Chapter 11 · Three Dimensional Geometry

11.3 Angle between Two Lines and Shortest Distance

Two lines in space may meet, run parallel, or pass each other without ever meeting. A dot product gives the angle between them; a cross product followed by a dot product gives the shortest distance between them.

In a factory, a hospital or a metro station, the ceiling space is crowded with straight runs of ducts, cables, sprinkler pipes and conveyor rails. Designers must answer two questions about every pair of them. At what angle do they cross? — a pipe joint or a support bracket is made for one particular angle. And how close do they come? — a hot-water pipe must keep a minimum distance from an electric cable. On a flat drawing two straight lines that are not parallel always meet. In a building they usually do not: one runs higher than the other, and they pass each other without touching. Such lines are called skew lines.

With the equations of Section 11.2 both questions have short answers. The angle between two lines is the angle between their direction vectors, found with a scalar product; it gives quick tests for perpendicular and parallel lines. The shortest distance between two lines is measured along their common perpendicular; for skew lines it comes from the vector product $\vec b_1\times\vec b_2$ followed by a scalar product, and for parallel lines from a single vector product. A shortest distance of zero tells you that two non-parallel lines intersect.

The Angle between Two Lines

Two lines through a common point form two pairs of opposite angles, $\phi$ and $180^\circ-\phi$. By convention we report the one that is not obtuse. If the lines do not meet, we first slide one of them, without turning it, until it meets the other; sliding does not change directions.

📘 Definition 11.4 — Angle between two lines

Let $L_1$ and $L_2$ be lines with direction vectors $\vec b_1$ and $\vec b_2$. The angle between the lines is the angle $\theta$ with $0\le\theta\le\tfrac\pi2$ between lines through one point parallel to $L_1$ and $L_2$: it is the angle $\phi$ between $\vec b_1$ and $\vec b_2$ if $\phi\le\tfrac\pi2$, and $\pi-\phi$ otherwise.

📐 Theorem 11.8 — The angle between two lines

If the lines have direction vectors $\vec b_1$ and $\vec b_2$ — direction ratios $a_1,b_1,c_1$ and $a_2,b_2,c_2$ — then the angle $\theta$ between them satisfies $$\cos\theta=\frac{|\vec b_1\cdot\vec b_2|}{|\vec b_1|\,|\vec b_2|},$$ $$\cos\theta=\frac{|a_1a_2+b_1b_2+c_1c_2|}{\sqrt{a_1^2+b_1^2+c_1^2}\,\sqrt{a_2^2+b_2^2+c_2^2}}.$$ If $l_1,m_1,n_1$ and $l_2,m_2,n_2$ are direction cosines of the lines, this becomes $\cos\theta=|l_1l_2+m_1m_2+n_1n_2|$.

Proof. By the scalar product (Section 10.3), $\vec b_1\cdot\vec b_2=|\vec b_1|\,|\vec b_2|\cos\phi$, where $\phi\in[0,\pi]$ is the angle between the vectors. By Definition 11.4, $\theta=\phi$ or $\theta=\pi-\phi$, and in both cases $\cos\theta=|\cos\phi|$ because $\cos(\pi-\phi)=-\cos\phi$ and $\cos\theta\ge0$. So $\cos\theta=\dfrac{|\vec b_1\cdot\vec b_2|}{|\vec b_1||\vec b_2|}$. Writing the vectors in components gives the second form; with direction cosines both lengths are $1$. $\square$

📐 Theorem 11.9 — Perpendicular and parallel lines

Two lines with direction ratios $a_1,b_1,c_1$ and $a_2,b_2,c_2$ are

(a) perpendicular if and only if $a_1a_2+b_1b_2+c_1c_2=0$;

(b) parallel (or the same line) if and only if their direction ratios are proportional, $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$ (with the usual reading when a ratio is $0$).

(a) is the case $\cos\theta=0$ of Theorem 11.8. For (b), the lines are parallel exactly when $\theta=0$, i.e. when $\vec b_1$ and $\vec b_2$ are collinear vectors: $\vec b_2=t\vec b_1$ for some $t\ne0$. Note that “perpendicular lines” in space need not meet — they only have perpendicular directions.

✏️ Example 1 — Angles between lines

Find the angle between the lines (a) $\dfrac{x-1}{1}=\dfrac{y+2}{2}=\dfrac z2$ and $\dfrac{x+3}{2}=\dfrac{y-1}{-1}=\dfrac{z-4}{2}$; (b) $\vec r=\lambda(\hat i-\hat j)$ and $\vec r=\hat k+\mu(\hat j+\hat k)$.

💡 Show solution▶

(a) Ratios $1,2,2$ and $2,-1,2$. Scalar product: $1\cdot2+2\cdot(-1)+2\cdot2=4$. Lengths: $\sqrt{1+4+4}=3$ and $\sqrt{4+1+4}=3$.

(a) Angle $\cos\theta=\dfrac{|4|}{3\cdot3}=\dfrac49$, so $\theta=\cos^{-1}\dfrac49\approx63.61^\circ$.

(b) Ratios $1,-1,0$ and $0,1,1$. Scalar product: $0-1+0=-1$; lengths $\sqrt2$ and $\sqrt2$.

(b) Angle $\cos\theta=\dfrac{|-1|}{\sqrt2\cdot\sqrt2}=\dfrac12$, so $\theta=60^\circ$. The direction vectors make $120^\circ$, but the angle between the lines is the acute one.

✏️ Example 2 — Perpendicular, parallel, or neither?

(a) Show that the lines $\vec r=(\hat i+\hat j)+\lambda(3\hat i-6\hat j+2\hat k)$ and $\vec r=(2\hat i-\hat k)+\mu(2\hat i+3\hat j+6\hat k)$ are perpendicular. (b) Find $k$ so that the lines with direction ratios $k,2,3$ and $2,k,-4$ are perpendicular. Can they ever be parallel?

💡 Show solution▶

(a) $(3)(2)+(-6)(3)+(2)(6)$ $=6-18+12=0$, so the lines are perpendicular by Theorem 11.9(a).

(b) Perpendicular $2k+2k-12=0$, so $k=3$.

(b) Parallel? We would need $\dfrac k2=\dfrac2k=\dfrac3{-4}$. The last ratio is negative, so $\dfrac k2=-\dfrac34$ gives $k=-\tfrac32$; but then $\dfrac2k=-\dfrac43\ne-\dfrac34$. So the lines are never parallel.

Check With $k=3$: $3\cdot2+2\cdot3+3\cdot(-4)=0$ ✓.

Both lines are drawn through the origin (sliding a line does not change the angle). The red line has a fixed direction $\vec b_1$ chosen from the list; set the direction ratios of the blue line with the sliders. The dashed arc is the angle $\phi$ between the two direction vectors; the green arc is the angle $\theta$ between the lines, which is never obtuse. Look for the settings that make the lines perpendicular ($\vec b_1\cdot\vec b_2=0$) or parallel. Drag the picture (or press Play) to turn it.

$\vec b_1$:$a_2$:$b_2$:$c_2$:

$\vec b_1\cdot\vec b_2$—

$|\vec b_1|\,|\vec b_2|$—

$\phi$ (vectors)—

$\cos\theta$, $\theta$ (lines)—

relation—

Intersecting, Parallel and Skew Lines

In a plane, two different lines either meet in one point or are parallel. In space there is a third possibility. Think of the edge where the floor of a room meets the front wall, and the edge where the ceiling meets a side wall: they are not parallel, yet they never meet.

📘 Definition 11.5 — Skew lines

Two lines in space are skew lines if they are neither parallel nor intersecting. (Skew lines cannot be drawn together on one flat sheet; in two dimensions they do not exist.)

🧭 Problem-solving strategy — deciding how two lines lie

For $L_1:\ \vec r=\vec a_1+\lambda\vec b_1$ and $L_2:\ \vec r=\vec a_2+\mu\vec b_2$:

Compare directions. If $\vec b_1$ and $\vec b_2$ are proportional, the lines are parallel — and they are the same line if, in addition, a point of $L_2$ lies on $L_1$.
Otherwise, look for a common point. Equate the general points: three equations in the two unknowns $\lambda,\mu$. Solve two of them.
Test the third equation. If it holds, the lines intersect, and substituting $\lambda$ gives the point of intersection. If it fails, the lines are skew.

✏️ Example 3 — Do the lines meet?

Let $L_1:\ \dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac z1$. Decide whether $L_1$ meets (a) $L_2:\ \dfrac{x-2}{1}=\dfrac{y-4}{-2}=\dfrac{z+1}{2}$ (b) $L_3:\ \dfrac{x-2}{1}=\dfrac{y-4}{-2}=\dfrac{z}{2}$, and find any point of intersection.

💡 Show solution▶

Directions $2,3,1$ and $1,-2,2$ are not proportional, so neither pair is parallel.

(a) Equate General points $(1+2\lambda,\ -1+3\lambda,\ \lambda)$ and $(2+\mu,\ 4-2\mu,\ -1+2\mu)$. From $z$: $\lambda=2\mu-1$. Then $x$: $1+2(2\mu-1)=2+\mu$, so $3\mu=3$, $\mu=1$ and $\lambda=1$.

(a) Test $y$: $-1+3(1)=2$ and $4-2(1)=2$ ✓. The lines intersect at $(1+2,\ -1+3,\ 1)=(3,2,1)$.

(b) Equate Now $(2+\mu,\ 4-2\mu,\ 2\mu)$. From $z$: $\lambda=2\mu$; from $x$: $1+4\mu=2+\mu$, so $\mu=\tfrac13$, $\lambda=\tfrac23$.

(b) Test $y$: $-1+3\cdot\tfrac23=1$ but $4-2\cdot\tfrac13=\tfrac{10}3$ ✗. No common point and not parallel: $L_1$ and $L_3$ are skew. (Moving the point of $L_2$ by just one unit in $z$ was enough to make the lines miss each other.)

Shortest Distance between Two Lines

For skew lines the natural question is how close they come. Among all segments joining a point of $L_1$ to a point of $L_2$ there is a shortest one, and it meets both lines at right angles (if it were not perpendicular to $L_1$, say, sliding its end along $L_1$ would shorten it, as in the foot-of-perpendicular argument of Section 11.2).

📘 Definition 11.6 — Shortest distance and common perpendicular

The shortest distance between two lines is the least possible distance between a point of one line and a point of the other. For skew lines it is the length of the segment $PQ$, with $P$ on $L_1$ and $Q$ on $L_2$, that is perpendicular to both lines; this segment lies on the common perpendicular of the two lines. Intersecting lines have shortest distance $0$.

📐 Theorem 11.10 — Shortest distance between two non-parallel lines

Let $L_1:\ \vec r=\vec a_1+\lambda\vec b_1$ and $L_2:\ \vec r=\vec a_2+\mu\vec b_2$ be lines that are not parallel (so $\vec b_1\times\vec b_2\ne\vec 0$). Their shortest distance is $$d=\left|\frac{(\vec b_1\times\vec b_2)\cdot(\vec a_2-\vec a_1)}{|\vec b_1\times\vec b_2|}\right|.$$ In particular the lines intersect if and only if $(\vec b_1\times\vec b_2)\cdot(\vec a_2-\vec a_1)=0$; otherwise they are skew.

Proof. Let $A_1$ and $A_2$ be the points with position vectors $\vec a_1$ and $\vec a_2$, and let $PQ$ be the common perpendicular ($P$ on $L_1$, $Q$ on $L_2$). The vector $\vec b_1\times\vec b_2$ is perpendicular to both $\vec b_1$ and $\vec b_2$ (Section 10.4), and so is $\overrightarrow{PQ}$; hence $\overrightarrow{PQ}$ is parallel to the unit vector $\hat n=\dfrac{\vec b_1\times\vec b_2}{|\vec b_1\times\vec b_2|}$, and $\overrightarrow{PQ}\cdot\hat n=\pm PQ$. Now go from $A_1$ to $A_2$ the long way round: $$\overrightarrow{A_1A_2}=\overrightarrow{A_1P}+\overrightarrow{PQ}+\overrightarrow{QA_2}.$$ Here $\overrightarrow{A_1P}$ is a multiple of $\vec b_1$ and $\overrightarrow{QA_2}$ a multiple of $\vec b_2$, so both have scalar product $0$ with $\hat n$. Taking the scalar product of the equation with $\hat n$ leaves $(\vec a_2-\vec a_1)\cdot\hat n=\overrightarrow{PQ}\cdot\hat n=\pm PQ$. Taking absolute values gives the formula. Finally, the lines meet exactly when the shortest distance is $0$. $\square$

The quantity on top is an ordinary scalar product of two vectors: first find the vector $\vec b_1\times\vec b_2$, then take its scalar product with $\vec a_2-\vec a_1$. In Cartesian form, with $L_1$ through $(x_1,y_1,z_1)$ with ratios $a_1,b_1,c_1$ and $L_2$ through $(x_2,y_2,z_2)$ with ratios $a_2,b_2,c_2$, the same arithmetic can be laid out as a $3\times3$ determinant (expand it as in Section 4.1):

$$d=\frac{\left|\begin{vmatrix}x_2-x_1&y_2-y_1&z_2-z_1\\a_1&b_1&c_1\\a_2&b_2&c_2\end{vmatrix}\right|}{\sqrt{D}},$$ $$\begin{aligned}D&=(b_1c_2-b_2c_1)^2+(c_1a_2-c_2a_1)^2\\&\quad+(a_1b_2-a_2b_1)^2.\end{aligned}$$

✏️ Example 4 — Shortest distance between skew lines

Find the shortest distance between $L_1:\ \vec r=(\hat i+2\hat k)+\lambda(\hat i+2\hat j+\hat k)$ and $L_2:\ \vec r=(2\hat i+3\hat j+\hat k)+\mu(2\hat i+\hat j-\hat k)$.

💡 Show solution▶

Read off $\vec a_1=\hat i+2\hat k$, $\vec b_1=\hat i+2\hat j+\hat k$, $\vec a_2=2\hat i+3\hat j+\hat k$, $\vec b_2=2\hat i+\hat j-\hat k$.

Cross product $$\vec b_1\times\vec b_2=\begin{vmatrix}\hat i&\hat j&\hat k\\1&2&1\\2&1&-1\end{vmatrix}=-3\hat i+3\hat j-3\hat k,$$ so $|\vec b_1\times\vec b_2|=\sqrt{27}=3\sqrt3$.

Join $\vec a_2-\vec a_1=\hat i+3\hat j-\hat k$.

Scalar product $(-3)(1)+(3)(3)+(-3)(-1)$ $=-3+9+3=9$.

Answer $d=\dfrac{|9|}{3\sqrt3}=\dfrac3{\sqrt3}=\sqrt3\approx1.73$ units. It is not $0$, so the lines are skew.

✏️ Example 5 — The formula confirms Example 3

Use the Cartesian form of Theorem 11.10 to find the shortest distance between $L_1$ and $L_2$, and between $L_1$ and $L_3$, in Example 3.

💡 Show solution▶

Denominator Ratios $2,3,1$ and $1,-2,2$: $b_1c_2-b_2c_1=6+2=8$, $c_1a_2-c_2a_1=1-4=-3$, $a_1b_2-a_2b_1=-4-3=-7$. So $\sqrt D=\sqrt{64+9+49}=\sqrt{122}$.

$L_1$, $L_2$ Points $(1,-1,0)$ and $(2,4,-1)$, so the top row is $1,5,-1$: $$\begin{vmatrix}1&5&-1\\2&3&1\\1&-2&2\end{vmatrix}=0,$$ since along the top row it is $1(8)-5(3)+(-1)(-7)$ $=0$. Shortest distance $0$: the lines intersect ✓ (at $(3,2,1)$).

$L_1$, $L_3$ Now the top row is $1,5,0$: the determinant is $8-15+0=-7$, so $d=\dfrac{7}{\sqrt{122}}=\dfrac{7\sqrt{122}}{122}\approx0.63$ units — skew ✓.

Note The expansion along the top row is exactly the scalar product of $(x_2-x_1,\,y_2-y_1,\,z_2-z_1)$ with the vector $(8,-3,-7)=\vec b_1\times\vec b_2$.

When the lines are parallel, $\vec b_1\times\vec b_2=\vec 0$ and Theorem 11.10 cannot be used. Instead, every point of $L_2$ is at the same distance from $L_1$, so we measure from one point.

📐 Theorem 11.11 — Distance between parallel lines

The distance between the parallel lines $\vec r=\vec a_1+\lambda\vec b$ and $\vec r=\vec a_2+\mu\vec b$ is $$d=\frac{|\vec b\times(\vec a_2-\vec a_1)|}{|\vec b|}.$$

Proof. Let $A_1$, $A_2$ be the points with position vectors $\vec a_1$, $\vec a_2$, let $T$ be the foot of the perpendicular from $A_2$ to $L_1$, and let $\theta$ be the angle between $\overrightarrow{A_1A_2}$ and $\vec b$. In the right triangle $A_1TA_2$, $d=A_2T=A_1A_2\sin\theta$. Since $|\vec b\times(\vec a_2-\vec a_1)|=|\vec b|\,A_1A_2\sin\theta$ (Section 10.4), dividing by $|\vec b|$ gives the formula. $\square$

✏️ Example 6 — Distance between parallel lines

Show that $\vec r=(\hat i-\hat k)+\lambda(\hat i+2\hat j+2\hat k)$ and $\vec r=(4\hat i+3\hat j-\hat k)+\mu(-2\hat i-4\hat j-4\hat k)$ are parallel, and find the distance between them.

💡 Show solution▶

Parallel $-2\hat i-4\hat j-4\hat k=-2(\hat i+2\hat j+2\hat k)$: the directions are proportional. Use $\vec b=\hat i+2\hat j+2\hat k$ for both lines (this does not change the second line).

Join $\vec a_2-\vec a_1=3\hat i+3\hat j+0\hat k$.

Cross product $$\vec b\times(\vec a_2-\vec a_1)=\begin{vmatrix}\hat i&\hat j&\hat k\\1&2&2\\3&3&0\end{vmatrix}=-6\hat i+6\hat j-3\hat k,$$ of length $\sqrt{36+36+9}=9$.

Answer $d=\dfrac{9}{|\vec b|}=\dfrac93=3$ units.

Check Foot of the perpendicular from $(4,3,-1)$ to the first line (Section 11.2): $\lambda=\dfrac{(3,3,0)\cdot(1,2,2)}{9}=1$, foot $(2,2,1)$, and the distance from $(4,3,-1)$ to $(2,2,1)$ is $\sqrt{4+1+4}=3$ ✓.

Choose a pair of lines. Move $\lambda$ and $\mu$ to slide $P$ along the red line and $Q$ along the blue one, and watch the gap $PQ$. Press Play: $P$ and $Q$ slide “downhill” until $PQ$ cannot get any shorter. The lab classifies the pair, computes the shortest distance from the formula, and — as an independent check — by searching over thousands of pairs $(\lambda,\mu)$. The green segment is the common perpendicular. Drag the picture to turn it.

lines:$\lambda$:$\mu$:

type—

gap $PQ$—

$|\vec b_1\times\vec b_2|$—

shortest distance (formula)—

search minimum—

⚠️ Common mistakes when studying alone

1. Forgetting the absolute value. The angle between two lines is never obtuse, and a distance is never negative. Write $|\ldots|$ in both formulas.

2. Using points instead of directions. In $\cos\theta$ use the direction ratios (the denominators of the Cartesian form), not the points (the numbers in the numerators).

3. The skew-lines formula on parallel lines. If $\vec b_1\times\vec b_2=\vec 0$ the formula gives $\tfrac00$. Check for parallel directions first and use Theorem 11.11.

4. Wrong sign in the cross product. The middle component of $\vec b_1\times\vec b_2$ is $c_1a_2-a_1c_2$ (note the order). A slip here changes the answer completely; check that your $\vec b_1\times\vec b_2$ has scalar product $0$ with both $\vec b_1$ and $\vec b_2$.

5. “Not parallel, so they meet.” That is true in two dimensions but false in space. Two equations may give $\lambda$ and $\mu$; always test the third.

🚀 Beyond the board syllabus (JEE Advanced)

In JEE Advanced 2026; not in the CBSE Class 12 syllabus or JEE Main. The 🚀 box of Section 11.2 described the plane $ax+by+cz+d=0$ with normal $\vec n=a\hat i+b\hat j+c\hat k$. The perpendicular distance of a point $P(x_1,y_1,z_1)$ from it is $$\frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}}.$$ Why: the perpendicular from $P$ runs along $\vec n$, so its foot is $P+t\vec n$ for the $t$ that puts it on the plane: $ax_1+by_1+cz_1+d+t(a^2+b^2+c^2)=0$. The distance is $|t|\,|\vec n|$, which simplifies to the formula. Worked example. The distance of $(2,1,-1)$ from $x+2y-2z+6=0$ is $\dfrac{|2+2+2+6|}{\sqrt{1+4+4}}=\dfrac{12}3=4$ units. Check: $t=-\tfrac{12}{9}=-\tfrac43$, and $|t|\,|\vec n|=\tfrac43\cdot3=4$ ✓.

Exercises

Angles between lines; perpendicular and parallel lines.

1 Find the angle between the lines with direction ratios $2,2,1$ and $4,1,8$.
2 Find the angle between the lines with direction ratios $1,-1,0$ and $1,0,-1$.
3 Find the angle between $\vec r=(2\hat i-\hat j)+\lambda(3\hat i+2\hat j+6\hat k)$ and $\vec r=(\hat i+\hat k)+\mu(2\hat i-6\hat j+3\hat k)$.
4 Find the angle between $\dfrac{x-1}{2}=\dfrac{y+2}{-1}=\dfrac z2$ and the line $\dfrac{x+3}{1}=\dfrac{y-1}{2},\ z=4$.
5 Show that the lines with direction ratios $1,-2,3$ and $4,5,2$ are perpendicular.
6 Find $k$ so that the lines with direction ratios $k,1,-2$ and $3,k,4$ are perpendicular.
7 Find $k$ so that the lines with direction ratios $2,k,-1$ and $-4,6,2$ are parallel.
8 In a cube, find the angle between a body diagonal and a face diagonal starting at the same corner, and the angle between two body diagonals.

💡 Show solutions▶

1 $8+2+8=18$; lengths $3$ and $9$: $\cos\theta=\tfrac{18}{27}=\tfrac23$, $\theta\approx48.19^\circ$.

2 $1+0+0=1$; lengths $\sqrt2,\sqrt2$: $\cos\theta=\tfrac12$, $\theta=60^\circ$.

3 $6-12+18=12$; lengths $7$ and $7$: $\cos\theta=\tfrac{12}{49}$, $\theta\approx75.82^\circ$.

4 The second line has ratios $1,2,0$. $2-2+0=0$, so $\theta=90^\circ$.

5 $4-10+6=0$ ✓ (Theorem 11.9(a)).

6 $3k+k-8=0$, so $k=2$.

7 $\tfrac2{-4}=\tfrac{-1}{2}=-\tfrac12$, so $\tfrac k6=-\tfrac12$: $k=-3$.

8 Body diagonal $1,1,1$ and face diagonal $1,1,0$: $\cos\theta=\tfrac2{\sqrt3\sqrt2}=\sqrt{\tfrac23}$, $\theta\approx35.26^\circ$. Two body diagonals $1,1,1$ and $-1,1,1$: $\cos\theta=\tfrac13$, $\theta\approx70.53^\circ$.

Intersecting, parallel or skew? In parts 1–3, $L_1$ is $\vec r=(\hat i+\hat j)+\lambda(\hat i-\hat j+\hat k)$.

1 Decide how $L_1$ and $\vec r=(4\hat i+\hat j+2\hat k)+\mu(2\hat i+\hat j+\hat k)$ lie; find any common point.
2 The same for $L_1$ and $\vec r=(2\hat j+3\hat k)+\mu(-2\hat i+2\hat j-2\hat k)$.
3 The same for $L_1$ and $\vec r=\hat k+\mu(\hat i+\hat j)$.
4 Show that $\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}$ and $\dfrac{x-4}{5}=\dfrac{y-1}{2}=\dfrac z1$ intersect, and find the point of intersection.
5 For which $k$ do $\vec r=(\hat i+2\hat j+3\hat k)+\lambda(\hat i+\hat k)$ and $\vec r=k\hat i+\mu(\hat j+\hat k)$ intersect? Find the point.
6 Two different lines are both perpendicular to the $z$-axis. Must they be parallel? Give an example.

💡 Show solutions▶

1 Not parallel. $1+\lambda=4+2\mu$, $1-\lambda=1+\mu$, $\lambda=2+\mu$. The second gives $\lambda=-\mu$; with the third, $\mu=-1$, $\lambda=1$; the first: $2=4-2$ ✓. They intersect at $(2,0,1)$.

2 $-2\hat i+2\hat j-2\hat k=-2(\hat i-\hat j+\hat k)$: parallel. $(0,2,3)-(1,1,0)=(-1,1,3)$ is not a multiple of $(1,-1,1)$, so they are distinct parallel lines.

3 Not parallel. $1+\lambda=\mu$, $1-\lambda=\mu$, $\lambda=1$: the third gives $\lambda=1$, then $\mu=2$ from the first but $\mu=0$ from the second ✗. Skew.

4 $(1+2\lambda,\ 2+3\lambda,\ 3+4\lambda)$ $=(4+5\mu,\ 1+2\mu,\ \mu)$. From $z$: $\mu=3+4\lambda$; then $x$: $1+2\lambda=19+20\lambda$, $\lambda=-1$, $\mu=-1$; $y$: $-1=-1$ ✓. Point $(-1,-1,-1)$.

5 $(1+\lambda,\ 2,\ 3+\lambda)=(k,\ \mu,\ \mu)$: $\mu=2$, $\lambda=-1$, and $k=1+\lambda=0$. Point $(0,2,2)$.

6 No. The $x$-axis and the line through $(0,0,1)$ parallel to the $y$-axis are both perpendicular to the $z$-axis, but they are not parallel and do not meet (all points of the first have $z=0$, of the second $z=1$): they are skew.

III

Shortest distances.

1 Find the shortest distance between the $x$-axis and the line $\vec r=2\hat k+\mu(\hat i+\hat j)$.
2 Find the shortest distance between $\vec r=(\hat i+\hat j+\hat k)+\lambda(2\hat i-\hat j+\hat k)$ and $\vec r=(2\hat i+\hat j-\hat k)+\mu(3\hat i-5\hat j+2\hat k)$.
3 Find the shortest distance between $\dfrac{x-1}{2}=\dfrac{y+1}{1}=\dfrac{z}{-1}$ and $\dfrac{x-2}{1}=\dfrac{y}{-1}=\dfrac{z-3}{2}$.
4 Find the distance between $\vec r=(\hat i+\hat j)+\lambda(2\hat i-\hat j+2\hat k)$ and $\vec r=(2\hat i+\hat j-\hat k)+\mu(4\hat i-2\hat j+4\hat k)$.
5 Find the distance between $\dfrac{x-1}{1}=\dfrac{y-2}{2}=\dfrac{z-3}{2}$ and $\dfrac{x+1}{1}=\dfrac{y}{2}=\dfrac{z-1}{2}$.
6 Use Theorem 11.10 to confirm that the lines of Exercise II.4 intersect.
7 In a cube of side $1$ with a corner at $O$ and edges along the axes, find the shortest distance between the body diagonal $x=y=z$ and the edge $x=1,\ z=0$.
8 Ducts. In an illustrative factory plan (metres), one duct runs along $\vec r=3\hat k+\lambda\hat i$ and another along $\vec r=4\hat j+5\hat k+\mu(\hat j+\hat k)$. A safety rule needs at least $1.5$ m between them. Is it met?

Check: classify two lines and find their shortest distance

$\vec a_1$:

$\vec b_1$:

$\vec a_2$:

$\vec b_2$:

—

Whole numbers from −99 to 99. For a Cartesian equation, $\vec a$ is the point and $\vec b$ the direction ratios.

💡 Show solutions▶

1 $\vec b_1=\hat i$, $\vec b_2=\hat i+\hat j$: $\vec b_1\times\vec b_2=\hat k$. $\ \vec a_2-\vec a_1=2\hat k$, scalar product $2$: $d=\tfrac21=2$.

2 $\vec b_1\times\vec b_2=3\hat i-\hat j-7\hat k$, length $\sqrt{59}$; $\vec a_2-\vec a_1=\hat i-2\hat k$; scalar product $3+14=17$: $d=\tfrac{17}{\sqrt{59}}\approx2.21$.

3 Ratios $2,1,-1$ and $1,-1,2$: cross product $(1,-5,-3)$, length $\sqrt{35}$; points give $(1,1,3)$; scalar product $1-5-9=-13$: $d=\tfrac{13}{\sqrt{35}}\approx2.20$.

4 Parallel: $\vec b=2\hat i-\hat j+2\hat k$, $|\vec b|=3$; $\vec a_2-\vec a_1=\hat i-\hat k$; $\vec b\times(\vec a_2-\vec a_1)=\hat i+4\hat j+\hat k$, length $\sqrt{18}=3\sqrt2$: $d=\sqrt2$.

5 Parallel, $\vec b=(1,2,2)$, $|\vec b|=3$; $\vec a_2-\vec a_1=(-2,-2,-2)$; cross product $(0,-2,2)$, length $2\sqrt2$: $d=\tfrac{2\sqrt2}3\approx0.94$.

6 $\vec b_1\times\vec b_2=(2,3,4)\times(5,2,1)$ $=(-5,18,-11)$; $\vec a_2-\vec a_1=(3,-1,-3)$; scalar product $-15-18+33=0$, so $d=0$: they intersect ✓.

7 $\vec b_1=(1,1,1)$ through $O$; the edge passes through $(1,0,0)$ with $\vec b_2=(0,1,0)$. $\vec b_1\times\vec b_2=(-1,0,1)$, length $\sqrt2$; scalar product with $(1,0,0)$ is $-1$: $d=\tfrac1{\sqrt2}\approx0.71$.

8 $\hat i\times(\hat j+\hat k)=-\hat j+\hat k$, length $\sqrt2$; $\vec a_2-\vec a_1=4\hat j+2\hat k$; scalar product $-4+2=-2$: $d=\tfrac2{\sqrt2}=\sqrt2\approx1.41$ m. This is less than $1.5$ m, so the rule is not met.

Reasoning.

1 Show directly that if $L_1:\ \vec r=\vec a_1+\lambda\vec b_1$ and $L_2:\ \vec r=\vec a_2+\mu\vec b_2$ have a common point, then $(\vec b_1\times\vec b_2)\cdot(\vec a_2-\vec a_1)=0$.
2 Find $k$ so that $\dfrac{x-1}{1}=\dfrac{y-2}{2}=\dfrac{z-3}{3}$ and $\dfrac{x-2}{1}=\dfrac{y-k}{1}=\dfrac{z-4}{2}$ intersect, and find the point of intersection.
3 Show that the line through $(0,0,0)$ and $(1,1,1)$ and the line through $(1,0,0)$ and $(0,1,0)$ are perpendicular but do not meet, and find the shortest distance between them.
4 Two non-parallel lines have direction cosines $l_1,m_1,n_1$ and $l_2,m_2,n_2$. Show that a line with direction ratios $m_1n_2-m_2n_1,\ n_1l_2-n_2l_1,\ l_1m_2-l_2m_1$ is perpendicular to both.

💡 Show solutions▶

1 If $\vec a_1+\lambda_0\vec b_1=\vec a_2+\mu_0\vec b_2$, then $\vec a_2-\vec a_1=\lambda_0\vec b_1-\mu_0\vec b_2$. Since $\vec b_1\times\vec b_2$ is perpendicular to $\vec b_1$ and to $\vec b_2$, its scalar product with this vector is $0$.

2 $(1+\lambda,\ 2+2\lambda,\ 3+3\lambda)$ $=(2+\mu,\ k+\mu,\ 4+2\mu)$. From $x$: $\lambda=1+\mu$; from $z$: $6+3\mu=4+2\mu$, so $\mu=-2$, $\lambda=-1$. Then $y$: $0=k-2$, so $k=2$; the point is $(0,0,0)$.

3 Directions $(1,1,1)$ and $(-1,1,0)$: $-1+1+0=0$, perpendicular. Cross product $(-1,-1,2)$, length $\sqrt6$; $\vec a_2-\vec a_1=(1,0,0)$, scalar product $-1$: $d=\tfrac1{\sqrt6}\ne0$, so they do not meet — perpendicular skew lines.

4 The given ratios are the components of $\vec u_1\times\vec u_2$, where $\vec u_1=l_1\hat i+m_1\hat j+n_1\hat k$ and $\vec u_2=l_2\hat i+m_2\hat j+n_2\hat k$; they are not all $0$ because the lines are not parallel. A vector product is perpendicular to both factors, so the scalar product with each direction is $0$ — e.g. $l_1(m_1n_2-m_2n_1)+m_1(n_1l_2-n_2l_1)$ $+\,n_1(l_1m_2-l_2m_1)=0$ after cancelling — and Theorem 11.9(a) applies.

🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marksBoard · 5 marksJEE MainJEE Advanced

How this is examined: the angle between two lines and the perpendicular/parallel conditions appear as 1- and 2-mark questions (often “find $k$”); the shortest distance between two skew or parallel lines is a regular 3- or 5-mark question, frequently combined with writing the equation of a line through two points. JEE adds the feet of the common perpendicular and intersection conditions.

E1. Board · 1 mark If $\theta$ is the angle between the lines with direction ratios $1,-1,2$ and $2,1,-1$, then $\cos\theta$ equals (A) $\tfrac16$ (B) $-\tfrac16$ (C) $\tfrac1{\sqrt6}$ (D) $\tfrac56$

💡 Show solution▶

Idea $1\cdot2+(-1)\cdot1+2\cdot(-1)=-1$; both lengths are $\sqrt6$. So $\cos\theta=\dfrac{|-1|}{\sqrt6\cdot\sqrt6}=\dfrac16$.

Answer (A). Trap (B) forgets the absolute value — the angle between lines is never obtuse; (C) divides by only one length; (D) ignores the minus signs.

E2. Board · 2 marks Find $k$ so that the lines $\dfrac{x-2}{k}=\dfrac{y+1}{2}=\dfrac{z}{-3}$ and $\dfrac{x}{1}=\dfrac{y-3}{k}=\dfrac{z+1}{2}$ are perpendicular.

💡 Show solution▶

Condition Direction ratios $k,2,-3$ and $1,k,2$; perpendicular when $k\cdot1+2k+(-3)(2)=0$. (1 mark)

Answer $3k-6=0$, so $k=2$. (1 mark) Check: $2+4-6=0$ ✓.

E3. Board · 3 marks Find the distance between the lines $\vec r=(2\hat i-\hat j+\hat k)+\lambda(\hat i-2\hat j+2\hat k)$ and $\vec r=(\hat i+\hat j+3\hat k)+\mu(-2\hat i+4\hat j-4\hat k)$.

💡 Show solution▶

Parallel $-2\hat i+4\hat j-4\hat k=-2(\hat i-2\hat j+2\hat k)$, so take $\vec b=\hat i-2\hat j+2\hat k$, $|\vec b|=3$. (1 mark)

Cross product $\vec a_2-\vec a_1=-\hat i+2\hat j+2\hat k$; $$\vec b\times(\vec a_2-\vec a_1)=\begin{vmatrix}\hat i&\hat j&\hat k\\1&-2&2\\-1&2&2\end{vmatrix}$$ $=-8\hat i-4\hat j+0\hat k$, of length $\sqrt{80}=4\sqrt5$. (1½ marks)

Answer $d=\dfrac{4\sqrt5}{3}\approx2.98$ units. (½ mark)

E4. Board · 5 marks (i) Find the vector and Cartesian equations of the line $L_1$ through $A(1,-1,2)$ and $B(2,1,1)$. (ii) Find the shortest distance between $L_1$ and $L_2:\ \dfrac x2=\dfrac{y-1}{1}=\dfrac{z+2}{-2}$. (iii) Do $L_1$ and $L_2$ intersect?

💡 Show solution▶

(i) $\overrightarrow{AB}=\hat i+2\hat j-\hat k$: $\ \vec r=(\hat i-\hat j+2\hat k)+\lambda(\hat i+2\hat j-\hat k)$; $\ \dfrac{x-1}{1}=\dfrac{y+1}{2}=\dfrac{z-2}{-1}$. (2 marks)

(ii) Cross product $\vec b_1=(1,2,-1)$, $\vec b_2=(2,1,-2)$: $\vec b_1\times\vec b_2=(-4+1,\ -2+2,\ 1-4)$ $=(-3,0,-3)$, length $3\sqrt2$. (1 mark)

(ii) Distance $\vec a_2-\vec a_1=(0,1,-2)-(1,-1,2)$ $=(-1,2,-4)$; scalar product $3+0+12=15$. $d=\dfrac{15}{3\sqrt2}=\dfrac{5}{\sqrt2}=\dfrac{5\sqrt2}2\approx3.54$ units. (1 mark)

(iii) The directions are not proportional and $d\ne0$, so the lines do not intersect: they are skew. (1 mark)

E5. JEE Main The shortest distance between the lines $\vec r=(\hat i-\hat j)+\lambda(2\hat i+\hat k)$ and $\vec r=(2\hat i-\hat j)+\mu(\hat i+\hat j-\hat k)$ is (A) $\tfrac1{\sqrt{14}}$ (B) $\tfrac2{\sqrt5}$ (C) $\tfrac3{\sqrt{14}}$ (D) $0$

💡 Show solution▶

Method $\vec b_1\times\vec b_2=(2,0,1)\times(1,1,-1)$ $=(0-1,\ 1+2,\ 2-0)$ $=(-1,3,2)$, length $\sqrt{14}$. $\ \vec a_2-\vec a_1=(1,0,0)$, scalar product $-1$.

Answer (A) $\dfrac1{\sqrt{14}}$. Trap (D): the lines are not parallel, but that does not make them meet. (B) is the projection of $\vec a_2-\vec a_1$ on $\vec b_1$, which is not a distance between the lines; (C) uses the wrong component of the cross product.

⏱ Fast route With $\vec a_2-\vec a_1=\hat i$, the scalar product is just the $\hat i$-component of $\vec b_1\times\vec b_2$, i.e. $b_1c_2-b_2c_1=0\cdot(-1)-1\cdot1=-1$.

E6. JEE Advanced (One or more correct options.) For $L_1:\ \vec r=(\hat i+2\hat k)+\lambda(\hat i+2\hat j+\hat k)$ and $L_2:\ \vec r=(2\hat i+3\hat j+\hat k)+\mu(2\hat i+\hat j-\hat k)$ (Example 4), let $PQ$ be the common perpendicular, $P$ on $L_1$, $Q$ on $L_2$. Then (A) $PQ=\sqrt3$ (B) $P=\left(\tfrac53,\tfrac43,\tfrac83\right)$ (C) $PQ$ has direction ratios $1,-1,1$ (D) $Q=(2,3,1)$

💡 Show solution▶

Set up $P=(1+\lambda,\ 2\lambda,\ 2+\lambda)$, $Q=(2+2\mu,\ 3+\mu,\ 1-\mu)$, $\overrightarrow{PQ}=(1+2\mu-\lambda,$ $3+\mu-2\lambda,$ $-1-\mu-\lambda)$.

Perpendicular to both $\overrightarrow{PQ}\cdot\vec b_1=6+3\mu-6\lambda=0$ and $\overrightarrow{PQ}\cdot\vec b_2=6+6\mu-3\lambda=0$. Solving: $\lambda=\tfrac23$, $\mu=-\tfrac23$.

Feet $P=\left(\tfrac53,\tfrac43,\tfrac83\right)$, $Q=\left(\tfrac23,\tfrac73,\tfrac53\right)$, $\overrightarrow{PQ}=(-1,1,-1)$, so $PQ=\sqrt3$ with direction ratios $1,-1,1$.

Answer (A), (B) and (C). (D) is the point $\vec a_2$ of $L_2$, not the foot.

⏱ Fast route (A) and (C) need no feet: $\vec b_1\times\vec b_2=-3(1,-1,1)$ fixes the direction of $PQ$, and $PQ=d=\sqrt3$ from Example 4.

🔬 Research–Teaching NexusMini project · about 90 minutes

Research question. Two aircraft cruising near a busy airport follow nearly straight paths. How close do their paths come — and is that the same as how close the aircraft come?

Why it matters. Air-traffic control keeps aircraft apart by fixed separation standards (a minimum vertical gap or a minimum horizontal gap). The shortest distance between two straight flight paths is a first, purely geometric, safety check; the actual closest approach also depends on when each aircraft reaches each point.

What to do.

From a public flight-tracking website, record two positions (latitude, longitude, altitude), about one minute apart, for each of two aircraft in the same area. (If you cannot get live data, invent realistic positions and label them as illustrative.)
Convert to local coordinates in km: $x=\Delta(\text{longitude})\times111.3\cos(\text{latitude})$, $y=\Delta(\text{latitude})\times110.6$, $z=$ altitude in km, measured from a reference point near both aircraft.
Write the vector equation of each path, decide whether the paths are parallel, intersecting or skew, and compute the shortest distance between them with Theorem 11.10 or 11.11.
Now use time as the parameter: position $=$ first point $+\,t\times$(displacement per minute). Find, with a spreadsheet, the time at which the two aircraft are closest, and that distance.

Tools: A flight-tracking website, spreadsheet (RADIANS, COS, SQRT), a calculator for the cross product.

Hand in: Your data table, both equations, the shortest distance between the paths, the closest approach of the aircraft, and a paragraph explaining why the two numbers differ.

Going further. Find out what the vertical and horizontal separation standards are for aircraft near an airport, and whether your pair would have needed an instruction from the controller.

Instructor note: the closest approach of the aircraft is never less than the shortest distance between the paths, and equals it only if both aircraft reach the feet of the common perpendicular at the same moment. With positions $\vec p_1+t\vec v_1$ and $\vec p_2+t\vec v_2$, the squared gap is a quadratic in $t$, minimised at $t=-\dfrac{(\vec p_2-\vec p_1)\cdot(\vec v_2-\vec v_1)}{|\vec v_2-\vec v_1|^2}$.

🧪 Self-check quiz

0/5 correct

Q1. The angle between the lines with direction ratios $1,0,0$ and $0,1,1$ is

Q2. Two different lines in space that do not intersect must be

Q3. Two non-parallel lines satisfy $(\vec b_1\times\vec b_2)\cdot(\vec a_2-\vec a_1)=0$. Then the lines

Q4. The distance between the parallel lines $\vec r=\vec a_1+\lambda\vec b$ and $\vec r=\vec a_2+\mu\vec b$ is

Q5. Lines with direction ratios $2,-1,3$ and $-4,2,-6$ are

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