, $C:1-1+1=1$ ✓.

(b) Distance from $P=(3,3,3)$ using $\frac{|\mathbf{p}\cdot \mathbf{n}-d|}{|\mathbf{n}|}$ with $d=1$: $\mathbf{p}\cdot \mathbf{n}=3-3+3=3,|\mathbf{n}|=\sqrt{1+1+1}=\sqrt{3}.$ $\text{dist}=\frac{|3-1|}{\sqrt{3}}=\frac{2}{\sqrt{3}}=\frac{2\sqrt{3}}{3}\approx \mathrm{1.155.}$

Use $\text{dist}=\frac{|(\mathbf{p}-\mathbf{a})\times \mathbf{d}|}{|\mathbf{d}|}$. Why the cross product gives the perpendicular distance: picture the parallelogram with sides $\mathbf{p}-\mathbf{a}$ (from the line's point $\mathbf{a}$ to the target $P$) and $\mathbf{d}$ (the line's direction). The magnitude of a cross product is the area of that parallelogram: $|(\mathbf{p}-\mathbf{a})\times \mathbf{d}|=|\mathbf{p}-\mathbf{a}||\mathbf{d}|\sin \theta$, where $\theta$ is the angle between the two sides. But area also equals $\text{base}\times \text{height}$, with base $|\mathbf{d}|$ and height = the perpendicular gap from $P$ to the line. So dividing the area by the base $|\mathbf{d}|$ leaves exactly that perpendicular height — precisely the distance we want. $\mathbf{p}-\mathbf{a}=(1,1,1),\mathbf{d}=(1,0,0).$ $(\mathbf{p}-\mathbf{a})\times \mathbf{d}=(1\cdot 0-1\cdot 0,1\cdot 1-1\cdot 0,1\cdot 0-1\cdot 1)=(0,1,-1).$ $|(0,1,-1)|=\sqrt{0+1+1}=\sqrt{2},|\mathbf{d}|=1.$ $\text{dist}=\frac{\sqrt{2}}{1}=\sqrt{2}\approx \mathrm{1.414.}$ Sanity: the closest point on the $x$-axis to $(1,1,1)$ is $(1,0,0)$; the gap is $(0,1,1)$ of length $\sqrt{2}$ ✓.

(a) Intersection. A line point is $(1+2t,2-t,3+2t)$. Substitute into $\Pi$: $2(1+2t)+(2-t)-2(3+2t)=3.$ $2+4t+2-t-6-4t=3\Rightarrow (-2)+(-t)=3\Rightarrow -t=5\Rightarrow t=-5.$ Point: $(1+2(-5),2-(-5),3+2(-5))=(-9,7,-7)$. Check: $2(-9)+7-2(-7)=-18+7+14=3$ ✓.

(b) Angle between line and plane. First, what does this angle mean? Let $\varphi$ be the acute angle between the line and the plane — that is, tilt the line down toward the surface and measure the small wedge between the line and its own shadow on the plane. When $\varphi =0^{\circ }$ the line lies flat in the plane; when $\varphi =90^{\circ }$ the line stabs straight through, parallel to the normal. That is the quantity we are computing.

A line's arrow points along it; a plane's normal points across it (parent mistake (d) warns not to mix these). The angle $\varphi$ satisfies $\sin \varphi =\frac{|\mathbf{d}\cdot \mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}.$ Why $\sin$, not $\cos$? The dot product gives the angle between $\mathbf{d}$ and the normal $\mathbf{n}$. But the plane sits $90^{\circ }$ from its normal, so the line-to-plane angle $\varphi$ is the complement of the line-to-normal angle — and $\cos$ of an angle equals $\sin$ of its complement. $\mathbf{d}=(2,-1,2),\mathbf{n}=(2,1,-2):\mathbf{d}\cdot \mathbf{n}=4-1-4=-1.$ $|\mathbf{d}|=\sqrt{4+1+4}=3,|\mathbf{n}|=\sqrt{4+1+4}=3.$ $\sin \varphi =\frac{|-1|}{3\cdot 3}=\frac{1}{9}\Rightarrow \varphi =\arcsin \frac{1}{9}\approx 6.38^{\circ }.$

(c) Foot of perpendicular from $O$ and distance. Drop a line from $O$ along the normal $\mathbf{n}$: $\mathbf{r}=(0,0,0)+\lambda (2,1,-2)$. It hits $\Pi$ when $2(2\lambda )+(1\lambda )-2(-2\lambda )=3\Rightarrow 4\lambda +\lambda +4\lambda =3\Rightarrow 9\lambda =3\Rightarrow \lambda =\frac{1}{3}.$ Foot: $\left(\frac{2}{3},\frac{1}{3},-\frac{2}{3}\right)$. Distance = length of $O\to$foot $=|\lambda ||\mathbf{n}|=\frac{1}{3}\cdot 3=1$. Cross-check with the direct formula: $\frac{|\mathbf{p}\cdot \mathbf{n}-d|}{|\mathbf{n}|}=\frac{|0-3|}{3}=1$ ✓.

Substitute the line point $(1+t,t,1-t)$ into the plane: $(1+t)-(t)+(1-t)=2\Rightarrow 2-t=2\Rightarrow t=0.$ Wait — that gives a single $t$. Re-examine: $1+t-t+1-t=2-t$. Setting $2-t=2$ gives $t=0$, one point $(1,0,1)$.

But the interesting degenerate case is when the line's direction is perpendicular to the normal (so the line runs parallel to the plane). Check: $\mathbf{d}\cdot \mathbf{n}=(1,1,-1)\cdot (1,-1,1)=1-1-1=-1\ne 0$, so the line is not parallel — it pierces the plane once, at $(1,0,1)$.

The general rule (all cases):

• $\mathbf{d}\cdot \mathbf{n}\ne 0$ → line pierces plane in exactly one point.
• $\mathbf{d}\cdot \mathbf{n}=0$ and the start point is off the plane → line is parallel, never touches → no solution.
• $\mathbf{d}\cdot \mathbf{n}=0$ and the start point is on the plane → line lies inside the plane → infinitely many points.

Here the first case applies: one point, $(1,0,1)$.