Exercises — Equation of a line in 3D — vector, symmetric, parametric forms

WHAT to do: match each fraction to the template $\frac{x-x_1}{a}$. Numerators (the point). Rewrite so each numerator looks like "$x$ minus something": $x-4=x-(4),y+3=y-(-3),z-1=z-(1).$ So $x_1=4,y_1=-3,z_1=1$. Point $=(4,-3,1)$. Denominators (the direction). $(a,b,c)=(2,-5,6)$. Direction $=(2,-5,6)$. WHY the sign check matters: $y+3$ looks like the point has $y_1=+3$, but the template subtracts, so the point value is $-3$. Always force the "$x-x_1$" shape first.

At $\lambda =0$: the $\lambda \vec{b}$ term vanishes, so $\vec{r}=(1,0,-2)$. This is the anchor $A$. At $\lambda =2$: add two whole steps of $\vec{b}$: $\vec{r}=(1,0,-2)+2(3,3,-1)=(1+6,0+6,-2-2)=(7,6,-4).$ WHY: $\lambda$ counts steps of the direction vector; $\lambda =0$ means "haven't moved," $\lambda =2$ means "two full jumps forward."

WHAT: the DR for $x$ is $0$, so $x$ is frozen — it stays at $5$ forever. WHY you can't write $\frac{x-5}{0}$: division by zero is meaningless; a $0$ in the bottom is code for "this coordinate is constant." Correct symmetric form: $x=5,\frac{y-1}{4}=\frac{z-2}{-3}.$

Vector form (anchor $+\lambda \cdot$direction): $\vec{r}=(0,4,-1)+\lambda (5,-2,1).$ Parametric (read each coordinate separately, all sharing one $\lambda$): $x=5\lambda ,y=4-2\lambda ,z=-1+\lambda .$ Symmetric (solve each for $\lambda$ and set equal): $\frac{x-0}{5}=\frac{y-4}{-2}=\frac{z+1}{1}⟹\frac{x}{5}=\frac{y-4}{-2}=\frac{z+1}{1}.$ WHY they agree: they're the same equation in three costumes; nothing was added or lost, only rewritten.

Direction $=\overrightarrow{PQ}=(2-2,3-(-1),4-4)=(0,4,0)$. WHAT this tells us: two DRs are zero. Both $x$ and $z$ are frozen (they were equal at $P$ and $Q$). Only $y$ moves. Symmetric form: $x=2,z=4,(y\text{ is free}).$ The line is the set $\{(2,t,4):t\in \mathbb{R}\}$ — a straight vertical-in-$y$ line. WHY not write fractions: with $a=c=0$ we'd be dividing by zero twice; the honest statement is just "$x$ and $z$ are pinned."

Set the common value to $\lambda$: each fraction $=\lambda$. $\frac{x+2}{1}=\lambda \Rightarrow x=-2+\lambda ,\frac{y-5}{-4}=\lambda \Rightarrow y=5-4\lambda ,\frac{z}{2}=\lambda \Rightarrow z=2\lambda .$ Parametric: $x=-2+\lambda ,y=5-4\lambda ,z=2\lambda$. Vector: collect constants and $\lambda$-terms: $\vec{r}=(-2,5,0)+\lambda (1,-4,2).$ WHY: the "$=\lambda$" trick is the bridge — symmetric hides $\lambda$, parametric exposes it, vector bundles it back up.

Strategy: find $\lambda$ from one coordinate, then demand the same $\lambda$ works for the other two. From $x$: $7=1+3\lambda \Rightarrow \lambda =2$. Check $y$: $0+(-3)(2)=-6$ ✓. Check $z$: $2+(1)(2)=4$. But we need $5$. $4\ne 5$ ✗. Conclusion: the point is not on the line — one shared $\lambda$ cannot satisfy all three. WHY all three: a single $\lambda$ drives $x,y,z$ together; passing one test is a coincidence, not membership.

Use different letters $\lambda ,\mu$ — the two lines move independently. Set coordinates equal: $1+2\lambda =3+\mu (x),2+\lambda =3-\mu (y),3-\lambda =2+\mu (z).$ From $x$: $\mu =2\lambda -2$. Substitute into $y$: $2+\lambda =3-(2\lambda -2)=5-2\lambda \Rightarrow 3\lambda =3\Rightarrow \lambda =1$, then $\mu =0$. Verify with the leftover equation $z$: $3-\lambda =3-1=2$ and $2+\mu =2+0=2$ ✓. They intersect. Plug $\lambda =1$ into $L_1$: $(1,2,3)+(2,1,-1)=(3,3,2)$. Intersection point $=(3,3,2)$. WHY the third check is essential: two equations give a candidate $(\lambda ,\mu )$; the third equation decides whether the lines truly cross or just pass near each other (skew).

Directions: ${\vec{b}}_1=(2,4,6)$, ${\vec{b}}_2=(1,2,3)$. Since ${\vec{b}}_1=2{\vec{b}}_2$, they are parallel (scalar multiples). Same line? A point on $L_1$ is $A_1=(1,0,-1)$. Test whether $A_1$ lies on $L_2$ using ${\vec{b}}_2$: $\frac{1-4}{1}=-3,\frac{0-6}{2}=-3,\frac{-1-8}{3}=-3.$ All three equal $-3$, so $A_1$ is on $L_2$. Conclusion: same direction and a shared point ⇒ they are the identical line (written two ways). WHY the extra step: parallel lines can be different (never touching) or coincident (fully overlapping). Only a shared point distinguishes them.

Idea: the closest point $F$ is where the segment $OF$ is perpendicular to the direction $\vec{b}=(1,2,2)$. Perpendicular means their dot product is zero. A general point on $L$: $F=(1+\lambda ,2+2\lambda ,3+2\lambda )$. The vector from origin to $F$ is $\overrightarrow{OF}=F$ itself. Demand $\overrightarrow{OF}\cdot \vec{b}=0$: $(1+\lambda )(1)+(2+2\lambda )(2)+(3+2\lambda )(2)=0.$ Expand: $1+\lambda +4+4\lambda +6+4\lambda =11+9\lambda =0\Rightarrow \lambda =-\frac{11}{9}.$ Substitute back: $F=(1-\frac{11}{9},2-\frac{22}{9},3-\frac{22}{9})=(-\frac{2}{9},-\frac{4}{9},\frac{5}{9}).$ WHY the dot product: it answers "which direction is at right angles?" — the shortest link from a point to a line is always the perpendicular one (see Distance of a point from a line in 3D).

Direction $\overrightarrow{AB}=(4-1,7-1,10-1)=(3,6,9)$; simplify by dividing by $3$ to DRs $(1,2,3)$. Line (symmetric): $\frac{x-1}{1}=\frac{y-1}{2}=\frac{z-1}{3}$. Test $C(7,13,19)$: $\frac{7-1}{1}=6,\frac{13-1}{2}=6,\frac{19-1}{3}=6.$ All equal $6$, so $C$ is on the line — the three points are collinear. WHY simplifying DRs is safe: DRs are unique only up to a nonzero scalar, so $(3,6,9)$ and $(1,2,3)$ describe the same direction.

(a) Not parallel. ${\vec{b}}_1=(1,1,0)$, ${\vec{b}}_2=(0,1,1)$. Is one a scalar multiple of the other? If ${\vec{b}}_1=k{\vec{b}}_2$ then $1=k\cdot 0=0$, impossible. So they are not parallel. ✓

(b)–(c) Intersect? Match coordinates ($\lambda$ for $L_1$, $\mu$ for $L_2$): $2+\lambda =0(x),-1+\lambda =\mu (y),3=1+\mu (z).$ From $x$: $\lambda =-2$. From $z$: $\mu =2$. Check $y$: $-1+\lambda =-1+(-2)=-3$, but $\mu =2$. $-3\ne 2$ ✗. The three equations are inconsistent, and the lines are not parallel, so they are skew — they never meet.

(d) Direction perpendicular to both. We need a vector $\vec{n}$ with $\vec{n}\cdot {\vec{b}}_1=0$ and $\vec{n}\cdot {\vec{b}}_2=0$. Let $\vec{n}=(p,q,r)$: $p+q=0(\perp {\vec{b}}_1),q+r=0(\perp {\vec{b}}_2).$ From the first $p=-q$; from the second $r=-q$. Pick $q=1$: $\vec{n}=(-1,1,-1)$. Check: $(-1,1,-1)\cdot (1,1,0)=-1+1+0=0$ ✓; $(-1,1,-1)\cdot (0,1,1)=0+1-1=0$ ✓. DRs of the common perpendicular direction: $(-1,1,-1)$ (or any nonzero multiple, e.g. $(1,-1,1)$). WHY this matters: that mutual perpendicular is exactly the direction along which the shortest distance between skew lines is measured.

• Point from numerators, direction from denominators — rewrite as $x-x_1$ first.
• A zero DR ⇒ that coordinate is frozen; write it as $=\text{const}$, never over zero.
• Membership: one $\lambda$ must satisfy all three coordinates.
• Intersection: use different parameters; solve two equations, verify the third.
• Parallel: directions are scalar multiples. Same line: parallel and share a point.
• Closest point: connecting vector perpendicular to direction ($\cdot =0$).
• Skew: not parallel and no intersection.