Exercises — Skew lines — shortest distance

What we do: compare the two direction vectors ${\vec{b}}_1=(2,3,-1)$ and ${\vec{b}}_2=(4,6,-2)$. Why: two lines are parallel exactly when one direction is a scalar multiple of the other. Here ${\vec{b}}_2=(4,6,-2)=2(2,3,-1)=2{\vec{b}}_1$. Every component doubles. So ${\vec{b}}_1\parallel {\vec{b}}_2$ → the lines are parallel. But wait — parallel or the same line? Parallel lines can be genuinely apart, or they can be the identical line written two ways (coincident). To separate these, take the gap ${\vec{a}}_2-{\vec{a}}_1=(0-1,0-1,3-0)=(-1,-1,3)$ and ask: does it point along ${\vec{b}}_1$? If $(-1,-1,3)=t(2,3,-1)$, the first component gives $t=-\frac{1}{2}$, but then the second would be $-\frac{3}{2}\ne -1$. Not a multiple ⇒ the gap sticks out sideways ⇒ the lines are distinct parallel lines, not coincident. Answer: distinct parallel lines (positive shortest distance, computed by the parallel formula in L4).

What we do: test if $(1,-1,1)$ is a multiple of $(2,1,1)$. Why: if NOT parallel, the pair is either intersecting or skew — never parallel; we then need the triple-product test to separate those two. For now we just rule parallel in or out. Suppose $(1,-1,1)=k(2,1,1)$. From the first component $k=\frac{1}{2}$, but then the second would be $\frac{1}{2}\ne -1$. Contradiction. Answer: not parallel → the pair is intersecting or skew (decided in L3).

Step 1 — the gap. ${\vec{a}}_2-{\vec{a}}_1=(4-1,5-2,6-3)=(3,3,3).$ Why: this is the arrow between chosen points; we will keep only its along-perpendicular part. Step 2 — the common perpendicular. ${\vec{b}}_1\times {\vec{b}}_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -3& 2\\ 2& 3& 1\end{array}\right|=\hat{i}(-3-6)-\hat{j}(1-4)+\hat{k}(3+6)=(-9,3,9).$ Why: the cross product is at right angles to both lines — the only direction the shortest stick can point. Step 3 — length of that direction. $|{\vec{b}}_1\times {\vec{b}}_2|=\sqrt{81+9+81}=\sqrt{171}=3\sqrt{19}.$ Why: this is the denominator of the distance formula — it turns the raw cross product into a unit direction $\hat{n}$, so the projection we compute next comes out as a true length, not a scaled one. Step 4 — project the gap. $(3,3,3)\cdot (-9,3,9)=-27+9+27=9.$ Why: this dot product is the numerator — it measures how much of the gap arrow lies along the common-perpendicular direction. Only that component is the shortest distance; the rest of the gap runs parallel to the lines and doesn't count. Answer: $d=\frac{|9|}{3\sqrt{19}}=\frac{3}{\sqrt{19}}=\frac{3\sqrt{19}}{19}.$

Gap: ${\vec{a}}_2-{\vec{a}}_1=(-3-3,-7-8,6-3)=(-6,-15,3).$ Cross: ${\vec{b}}_1\times {\vec{b}}_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& -1& 1\\ -3& 2& 4\end{array}\right|=\hat{i}(-4-2)-\hat{j}(12+3)+\hat{k}(6-3)=(-6,-15,3).$ Length: $|{\vec{b}}_1\times {\vec{b}}_2|=\sqrt{36+225+9}=\sqrt{270}=3\sqrt{30}.$ Why this length: it is the denominator that rescales the cross product to a unit direction, so the projection below is a genuine distance. Numerator: $(-6,-15,3)\cdot (-6,-15,3)=36+225+9=270.$ Why this dot product: it extracts the part of the gap pointing along the common perpendicular — the only part that contributes to the shortest distance. Answer: $d=\frac{270}{3\sqrt{30}}=\frac{90}{\sqrt{30}}=\frac{90\sqrt{30}}{30}=3\sqrt{30}.$ Neat coincidence: here the gap happens to point exactly along the common perpendicular, so all of it counts.

Gap: ${\vec{a}}_2-{\vec{a}}_1=(4-1,0-1,-1+1)=(3,-1,0).$ Cross: ${\vec{b}}_1\times {\vec{b}}_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& -1& 0\\ 2& 0& 3\end{array}\right|=\hat{i}(-3-0)-\hat{j}(9-0)+\hat{k}(0+2)=(-3,-9,2).$ Triple product: $(3,-1,0)\cdot (-3,-9,2)=-9+9+0=0.$ Interpret (why this matters): the numerator is the Scalar Triple Product — the volume of the box built from the three arrows. Volume $=0$ means the three arrows lie in one plane, i.e. the two lines are coplanar. Coplanar and not parallel ⇒ they must cross. Answer: the lines INTERSECT ($d=0$), they are not skew.

Gap: ${\vec{a}}_2-{\vec{a}}_1=(1-2,2+1,0-1)=(-1,3,-1).$ Cross: ${\vec{b}}_1\times {\vec{b}}_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& -1\\ 1& -1& 2\end{array}\right|=\hat{i}(2-1)-\hat{j}(2+1)+\hat{k}(-1-1)=(1,-3,-2).$ Length: $\sqrt{1+9+4}=\sqrt{14}.$ Triple product: $(-1,3,-1)\cdot (1,-3,-2)=-1-9+2=-8\ne 0.$ Non-zero → box has volume → not coplanar → skew. Answer: $d=\frac{|-8|}{\sqrt{14}}=\frac{8}{\sqrt{14}}=\frac{8\sqrt{14}}{14}=\frac{4\sqrt{14}}{7}.$

Recognise first: ${\vec{b}}_2=(4,6,8)=2(2,3,4)=2{\vec{b}}_1$ → parallel. The cross product ${\vec{b}}_1\times {\vec{b}}_2=\vec{0}$, so the main formula gives $0/0$ and is useless here. Coincident check: gap ${\vec{a}}_2-{\vec{a}}_1=(1,2,2)$; is it a multiple of $\vec{b}=(2,3,4)$? If $(1,2,2)=t(2,3,4)$ then $t=\frac{1}{2}$ from the first component, but the second would be $\frac{3}{2}\ne 2$. Not a multiple ⇒ distinct parallel lines, so the distance is positive and worth computing. Switch tools: for parallel lines use the point-to-line distance with $\vec{b}=(2,3,4)$: $d=\frac{|\vec{b}\times ({\vec{a}}_2-{\vec{a}}_1)|}{|\vec{b}|}.$ Cross: $\vec{b}\times ({\vec{a}}_2-{\vec{a}}_1)=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 3& 4\\ 1& 2& 2\end{array}\right|=\hat{i}(6-8)-\hat{j}(4-4)+\hat{k}(4-3)=(-2,0,1).$ Lengths: $|(-2,0,1)|=\sqrt{5}$, $|\vec{b}|=\sqrt{4+9+16}=\sqrt{29}.$ Answer: $d=\frac{\sqrt{5}}{\sqrt{29}}=\sqrt{\frac{5}{29}}=\frac{\sqrt{145}}{29}.$

What we want: intersection ⇔ the three arrows are coplanar ⇔ triple product $=0$. So set it up and solve for $k$. Gap: ${\vec{a}}_2-{\vec{a}}_1=(2-1,3-1,k-1)=(1,2,k-1).$ Cross: ${\vec{b}}_1\times {\vec{b}}_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 3& 4\\ 3& 4& 5\end{array}\right|=\hat{i}(15-16)-\hat{j}(10-12)+\hat{k}(8-9)=(-1,2,-1).$ Triple product $=0$: $(1,2,k-1)\cdot (-1,2,-1)=-1+4-(k-1)=4-k=0.$ Answer: $k=4$. (For any other $k$ the lines are skew, since the directions aren't parallel.)

Part A — distance. Gap: ${\vec{a}}_2-{\vec{a}}_1=(15-8,29-(-9),5-10)=(7,38,-5).$ (Note the middle entry: $29-(-9)=29+9=38$ — subtracting a negative adds.) Why: this is the arrow between the two given base points; we will keep only the part of it pointing perpendicular to both lines. Cross: ${\vec{b}}_1\times {\vec{b}}_2=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& -16& 7\\ 3& 8& -5\end{array}\right|=\hat{i}(80-56)-\hat{j}(-15-21)+\hat{k}(24+48)=(24,36,72).$ Why: the cross product is, by definition, at right angles to both ${\vec{b}}_1$ and ${\vec{b}}_2$ — it is the direction the shortest stick $PQ$ must run along. Simplify direction: $(24,36,72)=12(2,3,6)$, so $|{\vec{b}}_1\times {\vec{b}}_2|=12\sqrt{4+9+36}=12\cdot 7=84.$ Numerator: $(7,38,-5)\cdot (24,36,72)=168+1368-360=1176.$ Why: this dot product measures how much of the gap arrow lies along the common-perpendicular direction — precisely the length that survives after the along-line parts are discarded. $d=\frac{1176}{84}=14.$

Part B — the touching points, step by step. Let $P={\vec{a}}_1+\lambda {\vec{b}}_1$ (a moving point on $L_1$) and $Q={\vec{a}}_2+\mu {\vec{b}}_2$ (a moving point on $L_2$). The stick $\overrightarrow{PQ}=Q-P$ must be perpendicular to both directions: $\overrightarrow{PQ}\cdot {\vec{b}}_1=0,\overrightarrow{PQ}\cdot {\vec{b}}_2=0.$ Why these two equations: perpendicular to both lines is exactly the defining property of the shortest segment (see the page-top figure, where the amber stick meets each line at a square corner). Two conditions, two unknowns $\lambda ,\mu$ — a clean $2\times 2$ system. Write the stick using the gap $\vec{g}={\vec{a}}_2-{\vec{a}}_1=(7,38,-5)$: $\overrightarrow{PQ}=\vec{g}+\mu {\vec{b}}_2-\lambda {\vec{b}}_1=(7+3\mu -3\lambda ,38+8\mu +16\lambda ,-5-5\mu -7\lambda ).$ Equation 1 — dot with ${\vec{b}}_1=(3,-16,7)$, term by term:

• $3\cdot (7+3\mu -3\lambda )=21+9\mu -9\lambda$
• $-16\cdot (38+8\mu +16\lambda )=-608-128\mu -256\lambda$
• $7\cdot (-5-5\mu -7\lambda )=-35-35\mu -49\lambda$

Add: $(21-608-35)+(9-128-35)\mu +(-9-256-49)\lambda =-622-154\mu -314\lambda =0.$ $\boxed{314\lambda +154\mu =-622.}(\text{i})$ Equation 2 — dot with ${\vec{b}}_2=(3,8,-5)$, term by term:

• $3\cdot (7+3\mu -3\lambda )=21+9\mu -9\lambda$
• $8\cdot (38+8\mu +16\lambda )=304+64\mu +128\lambda$
• $-5\cdot (-5-5\mu -7\lambda )=25+25\mu +35\lambda$

Add: $(21+304+25)+(9+64+25)\mu +(-9+128+35)\lambda =350+98\mu +154\lambda =0.$ $\boxed{154\lambda +98\mu =-350.}(\text{ii})$ Solve (i) & (ii). Eliminate $\mu$: multiply (i) by $49$ and (ii) by $77$, since $154\cdot 49=98\cdot 77=7546$. $(\text{i})\times 49:15386\lambda +7546\mu =-30478.$ $(\text{ii})\times 77:11858\lambda +7546\mu =-26950.$ Subtract the second from the first: $3528\lambda =-3528\Rightarrow \lambda =-1.$ Put $\lambda =-1$ into (ii): $154(-1)+98\mu =-350\Rightarrow -154+98\mu =-350\Rightarrow 98\mu =-196\Rightarrow \mu =-2.$ Feet (substitute back): $P={\vec{a}}_1+(-1){\vec{b}}_1=(8-3,-9+16,10-7)=(5,7,3),$ $Q={\vec{a}}_2+(-2){\vec{b}}_2=(15-6,29-16,5+10)=(9,13,15).$ Check the length: $\overrightarrow{PQ}=(9-5,13-7,15-3)=(4,6,12)$, and $|\overrightarrow{PQ}|=\sqrt{16+36+144}=\sqrt{196}=14.$ ✓ This matches Part A, so the feet are correct — and these are exactly the cyan point $P$ and white point $Q$ marked in the page-top figure. Answer: $d=14$, with $P=(5,7,3)$ on $L_1$ and $Q=(9,13,15)$ on $L_2$.

Read off (see Equation of a Line in 3D): point $(1,2,3)$, direction $(2,3,4)$; point $(2,4,5)$, direction $(3,4,5)$. Gap: $(2-1,4-2,5-3)=(1,2,2).$ Cross: $(2,3,4)\times (3,4,5)=(15-16,12-10,8-9)=(-1,2,-1).$ Length: $\sqrt{1+4+1}=\sqrt{6}.$ Numerator: $(1,2,2)\cdot (-1,2,-1)=-1+4-2=1.$ Answer: $d=\frac{1}{\sqrt{6}}=\frac{\sqrt{6}}{6}.$ This is the Cartesian determinant formula from the parent note, just done component-by-component.

Which quantity decides parallel vs not-parallel? ::: Whether ${\vec{b}}_1$ is a scalar multiple of ${\vec{b}}_2$. Which quantity decides intersecting vs skew (given not parallel)? ::: The scalar triple product $({\vec{a}}_2-{\vec{a}}_1)\cdot ({\vec{b}}_1\times {\vec{b}}_2)$ — zero ⇒ intersect, non-zero ⇒ skew. How do you tell distinct parallel lines from coincident ones? ::: Test if the gap ${\vec{a}}_2-{\vec{a}}_1$ is a multiple of $\vec{b}$; if yes, coincident ($d=0$); if no, distinct parallel. Distance formula for parallel lines? ::: $d=\frac{|\vec{b}\times ({\vec{a}}_2-{\vec{a}}_1)|}{|\vec{b}|}$. How do you find the actual feet of the common perpendicular? ::: Solve $\overrightarrow{PQ}\cdot {\vec{b}}_1=0$ and $\overrightarrow{PQ}\cdot {\vec{b}}_2=0$ for $\lambda ,\mu$, then check $|\overrightarrow{PQ}|$ equals the formula distance.