Chapter 11 · Three Dimensional Geometry

11.2 Equation of a Line in Space

In space a line has no slope-and-intercept equation. Instead, one point and one direction vector do the job: r = a + λb, where the single number λ slides a point along the whole line.

A delivery drone takes off and flies in a straight line towards a rooftop. Its flight computer must know, at every instant, exactly where the drone is: its east, north and height coordinates. In two-dimensional coordinate geometry you described a straight line by one equation such as $y=2x+3$. In space that cannot work: a single equation in $x,y,z$ describes a whole surface, not a line. What does fix a line in space is what the drone's computer actually stores — a starting point and a direction. Every point of the line is reached by starting at the point and moving some multiple $\lambda$ of the direction vector.

This section turns that idea into the vector equation of a line, $\vec r=\vec a+\lambda\vec b$, and its coordinate version, the Cartesian equation of a line. You will learn to write both forms for a line through a point in a given direction and for a line through two points, to switch between the forms, to find points on a line at a given distance, and to drop a perpendicular from a point to a line. The vector algebra of Chapter 10 — position vectors, collinear vectors and the scalar product — is all you need.

The Vector Equation of a Line

Through a given point there is exactly one line parallel to a given non-zero vector. Any vector along the line will do to describe its direction, so we give such vectors a name.

📘 Definition 11.3 — Direction vector of a line

A non-zero vector $\vec b$ is a direction vector of a line $L$ if $\vec b$ is parallel to $L$. If $\vec b=a\hat i+b\hat j+c\hat k$, then $a,b,c$ are direction ratios of $L$ (Definition 11.2); conversely, any set of direction ratios $a,b,c$ of $L$ gives a direction vector $a\hat i+b\hat j+c\hat k$. Every non-zero multiple $t\vec b$ is also a direction vector of $L$.

📐 Theorem 11.5 — Vector equation of a line

Let $A$ be a point with position vector $\vec a$ and let $\vec b$ be a non-zero vector. The line through $A$ parallel to $\vec b$ consists exactly of the points $P$ whose position vectors $\vec r$ satisfy $$\vec r=\vec a+\lambda\vec b\qquad\text{for some real number }\lambda.$$ Each real value of the parameter $\lambda$ gives exactly one point of the line, and each point of the line comes from exactly one $\lambda$.

Proof. Let $P$ be any point, with position vector $\vec r$. Then $\overrightarrow{AP}=\vec r-\vec a$. The point $P$ lies on the line through $A$ parallel to $\vec b$ exactly when $P=A$ or $\overrightarrow{AP}$ is parallel to $\vec b$, that is, exactly when $\overrightarrow{AP}$ is a scalar multiple of $\vec b$ (Theorem 10.5 in Section 10.2): $\vec r-\vec a=\lambda\vec b$ for some $\lambda$. This is $\vec r=\vec a+\lambda\vec b$. If $\vec a+\lambda_1\vec b=\vec a+\lambda_2\vec b$, then $(\lambda_1-\lambda_2)\vec b=\vec 0$, and since $\vec b\ne\vec 0$, $\lambda_1=\lambda_2$. $\square$

Read the equation as a recipe: start at $A$, then move $\lambda$ copies of $\vec b$. Positive $\lambda$ moves you along $\vec b$, negative $\lambda$ against it, and $\lambda=0$ gives $A$ itself. The distance moved is $AP=|\lambda|\,|\vec b|$.

✏️ Example 1 — Writing and using a vector equation

(a) Find the vector equation of the line through $A(2,-1,3)$ parallel to $\vec b=4\hat i-3\hat j+\hat k$. (b) Find the points of the line given by $\lambda=1$ and $\lambda=-2$. (c) Do the points $(10,-7,5)$ and $(14,-10,7)$ lie on the line?

💡 Show solution▶

(a) $\vec a=2\hat i-\hat j+3\hat k$, so $$\vec r=(2\hat i-\hat j+3\hat k)+\lambda(4\hat i-3\hat j+\hat k).$$

(b) Collecting components, $\vec r=(2+4\lambda)\hat i+(-1-3\lambda)\hat j+(3+\lambda)\hat k$. For $\lambda=1$: $(6,-4,4)$. For $\lambda=-2$: $(-6,5,1)$.

(c) First point $2+4\lambda=10$ gives $\lambda=2$. Then $-1-3(2)=-7$ ✓ and $3+2=5$ ✓. All three components agree with the same $\lambda$, so $(10,-7,5)$ is on the line.

(c) Second point $2+4\lambda=14$ gives $\lambda=3$; then $-1-9=-10$ ✓ but $3+3=6\ne7$ ✗. No single $\lambda$ works, so $(14,-10,7)$ is not on the line.

The Cartesian Equation of a Line

Write the vector equation in components. With $\vec r=x\hat i+y\hat j+z\hat k$, $\vec a=x_1\hat i+y_1\hat j+z_1\hat k$ and $\vec b=a\hat i+b\hat j+c\hat k$, the equation $\vec r=\vec a+\lambda\vec b$ says, by equality of vectors, $$x=x_1+\lambda a,\qquad y=y_1+\lambda b,\qquad z=z_1+\lambda c.$$ These are the parametric equations of the line. Solving each for $\lambda$ and equating the results removes the parameter.

📐 Theorem 11.6 — Cartesian equation of a line

The line through $(x_1,y_1,z_1)$ with direction ratios $a,b,c$ (all non-zero) has the Cartesian equation $$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}.$$ The common value of the three fractions is the parameter $\lambda$ of the point $(x,y,z)$. If $l,m,n$ are its direction cosines, the equation may also be written $\dfrac{x-x_1}{l}=\dfrac{y-y_1}{m}=\dfrac{z-z_1}{n}$; then the common value is the signed distance from $(x_1,y_1,z_1)$.

Zero direction ratios. If, say, $a=0$, the parametric equation is simply $x=x_1$ for every point; the line is then written $x=x_1,\ \dfrac{y-y_1}{b}=\dfrac{z-z_1}{c}$. (You may meet the shorthand $\dfrac{x-x_1}{0}$; it means only “$x-x_1=0$”, never division by $0$.)

🧭 Problem-solving strategy — reading a Cartesian equation

Make each numerator of the form $x-x_1$, $y-y_1$, $z-z_1$ with coefficient $+1$: rewrite $(3-x)$ as $-(x-3)$, and $(2y+1)$ as $2\left(y+\tfrac12\right)$.
Move any factor into the denominator: $\dfrac{-(x-3)}{5}=\dfrac{x-3}{-5}$ and $\dfrac{2(y+\frac12)}{4}=\dfrac{y+\frac12}{2}$.
Read off the point $(x_1,y_1,z_1)$ from the numerators and the direction ratios from the denominators.
Write the vector form $\vec r=(x_1\hat i+y_1\hat j+z_1\hat k)+\lambda(a\hat i+b\hat j+c\hat k)$ if it is asked for.

✏️ Example 2 — From vector form to Cartesian form

Write the line of Example 1 in Cartesian form and in parametric form.

💡 Show solution▶

Read off Point $(2,-1,3)$, direction ratios $4,-3,1$.

Cartesian $$\frac{x-2}{4}=\frac{y+1}{-3}=\frac{z-3}{1}.$$

Parametric $x=2+4\lambda$, $\ y=-1-3\lambda$, $\ z=3+\lambda$.

Check $(10,-7,5)$ from Example 1: $\tfrac84=\tfrac{-6}{-3}=\tfrac21=2$ ✓ — the common value is its parameter $\lambda=2$.

✏️ Example 3 — A Cartesian equation in disguise

A line is given by $\dfrac{1-x}{2}=\dfrac{2y+3}{4}=\dfrac{z-5}{-3}$. Find a point on it, its direction ratios and direction cosines, and its vector equation.

💡 Show solution▶

Numerator 1 $\dfrac{1-x}{2}=\dfrac{-(x-1)}{2}=\dfrac{x-1}{-2}$.

Numerator 2 $\dfrac{2y+3}{4}=\dfrac{2\left(y+\frac32\right)}{4}=\dfrac{y+\frac32}{2}$.

Standard form $$\frac{x-1}{-2}=\frac{y+\frac32}{2}=\frac{z-5}{-3}.$$ Point $\left(1,-\tfrac32,5\right)$; direction ratios $-2,2,-3$.

Cosines $\sqrt{4+4+9}=\sqrt{17}$, so $\pm\left(-\tfrac2{\sqrt{17}},\tfrac2{\sqrt{17}},-\tfrac3{\sqrt{17}}\right)$.

Vector form $$\vec r=\left(\hat i-\tfrac32\hat j+5\hat k\right)+\lambda(-2\hat i+2\hat j-3\hat k).$$ The tempting answer “direction ratios $2,4,-3$” ignores the minus sign in $1-x$ and the factor $2$ in $2y$.

The purple arrow is the position vector $\vec a$ of the fixed point $A$; the blue arrow is the direction vector $\vec b$, drawn from $A$. Move $\lambda$ (or press Play): the red point $P$ with $\vec r=\vec a+\lambda\vec b$ slides along the whole line. The readouts show that the three Cartesian fractions always share the same value — and that value is $\lambda$. Drag the picture to turn it.

line:$\lambda$:

$\lambda$—

point $P$—

Cartesian fractions—

$AP=|\lambda|\,|\vec b|$—

The Line through Two Points

If a line is given by two of its points $A$ and $B$, the vector $\overrightarrow{AB}=\vec b-\vec a$ is a direction vector of it (here $\vec b$ is the position vector of $B$). Theorem 11.5 then gives the equation at once.

📐 Theorem 11.7 — The line through two points

The line through two different points $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$, with position vectors $\vec a$ and $\vec b$, has vector equation $$\vec r=\vec a+\lambda(\vec b-\vec a),$$ and Cartesian equation $$\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}.$$ Here $\lambda=0$ gives $A$, $\lambda=1$ gives $B$, $0\lt\lambda\lt1$ gives the points between $A$ and $B$, and $\lambda=\tfrac12$ gives the mid-point.

(If one of the differences, say $x_2-x_1$, is $0$, the Cartesian form reads $x=x_1$ together with the other two fractions, as before.)

✏️ Example 4 — Through two points, and points at a given distance

(a) Find the vector and Cartesian equations of the line through $A(1,2,-1)$ and $B(4,-2,11)$. (b) Find the points of this line at a distance of $26$ units from $A$.

💡 Show solution▶

Direction $\overrightarrow{AB}=(4-1)\hat i+(-2-2)\hat j+(11+1)\hat k$ $=3\hat i-4\hat j+12\hat k$, with $|\overrightarrow{AB}|=\sqrt{9+16+144}=13$.

(a) Vector $$\vec r=(\hat i+2\hat j-\hat k)+\lambda(3\hat i-4\hat j+12\hat k).$$

(a) Cartesian $$\frac{x-1}{3}=\frac{y-2}{-4}=\frac{z+1}{12}.$$

(b) The point with parameter $\lambda$ is at distance $|\lambda|\cdot13$ from $A$, so $|\lambda|=2$. $\lambda=2$: $(1+6,\ 2-8,\ -1+24)=(7,-6,23)$. $\ \lambda=-2$: $(1-6,\ 2+8,\ -1-24)=(-5,10,-25)$.

Check From $A$ to $(7,-6,23)$: $\sqrt{36+64+576}=\sqrt{676}=26$ ✓.

✏️ Example 5 — A drone's straight flight

In an illustrative flight plan, a drone leaves a rooftop launch point $(0,0,10)$ and flies so that its position after $t$ seconds is $\vec r=10\hat k+t(12\hat i+4\hat j+3\hat k)$, in metres ($x$ east, $y$ north, $z$ height). (a) Write the path in Cartesian form. (b) Find the drone's speed. (c) When does it reach a height of $34$ m, where is it then, and how far has it flown?

💡 Show solution▶

(a) Point $(0,0,10)$, direction ratios $12,4,3$: $$\frac{x}{12}=\frac{y}{4}=\frac{z-10}{3}.$$

(b) Each second the drone moves by the vector $12\hat i+4\hat j+3\hat k$, of length $\sqrt{144+16+9}=\sqrt{169}=13$ m. Its speed is $13$ m/s.

Check $\sqrt{96^2+32^2+24^2}$ $=\sqrt{9216+1024+576}$ $=\sqrt{10816}=104$ ✓.

The Foot of the Perpendicular from a Point to a Line

How far is a point $P$ from a line $L$? The shortest segment from $P$ to $L$ is the one that meets $L$ at right angles; its end $F$ on $L$ is called the foot of the perpendicular, and $PF$ is the perpendicular distance of $P$ from $L$. (By Pythagoras, any other segment $PQ$ to the line is the hypotenuse of the right triangle $PFQ$, so it is longer.) The scalar product finds $F$ in one step.

🧭 Problem-solving strategy — foot of the perpendicular from $P$ to $\vec r=\vec a+\lambda\vec b$

General point. Write the point of the line with parameter $\lambda$: $Q=(x_1+\lambda a,\ y_1+\lambda b,\ z_1+\lambda c)$.
Join. Find $\overrightarrow{PQ}$ in terms of $\lambda$.
Perpendicular. Solve $\overrightarrow{PQ}\cdot\vec b=0$ — one linear equation in $\lambda$.
Substitute that $\lambda$ to get the foot $F$; the perpendicular distance is $|\overrightarrow{PF}|$.
Check $\overrightarrow{PF}\cdot\vec b=0$. For the image (reflection) $P'$ of $P$ in the line, use that $F$ is the mid-point of $PP'$: $P'=2F-P$.

✏️ Example 6 — Foot, distance and image

For the point $P(5,3,1)$ and the line $L:\ \vec r=(\hat i-\hat k)+\lambda(2\hat i-\hat j+2\hat k)$, find (a) the foot of the perpendicular from $P$ to $L$, (b) the perpendicular distance of $P$ from $L$, (c) the image of $P$ in $L$.

💡 Show solution▶

General point $Q=(1+2\lambda,\ -\lambda,\ -1+2\lambda)$, so $\overrightarrow{PQ}=(2\lambda-4)\hat i+(-\lambda-3)\hat j+(2\lambda-2)\hat k$.

Perpendicular $\overrightarrow{PQ}\cdot(2\hat i-\hat j+2\hat k)$ $=2(2\lambda-4)-(-\lambda-3)+2(2\lambda-2)$ $=9\lambda-9=0$, so $\lambda=1$.

(a) Foot $F=(3,-1,1)$.

(b) Distance $\overrightarrow{PF}=-2\hat i-4\hat j+0\hat k$, so $PF=\sqrt{4+16}=\sqrt{20}=2\sqrt5$ units.

Check $\overrightarrow{PF}\cdot\vec b=-4+4+0=0$ ✓, and the mid-point of $PP'$ is $(3,-1,1)=F$ ✓.

On the left, $P$ is a fixed point and $Q$ moves along the line as you change $\lambda$. On the right, the graph shows $PQ^2$ against $\lambda$ — a parabola. Press Play to sweep $\lambda$: the lab pauses at the lowest point of the parabola, where $\overrightarrow{PQ}\cdot\vec b=0$ and $PQ$ meets the line at right angles. That $Q$ is the foot $F$, and $PF$ is the perpendicular distance. Drag the 3-D picture to turn it.

case:$\lambda$:

$\lambda$—

$Q$ on the line—

$PQ$—

$\overrightarrow{PQ}\cdot\vec b$—

⚠️ Common mistakes when studying alone

1. Reading off the wrong point or ratios. In $\dfrac{x+2}{3}$ the point has $x_1=-2$, not $2$. In $\dfrac{4-y}{5}$ the direction ratio is $-5$, and in $\dfrac{3z-1}{6}$ it is $2$ (with $z_1=\tfrac13$). Always rewrite to the standard form first.

2. Mixing up $\vec a$ and $\vec b$. In $\vec r=\vec a+\lambda\vec b$, $\vec a$ is the position vector of a point and $\vec b$ is a direction. The line $\vec r=\vec b+\lambda\vec a$ is a different line.

3. Using different parameters. To test whether a point is on a line, all three coordinates must give the same $\lambda$ (Example 1(c)).

5. The foot condition. The condition is $\overrightarrow{PF}\cdot\vec b=0$ with the direction vector of the line — not $\overrightarrow{PF}\cdot\vec a=0$.

🚀 Beyond the board syllabus (JEE Advanced)

In JEE Advanced 2026; not in the CBSE Class 12 syllabus or JEE Main. Just as a point and a direction fix a line, a point and a perpendicular direction fix a flat surface called a plane. If $A$ has position vector $\vec a$ and $\vec n=a\hat i+b\hat j+c\hat k\ne\vec 0$, the points $P$ with $\overrightarrow{AP}\perp\vec n$ form the plane through $A$ with normal $\vec n$: $$(\vec r-\vec a)\cdot\vec n=0,$$ $$a(x-x_1)+b(y-y_1)+c(z-z_1)=0.$$ Worked example. The plane through $(2,-1,3)$ with normal $\hat i+2\hat j-2\hat k$ is $(x-2)+2(y+1)-2(z-3)=0$, that is $x+2y-2z+6=0$. Check: $2-2-6+6=0$ ✓. A line $\vec r=\vec a+\lambda\vec b$ is perpendicular to this plane exactly when $\vec b$ is parallel to $\vec n$ — for example, the line through $(2,-1,3)$ with direction ratios $1,2,-2$.

Exercises

Vector equations.

1 Find the vector equation of the line through $(1,-2,3)$ parallel to $2\hat i+\hat j-5\hat k$.
2 Find the vector equation of the line through the origin parallel to $3\hat i-\hat j+2\hat k$.
3 Find the vector equation of the line through $(4,0,-1)$ parallel to the $z$-axis.
4 Find the vector equation of the line through $(-1,2,0)$ parallel to the line $\vec r=(\hat i+\hat j)+\mu(2\hat i-3\hat j+6\hat k)$.
5 For the line $\vec r=(3\hat i-\hat j+2\hat k)+\lambda(\hat i+2\hat j-2\hat k)$, find the points with $\lambda=-1$, $0$ and $2$.
6 Do the points $(6,5,-4)$ and $(1,-5,5)$ lie on the line of part 5?
7 Find the point of the line of part 5 at which $z=0$.

Check: is the point on the line?

point $A$:

ratios $a,b,c$:

test point:

—

Numbers such as 4, −2.5 or 3/4. The lab solves each coordinate equation for $\lambda$ and compares.

💡 Show solutions▶

1 $\vec r=(\hat i-2\hat j+3\hat k)+\lambda(2\hat i+\hat j-5\hat k)$.

2 $\vec a=\vec 0$: $\ \vec r=\lambda(3\hat i-\hat j+2\hat k)$.

3 The $z$-axis has direction $\hat k$: $\ \vec r=(4\hat i-\hat k)+\lambda\hat k$ (every point has $x=4$, $y=0$).

4 Parallel lines share a direction vector: $\ \vec r=(-\hat i+2\hat j)+\lambda(2\hat i-3\hat j+6\hat k)$.

5 $\vec r=(3+\lambda)\hat i+(-1+2\lambda)\hat j+(2-2\lambda)\hat k$: $\lambda=-1$ gives $(2,-3,4)$; $\lambda=0$ gives $(3,-1,2)$; $\lambda=2$ gives $(5,3,-2)$.

6 $(6,5,-4)$: $3+\lambda=6$ gives $\lambda=3$; then $-1+6=5$ ✓, $2-6=-4$ ✓ — yes. $\ (1,-5,5)$: $\lambda=-2$; $-1-4=-5$ ✓ but $2+4=6\ne5$ — no.

7 $2-2\lambda=0$ gives $\lambda=1$: the point $(4,1,0)$.

Cartesian equations and conversions.

1 Write the line of Exercise I.1 in Cartesian form.
2 Write $\dfrac{x+3}{2}=\dfrac{y-1}{-4}=\dfrac{z-6}{5}$ in vector form.
3 Find a point, the direction ratios and the direction cosines of $\dfrac{2x-1}{4}=\dfrac{3-y}{2}=\dfrac{z+2}{3}$.
4 Find the direction cosines of the line $\dfrac{x-4}{2}=\dfrac{y}{6}=\dfrac{1-z}{3}$.
5 Write the Cartesian and vector equations of the line through $(2,-1,4)$ with direction ratios $0,3,-2$.
6 Write vector and Cartesian equations of the $x$-axis.
7 Find $k$ so that $(k,-1,7)$ lies on $\dfrac{x-1}{2}=\dfrac{y+3}{1}=\dfrac{z-3}{2}$.
8 Show that $\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}$ and $\dfrac{x-3}{-4}=\dfrac{y-5}{-6}=\dfrac{z-7}{-8}$ describe the same line.

💡 Show solutions▶

1 $\dfrac{x-1}{2}=\dfrac{y+2}{1}=\dfrac{z-3}{-5}$.

2 Point $(-3,1,6)$, ratios $2,-4,5$: $\ \vec r=(-3\hat i+\hat j+6\hat k)+\lambda(2\hat i-4\hat j+5\hat k)$.

3 $\dfrac{x-\frac12}{2}=\dfrac{y-3}{-2}=\dfrac{z+2}{3}$: point $\left(\tfrac12,3,-2\right)$, ratios $2,-2,3$, cosines $\pm\left(\tfrac2{\sqrt{17}},-\tfrac2{\sqrt{17}},\tfrac3{\sqrt{17}}\right)$.

4 $\dfrac{1-z}{3}=\dfrac{z-1}{-3}$, so the ratios are $2,6,-3$; $\sqrt{4+36+9}=7$: $\pm\left(\tfrac27,\tfrac67,-\tfrac37\right)$.

5 $x=2,\ \dfrac{y+1}{3}=\dfrac{z-4}{-2}$; $\ \vec r=(2\hat i-\hat j+4\hat k)+\lambda(3\hat j-2\hat k)$.

6 Through $O$ with direction $\hat i$: $\vec r=\lambda\hat i$; Cartesian: $y=0,\ z=0$ (that is, $\tfrac x1=\tfrac y0=\tfrac z0$ in the shorthand).

7 $\dfrac{-1+3}{1}=2$ and $\dfrac{7-3}{2}=2$ agree, so $\lambda=2$ and $\dfrac{k-1}{2}=2$: $k=5$.

8 Ratios $-4,-6,-8=-2(2,3,4)$: parallel. The point $(3,5,7)$ of the second line satisfies the first: $\tfrac22=\tfrac33=\tfrac44=1$. Parallel lines with a common point coincide.

III

Lines through two points; points on a line.

1 Find the vector and Cartesian equations of the line through $(2,-1,1)$ and $(5,3,-1)$.
2 Find the vector and Cartesian equations of the line through $(1,0,-2)$ and $(1,4,1)$.
3 Find the points of the line of part 2 at a distance of $10$ units from $(1,0,-2)$.
4 Find the equations of the line through the origin and $(-2,6,-3)$, and the points on it at a distance of $14$ from the origin.
5 Find the point at which the line through $(2,3,-1)$ and $(4,7,3)$ reaches $z=0$.
6 Does $(8,7,-3)$ lie on the line of part 1?
7 For $\vec r=\vec a+\lambda(\vec b-\vec a)$ through $A$ and $B$, which values of $\lambda$ give: $A$; $B$; the mid-point of $AB$; the point dividing $AB$ internally in the ratio $1:2$; points beyond $B$?

💡 Show solutions▶

1 Ratios $3,4,-2$: $\ \vec r=(2\hat i-\hat j+\hat k)+\lambda(3\hat i+4\hat j-2\hat k)$; $\ \dfrac{x-2}{3}=\dfrac{y+1}{4}=\dfrac{z-1}{-2}$.

2 Ratios $0,4,3$: $\ \vec r=(\hat i-2\hat k)+\lambda(4\hat j+3\hat k)$; $\ x=1,\ \dfrac y4=\dfrac{z+2}{3}$.

3 $|\vec b|=\sqrt{0+16+9}=5$, so $|\lambda|=2$: $(1,8,4)$ and $(1,-8,-8)$.

4 $\vec r=\lambda(-2\hat i+6\hat j-3\hat k)$; $\ \dfrac x{-2}=\dfrac y6=\dfrac z{-3}$. $|\vec b|=7$, so $\lambda=\pm2$: $(-4,12,-6)$ and $(4,-12,6)$.

5 Ratios $2,4,4$: $(2+2\lambda,\ 3+4\lambda,\ -1+4\lambda)$; $-1+4\lambda=0$ gives $\lambda=\tfrac14$: $\left(\tfrac52,4,0\right)$.

6 $2+3\lambda=8$ gives $\lambda=2$; $-1+8=7$ ✓, $1-4=-3$ ✓. Yes.

7 $A$: $\lambda=0$; $B$: $\lambda=1$; mid-point: $\lambda=\tfrac12$; ratio $1:2$: $\lambda=\tfrac13$ (one third of the way from $A$); beyond $B$: $\lambda\gt1$ (and beyond $A$: $\lambda\lt0$).

Perpendiculars and applications.

1 Find the foot of the perpendicular from $P(6,3,0)$ to the line $\vec r=\lambda(\hat i+2\hat j+2\hat k)$, and the perpendicular distance.
2 Find the foot of the perpendicular from $P(7,0,-1)$ to the line $\dfrac x2=\dfrac{y-1}{-2}=\dfrac{z+3}{1}$, and the perpendicular distance.
3 Find the image of the point $P$ of part 2 in that line.
4 Pipeline. In an illustrative site plan (metres), a straight pipeline runs from $O(0,0,0)$ to $B(12,6,4)$, and a pump stands at $P(8,-1,2)$. A connecting pipe from $P$ must meet the pipeline at right angles. Where does it join, and how long is it?
5 A ball moves so that its position after $t$ seconds is $\vec r=(\hat i+2\hat j)+t(2\hat i-\hat j+2\hat k)$ metres. Find its speed, and its position when it has travelled $12$ m.
6 Show that $x-1=2(y+3)=z$ is the equation of a line, and find a point on it and its direction cosines.

💡 Show solutions▶

1 $Q=(\lambda,2\lambda,2\lambda)$, $\overrightarrow{PQ}=(\lambda-6,\ 2\lambda-3,\ 2\lambda)$; $\ \overrightarrow{PQ}\cdot(1,2,2)=9\lambda-12=0$, $\lambda=\tfrac43$. Foot $\left(\tfrac43,\tfrac83,\tfrac83\right)$; $\overrightarrow{FP}=\left(\tfrac{14}3,\tfrac13,-\tfrac83\right)$, distance $\sqrt{\tfrac{196+1+64}{9}}=\sqrt{29}$.

2 $Q=(2\lambda,\ 1-2\lambda,\ -3+\lambda)$, $\overrightarrow{PQ}=(2\lambda-7,\ 1-2\lambda,\ \lambda-2)$; dot with $(2,-2,1)$: $9\lambda-18=0$, $\lambda=2$. Foot $(4,-3,-1)$; $\overrightarrow{FP}=(3,3,0)$, distance $3\sqrt2$.

3 $P'=2F-P$ $=(8-7,\ -6-0,\ -2+1)$ $=(1,-6,-1)$.

4 $Q=\lambda(12,6,4)$, $\overrightarrow{PQ}=(12\lambda-8,\ 6\lambda+1,\ 4\lambda-2)$; dot with $(12,6,4)$: $196\lambda-98=0$, $\lambda=\tfrac12$. It joins at $(6,3,2)$, the mid-point of $OB$; length $\sqrt{2^2+4^2+0^2}=2\sqrt5\approx4.47$ m.

5 Speed $|2\hat i-\hat j+2\hat k|=3$ m/s, so $12$ m takes $t=4$ s: position $(9,-2,8)$.

6 $2(y+3)=\dfrac{y+3}{1/2}$, so the equation is $\dfrac{x-1}{1}=\dfrac{y+3}{1/2}=\dfrac{z}{1}$: a line through $(1,-3,0)$ with ratios $1,\tfrac12,1$, i.e. $2,1,2$. Cosines $\pm\left(\tfrac23,\tfrac13,\tfrac23\right)$.

🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marksBoard · Case studyJEE Main

How this is examined: Board papers ask you to read the direction ratios or a point from a Cartesian equation (1 mark, often with a $1-x$ trap), to write the vector and Cartesian equations of a line through two points (2 marks), and to find the foot of a perpendicular and a perpendicular distance (3 marks); case studies put a line into a drone, laser or pipeline story. JEE favours images of points in lines.

E1. Board · 1 mark The direction ratios of the line $\dfrac{x-3}{2}=\dfrac{1-y}{4}=\dfrac{z+2}{-1}$ are (A) $2,4,-1$ (B) $2,-4,-1$ (C) $3,1,-2$ (D) $3,-1,-2$

💡 Show solution▶

Idea $\dfrac{1-y}{4}=\dfrac{y-1}{-4}$, so the standard form is $\dfrac{x-3}{2}=\dfrac{y-1}{-4}=\dfrac{z+2}{-1}$.

Answer (B). Trap (A) reads the denominator $4$ without fixing the sign of $1-y$; (C) and (D) are built from the point $(3,1,-2)$, not the direction.

E2. Board · 1 mark Assertion–Reason. Assertion (A): the line $\vec r=(\hat i-\hat j)+\lambda(2\hat i+\hat k)$ passes through $(3,-1,1)$. Reason (R): a point with position vector $\vec r$ lies on the line $\vec r=\vec a+\lambda\vec b$ if and only if $\vec r-\vec a$ is a scalar multiple of $\vec b$. (A) Both A and R are true and R is the correct explanation of A (B) Both A and R are true but R is not the correct explanation of A (C) A is true but R is false (D) A is false but R is true

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Idea R is Theorem 11.5. $(3\hat i-\hat j+\hat k)-(\hat i-\hat j)=2\hat i+\hat k=1\cdot\vec b$, so by R the point is on the line ($\lambda=1$).

Answer (A).

E3. Board · 2 marks Find the vector and Cartesian equations of the line through $(-1,0,2)$ and $(3,4,6)$.

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Direction $(3+1,\ 4-0,\ 6-2)=(4,4,4)$; the simpler ratios $1,1,1$ describe the same direction. (½ mark)

Vector $\vec r=(-\hat i+2\hat k)+\lambda(\hat i+\hat j+\hat k)$. (1 mark)

Cartesian $\dfrac{x+1}{1}=\dfrac{y}{1}=\dfrac{z-2}{1}$. (½ mark) Check: $(3,4,6)$ gives $4=4=4$ ✓.

E4. Board · 3 marks Find the coordinates of the foot of the perpendicular from $P(4,1,1)$ to the line $\dfrac{x-1}{1}=\dfrac{y+2}{2}=\dfrac{z-1}{-2}$, and the perpendicular distance of $P$ from the line.

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General point $Q=(1+\lambda,\ -2+2\lambda,\ 1-2\lambda)$, so $\overrightarrow{PQ}=(\lambda-3,\ 2\lambda-3,\ -2\lambda)$. (1 mark)

Perpendicular $\overrightarrow{PQ}\cdot(1,2,-2)$ $=\lambda-3+4\lambda-6+4\lambda$ $=9\lambda-9=0$, so $\lambda=1$. (1 mark)

Answer Foot $F=(2,0,-1)$; $\overrightarrow{FP}=(2,1,2)$, so the distance is $\sqrt{4+1+4}=3$ units. (1 mark) Check: $(2,1,2)\cdot(1,2,-2)=0$ ✓.

E5. Board · Case study In a physics lab (illustrative set-up, lengths in cm), a laser source at $S(1,-2,0)$ sends a straight beam in the direction $2\hat i+2\hat j+\hat k$. (i) Write the vector and Cartesian equations of the beam. (2 marks) (ii) Find the point the beam reaches after travelling $9$ cm. (1 mark) (iii) Will the beam hit a sensor placed at $(5,2,3)$? (1 mark)

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(i) $\vec r=(\hat i-2\hat j)+\lambda(2\hat i+2\hat j+\hat k)$, $\lambda\ge0$ for the beam; $\ \dfrac{x-1}{2}=\dfrac{y+2}{2}=\dfrac{z}{1}$.

(ii) $|\vec b|=\sqrt{4+4+1}=3$, so $9$ cm needs $\lambda=3$: the point $(7,4,3)$.

(iii) $\dfrac{5-1}{2}=2$, $\dfrac{2+2}{2}=2$, but $\dfrac{3}{1}=3\ne2$. No single $\lambda$ works, so the beam misses the sensor (it passes through $(5,2,2)$, $1$ cm below it).

E6. JEE Main The image of the point $P(9,1,-1)$ in the line $\dfrac{x-2}{3}=\dfrac{y+1}{2}=\dfrac{z-3}{-6}$ is (A) $(1,1,-5)$ (B) $(5,1,-3)$ (C) $(1,-1,-5)$ (D) $(13,1,1)$

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Method $Q=(2+3\lambda,\ -1+2\lambda,\ 3-6\lambda)$, $\overrightarrow{PQ}=(3\lambda-7,\ 2\lambda-2,\ 4-6\lambda)$. Then $\overrightarrow{PQ}\cdot(3,2,-6)$ $=49\lambda-49=0$, so $\lambda=1$ and the foot is $F=(5,1,-3)$.

Image $P'=2F-P$ $=(10-9,\ 2-1,\ -6+1)$ $=(1,1,-5)$.

Answer (A). Trap (B) is the foot, not the image; (D) $=P+(P-F)$ reflects away from the line; (C) has a sign slip.

⏱ Fast route Once you have $\lambda$, the image is $P+2(F-P)$; with $F-P=(-4,0,-2)$ this is $(1,1,-5)$ at once.

🔬 Research–Teaching NexusMini project · about 60 minutes

Research question. If you know two points of a stretched string, can the equation of the line predict exactly where the string passes over any chosen spot in the room — and how accurate is the prediction?

Why it matters. Builders lay out cables, ducts and conveyor belts from a few surveyed points; robot arms and 3-D printers move their tools along straight segments described by $\vec r=\vec a+\lambda\vec b$. Predicting intermediate points from the equation, and measuring how far reality differs, is everyday engineering.

What to do.

Choose one corner of the classroom floor as the origin, with $x$ and $y$ along two walls and $z$ up. Stretch a string tightly between two points $A$ and $B$ at different heights (for example, a hook on one wall and a chair leg across the room). Measure the coordinates of $A$ and $B$ with a tape measure, in cm.
Write the vector and Cartesian equations of the line $AB$.
Predict, from the equations, the point of the string at height $z=100$ cm, and the point of the string directly above a mark on the floor of your choice (pick its $x$ and work out $\lambda$).
Measure where the string actually is at those places and record the differences. Repeat with the string less tight.

Tools: Tape measure, string, masking tape, plumb line (a weight on a thread), spreadsheet.

Hand in: A labelled sketch with coordinates, your equations, a table of predicted and measured points with their differences, and a paragraph explaining the main sources of error (sag of the string, measuring errors).

Going further. Find the foot of the perpendicular from a ceiling light to your string: where would a short vertical cord from the light not meet the string at right angles, and why?

Instructor note: the point at height $h$ has $\lambda=\dfrac{h-z_A}{z_B-z_A}$, and then $x=x_A+\lambda(x_B-x_A)$, $y=y_A+\lambda(y_B-y_A)$. A tight string usually agrees with the prediction to within 1–2 cm; a slack string sags below the line, most near the middle ($\lambda\approx\tfrac12$).

🧪 Self-check quiz

0/5 correct

Q1. The vector equation of the line through $(1,0,-2)$ parallel to $3\hat i-\hat j+\hat k$ is

Q2. Which point lies on $\dfrac{x-2}{3}=\dfrac{y+1}{-2}=\dfrac{z}{5}$?

Q3. Direction ratios of the line $\dfrac{1-x}{2}=\dfrac{y-3}{1}=\dfrac{2z+1}{4}$ are

Q4. On the line $\vec r=\vec a+\lambda\vec b$, the point with $\lambda=0$ is

Q5. $F$ is the foot of the perpendicular from $P$ to the line $\vec r=\vec a+\lambda\vec b$. Which condition determines $F$?

← 11.1 Direction Cosines and Direction Ratios of a Line 11.3 Angle between Two Lines and Shortest Distance →