Chapter 10 · Vector Algebra

10.1 Vectors, Their Types and Components

Some quantities are fixed by one number; others also need a direction. A vector is an arrow with a length and a direction — and in three dimensions every vector can be written with just three numbers, its components along î, ĵ and k̂.

A delivery drone leaves its charging pad and hovers 30 m east, 40 m north and 120 m up. To describe where it is, “130 metres away” is not enough — 130 m in which direction? On the other hand, its battery holds 54 watt-hours, and that single number says everything there is to say about the charge. Physics and engineering are full of this split: mass, time, temperature and energy need only a number (with a unit), while displacement, velocity, force and acceleration need a number and a direction.

This section sets up the language for the second kind of quantity. A vector is drawn as an arrow: its length is the magnitude, and the way it points is the direction. You will describe directions in space by three angles and their cosines — the direction cosines — meet the special kinds of vectors (zero, unit, equal, negative, collinear, coinitial), and then see the key idea of the whole chapter: once the three unit vectors $\hat i,\hat j,\hat k$ along the axes are fixed, every vector in space is just $x\hat i+y\hat j+z\hat k$, three numbers you can calculate with.

Scalars, Vectors and Position Vectors

📘 Definition 10.1 — Scalars and vectors

A scalar quantity is one that is completely described by a single real number (its magnitude, with a unit where needed) — for example length, mass, time, speed, temperature, work, density.

A vector quantity is one that has both a magnitude and a direction — for example displacement, velocity, acceleration, force, momentum. Geometrically a vector is represented by a directed line segment $\overrightarrow{AB}$: $A$ is its initial point, $B$ its terminal point, the arrow points from $A$ to $B$, and the length $AB$ is its magnitude, written $|\overrightarrow{AB}|$. A vector is also named by a single letter with an arrow, $\vec a$, with magnitude $|\vec a|$ (or simply $a$). A magnitude is never negative.

✏️ Example 1 — Scalar or vector?

Classify each quantity as a scalar or a vector: (a) a mass of $40$ kg; (b) a wind of $15$ m/s from the west; (c) a temperature of $310$ K; (d) a force of $10$ N vertically downwards; (e) $5$ seconds; (f) a displacement of $3$ km towards the north-east; (g) a density of $1.2$ g/cm³; (h) the work done, $200$ J.

💡 Show solution▶

Test Ask: does the statement still make complete sense if the direction is thrown away? If yes, it is a scalar; if the direction is part of the information, it is a vector.

Scalars (a) mass, (c) temperature, (e) time, (g) density and (h) work — each is one number with a unit; none has a direction.

Vectors (b) the wind velocity (speed $15$ m/s, direction west-to-east), (d) the force, (f) the displacement — each needs its direction.

Note Work in (h) is a scalar even though it is made from two vectors (force and displacement): you will see in Section 10.3 that it is their scalar product.

To work with vectors in space we fix a rectangular coordinate system: an origin $O$ and three mutually perpendicular axes $OX$, $OY$, $OZ$ (Class 11 Mathematics, Ch. 11 — Introduction to Three Dimensional Geometry). We always use a right-handed system: curl the fingers of your right hand from $OX$ towards $OY$ and your thumb points along $OZ$.

📘 Definition 10.2 — Position vector

The position vector of a point $P(x,y,z)$ with respect to the origin $O$ is the vector $\vec r=\overrightarrow{OP}$, with initial point $O$ and terminal point $P$. Its magnitude is the distance $OP$: $$|\overrightarrow{OP}|=\sqrt{x^2+y^2+z^2}.$$

Why the formula holds. Drop a perpendicular from $P$ to the $xy$-plane, meeting it at $Q(x,y,0)$. In the $xy$-plane, $OQ^2=x^2+y^2$ (Pythagoras). The segment $QP$ is vertical, of length $|z|$, and perpendicular to $OQ$, so triangle $OQP$ is right-angled at $Q$ and $OP^2=OQ^2+QP^2=x^2+y^2+z^2$ — Pythagoras twice.

Now for the direction. The arrow $\overrightarrow{OP}$ makes an angle with each positive axis, and these three angles pin down which way it points.

📘 Definition 10.3 — Direction angles, direction cosines and direction ratios

Let $\vec r=\overrightarrow{OP}$ be a non-zero vector, and let $\alpha,\beta,\gamma$ be the angles it makes with the positive directions of the $x$-, $y$- and $z$-axes, each taken in $[0,\pi]$. The numbers $$l=\cos\alpha,\quad m=\cos\beta,\quad n=\cos\gamma$$ are the direction cosines of $\vec r$. Any three numbers $a,b,c$ proportional to them — $a=\lambda l$, $b=\lambda m$, $c=\lambda n$ for some $\lambda\ne0$ — are called direction ratios of $\vec r$.

📐 Theorem 10.1 — Direction cosines of a position vector

Let $P(x,y,z)$ be a point other than $O$, and let $r=|\overrightarrow{OP}|$. Then the direction cosines of $\overrightarrow{OP}$ are $$l=\frac xr,\qquad m=\frac yr,\qquad n=\frac zr,$$ so the coordinates $x,y,z$ themselves are direction ratios of $\overrightarrow{OP}$, and $$l^2+m^2+n^2=1.$$

Proof. Let $A$ be the foot of the perpendicular from $P$ to the $x$-axis; then $A=(x,0,0)$ and triangle $OAP$ has a right angle at $A$. The angle at $O$ is $\alpha$, so the signed length $OA$ along the positive $x$-axis is $r\cos\alpha$, that is $x=r\cos\alpha$. (If $x\lt0$ the angle $\alpha$ is obtuse and both sides are negative; if $x=0$, $\alpha=\tfrac\pi2$.) In the same way $y=r\cos\beta$ and $z=r\cos\gamma$. Dividing by $r$ gives $l,m,n$, and $$l^2+m^2+n^2=\frac{x^2+y^2+z^2}{r^2}=\frac{r^2}{r^2}=1.\ \square$$

So the direction cosines are not three free numbers: once two of them are known, the third is fixed up to its sign. And from any direction ratios $a,b,c$ (not all zero) you recover the direction cosines by dividing by $\sqrt{a^2+b^2+c^2}$: $$l=\pm\frac{a}{\sqrt{a^2+b^2+c^2}},\ \ m=\pm\frac{b}{\sqrt{a^2+b^2+c^2}},\ \ n=\pm\frac{c}{\sqrt{a^2+b^2+c^2}},$$ the same sign being taken throughout. The two choices of sign give two opposite directions along the same line.

✏️ Example 2 — Magnitude and direction cosines of a position vector

Find the magnitude and the direction cosines of the position vector of $P(2,-1,2)$, and the angles it makes with the axes.

💡 Show solution▶

Magnitude $|\overrightarrow{OP}|=\sqrt{4+1+4}=\sqrt9=3$.

Direction cosines By Theorem 10.1, $l=\tfrac23$, $m=-\tfrac13$, $n=\tfrac23$.

Angles $\alpha=\gamma=\cos^{-1}\tfrac23\approx48.19^\circ$ and $\beta=\cos^{-1}\!\left(-\tfrac13\right)\approx109.47^\circ$ — obtuse, because $P$ lies on the negative-$y$ side.

Check $l^2+m^2+n^2=\tfrac49+\tfrac19+\tfrac49=1$ ✓.

✏️ Example 3 — The third angle

A vector makes an angle of $45^\circ$ with the $x$-axis and $60^\circ$ with the $y$-axis. Find the angle it makes with the $z$-axis.

💡 Show solution▶

Use the identity $l=\cos45^\circ=\tfrac1{\sqrt2}$ and $m=\cos60^\circ=\tfrac12$, so $n^2=1-\tfrac12-\tfrac14=\tfrac14$.

Both signs $n=\tfrac12$ gives $\gamma=60^\circ$; $n=-\tfrac12$ gives $\gamma=120^\circ$.

Answer $\gamma=60^\circ$ or $120^\circ$. There are two such vectors, mirror images in the $xy$-plane — one pointing upwards, one downwards. If the question says the angle with $OZ$ is acute, only $60^\circ$ remains.

The red arrow is $\vec a=\overrightarrow{OP}$. Set its components with the sliders, or drag its tip to move $P$ parallel to the $xy$-plane; drag anywhere else (or press Play) to turn the axes. The dashed box shows the components; the coloured arcs are the angles $\alpha,\beta,\gamma$ with the axes. However you change $\vec a$, the three squared direction cosines always add up to exactly $1$.

$x$:$y$:$z$:

$\vec a$—

$|\vec a|$—

$l,\ m,\ n$—

$\alpha,\ \beta,\ \gamma$—

$l^2+m^2+n^2$—

Types of Vectors

In this chapter a vector is fixed by its magnitude and direction only; where it starts does not matter. Sliding an arrow parallel to itself does not change the vector. (Physicists call these free vectors. A force acting on a door at the hinge or at the handle has a different effect, but that is extra information about the point of application, not about the vector.)

📘 Definition 10.4 — Special kinds of vectors

Zero vector. A vector whose initial and terminal points coincide, such as $\overrightarrow{AA}$. Its magnitude is $0$ and it has no definite direction (it may be regarded as having any direction). It is written $\vec 0$.

Unit vector. A vector of magnitude $1$. The unit vector in the direction of a non-zero vector $\vec a$ is written $\hat a$ (“$a$ hat”).

Coinitial vectors. Two or more vectors with the same initial point, such as $\overrightarrow{OA}$, $\overrightarrow{OB}$, $\overrightarrow{OC}$.

Collinear vectors. Two or more vectors parallel to the same line, whatever their magnitudes and whichever way along the line they point. (Also called parallel vectors.)

Equal vectors. $\vec a=\vec b$ when they have the same magnitude and the same direction, regardless of their initial points.

Negative of a vector. The vector with the same magnitude as $\vec a$ but the opposite direction, written $-\vec a$. In particular $\overrightarrow{BA}=-\overrightarrow{AB}$.

✏️ Example 4 — Naming vectors in a parallelogram

$ABCD$ is a parallelogram whose diagonals meet at $O$. Using only the points $A,B,C,D,O$: (a) name a vector equal to $\overrightarrow{AB}$ and one equal to $\overrightarrow{AO}$; (b) name the negative of $\overrightarrow{AB}$; (c) name three vectors collinear with $\overrightarrow{AB}$; (d) name four coinitial vectors; (e) is $\overrightarrow{OA}=\overrightarrow{OC}$?

💡 Show solution▶

(a) Opposite sides of a parallelogram are parallel and equal in length, and $\overrightarrow{AB}$, $\overrightarrow{DC}$ point the same way: $\overrightarrow{DC}=\overrightarrow{AB}$. The diagonals bisect each other, so $AO=OC$ with the same direction: $\overrightarrow{OC}=\overrightarrow{AO}$.

(b) $\overrightarrow{BA}$, and also $\overrightarrow{CD}$ (same length as $AB$, opposite direction): $\overrightarrow{CD}=-\overrightarrow{AB}$.

(c) $\overrightarrow{BA}$, $\overrightarrow{DC}$ and $\overrightarrow{CD}$ are all parallel to the line $AB$, so all are collinear with $\overrightarrow{AB}$ — even though $DC$ is a different line, parallel to $AB$.

(d) $\overrightarrow{OA}$, $\overrightarrow{OB}$, $\overrightarrow{OC}$, $\overrightarrow{OD}$ all start at $O$.

(e) No. They have equal magnitudes but opposite directions: $\overrightarrow{OC}=-\overrightarrow{OA}$. They are collinear and coinitial, but not equal.

⚠️ Equal, collinear, same length — keep them apart

1. Same length is not enough. Two vectors of magnitude $5$ pointing different ways are not equal. Equality needs magnitude and direction.

2. Collinear does not mean “on the same line”. Collinear vectors only need to be parallel to one line; $\overrightarrow{AB}$ and $\overrightarrow{DC}$ in a parallelogram lie on different lines and are collinear. They may also point in opposite senses.

3. Equal vectors may start anywhere. $\overrightarrow{AB}=\overrightarrow{DC}$ although $A\ne D$. Do not reject equality because the initial points differ.

4. $\vec 0$ is not the number $0$. Write $\overrightarrow{AA}=\vec 0$ (a vector) but $|\overrightarrow{AA}|=0$ (a number).

The red vector $\vec a$ is fixed. Choose a blue vector $\vec b$ from the list, or drag its tail (to slide it) and its tip (to change it). Press Play to slide $\vec b$ around the plane: its components never change, so it stays the same vector wherever it starts. The lab names every relation that holds between $\vec a$ and $\vec b$.

$\vec b$:

$\vec a$—

$\vec b$—

$|\vec a|,\ |\vec b|$—

relations—

Components of a Vector

Here is the idea that turns geometry into arithmetic. Put a unit vector along each positive axis: $\hat i$ along $OX$, $\hat j$ along $OY$, $\hat k$ along $OZ$ — so $\hat i$ is the position vector of $(1,0,0)$, $\hat j$ of $(0,1,0)$, $\hat k$ of $(0,0,1)$. To reach $P(x,y,z)$ from $O$ you can walk $x$ units along $\hat i$, then $y$ units along $\hat j$, then $z$ units along $\hat k$. In Section 10.2 you will see that this “walk” is exactly a sum of vectors; for now we use it as a way of writing the vector.

📘 Definition 10.5 — Component form

The position vector of $P(x,y,z)$ is written $$\overrightarrow{OP}=x\hat i+y\hat j+z\hat k.$$ This is the component form of the vector. The numbers $x,y,z$ are its scalar components (or simply components), and $x\hat i$, $y\hat j$, $z\hat k$ are its vector components along the three axes. Since a vector may be slid to start at $O$, every vector $\vec a$ in space can be written as $\vec a=a_1\hat i+a_2\hat j+a_3\hat k$.

📐 Theorem 10.2 — Working with components

Let $\vec a=a_1\hat i+a_2\hat j+a_3\hat k$ and $\vec b=b_1\hat i+b_2\hat j+b_3\hat k$.

(a) $|\vec a|=\sqrt{a_1^2+a_2^2+a_3^2}$.

(b) $\vec a=\vec b$ if and only if $a_1=b_1$, $a_2=b_2$ and $a_3=b_3$.

(c) If $\vec a\ne\vec 0$, its direction cosines are $\dfrac{a_1}{|\vec a|},\dfrac{a_2}{|\vec a|},\dfrac{a_3}{|\vec a|}$; its components $a_1,a_2,a_3$ are direction ratios; and the unit vector in its direction is

$$\hat a=\frac{a_1}{|\vec a|}\hat i+\frac{a_2}{|\vec a|}\hat j+\frac{a_3}{|\vec a|}\hat k=l\hat i+m\hat j+n\hat k.$$

Why. Slide $\vec a$ so that it starts at $O$; it then ends at the point $(a_1,a_2,a_3)$. Part (a) is Definition 10.2 and part (c) is Theorem 10.1 for that point; the vector $l\hat i+m\hat j+n\hat k$ is the position vector of $(l,m,n)$, whose magnitude is $\sqrt{l^2+m^2+n^2}=1$ and whose direction is that of $\vec a$. For (b): two vectors are equal exactly when, both slid to start at $O$, they end at the same point — and two points are the same exactly when their coordinates agree. $\square$

✏️ Example 5 — Unit vector and a vector of given length

Let $\vec a=2\hat i-3\hat j+6\hat k$. Find (a) $|\vec a|$; (b) the direction cosines of $\vec a$; (c) the unit vector $\hat a$; (d) the vector of magnitude $14$ in the direction of $\vec a$.

💡 Show solution▶

(a) $|\vec a|=\sqrt{4+9+36}=\sqrt{49}=7$.

(b) $l=\tfrac27$, $m=-\tfrac37$, $n=\tfrac67$.

(d) A vector of magnitude $14$ in the same direction has the same direction cosines, so its components are $14l,14m,14n$: $\ 4\hat i-6\hat j+12\hat k$. Check: $\sqrt{16+36+144}=\sqrt{196}=14$ ✓.

✏️ Example 6 — Equal vectors and direction ratios

(a) Find $x,y,z$ if $(x+1)\hat i+2y\hat j+(z-3)\hat k=3\hat i-4\hat j+\hat k$. (b) A vector has direction ratios $1,-2,2$. Find its direction cosines.

💡 Show solution▶

(a) By Theorem 10.2(b), compare components: $x+1=3$, $2y=-4$, $z-3=1$, so $x=2$, $y=-2$, $z=4$.

(b) $\sqrt{1+4+4}=3$, so the direction cosines are $\tfrac13,-\tfrac23,\tfrac23$ — or, for the opposite direction along the same line, $-\tfrac13,\tfrac23,-\tfrac23$.

Check $\tfrac19+\tfrac49+\tfrac49=1$ ✓. Direction ratios fix a line; the actual vector picks one of its two directions.

⚠️ Common mistakes when studying alone

1. Adding instead of squaring. $|2\hat i-3\hat j+6\hat k|$ is $\sqrt{4+9+36}=7$, not $2-3+6=5$.

2. Forgetting the sign choice. From $l^2+m^2+n^2=1$ you get $n$ only up to sign. Keep both answers unless the question says the angle is acute or obtuse.

3. Direction ratios are not unique. $1,-2,2$ and $3,-6,6$ and $-\tfrac12,1,-1$ are all direction ratios of the same line. Direction cosines are unique for a given vector.

4. The angles do not add to $180^\circ$. The identity is about cosines squared: $\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1$. Equivalently $\sin^2\alpha+\sin^2\beta+\sin^2\gamma=2$. There is no rule for $\alpha+\beta+\gamma$.

5. The zero vector has no direction cosines — you would be dividing by $|\vec 0|=0$.

Exercises

Scalars, vectors and types of vectors.

1 Classify as scalar or vector: (a) $25$ kg; (b) $12$ m/s due north; (c) $300$ K; (d) $9.8$ m/s² vertically downwards; (e) $20$ J; (f) $4$ km to the south-west; (g) $1.2$ g/cm³.
2 Classify as scalar or vector: time period, displacement, speed, force, volume, momentum.
3 $ABCDEF$ is a regular hexagon with centre $O$, lettered anticlockwise. Using only the points $A,\dots,F,O$, name three vectors equal to $\overrightarrow{AB}$.
4 In the hexagon of part 3, name four vectors equal to $-\overrightarrow{AB}$.
5 True or false? (a) Two collinear vectors are always equal. (b) Two vectors with the same magnitude are equal. (c) Equal vectors are collinear. (d) The zero vector is collinear with every vector. (e) Coinitial vectors have the same initial point. (f) If $|\vec a|=|\vec b|$ then $\vec a=\pm\vec b$.
6 Draw a vector representing a displacement of $40$ km, $30^\circ$ east of north, using a scale of $1$ cm $=10$ km. How long is your arrow?

💡 Show solutions▶

1 Scalars: (a), (c), (e), (g). Vectors: (b) velocity, (d) acceleration, (f) displacement.

2 Scalars: time period, speed, volume. Vectors: displacement, force, momentum.

3 The six triangles round $O$ are equilateral, so $OC$, $FO$ and $ED$ are parallel to $AB$, of the same length and pointing the same way: $\overrightarrow{FO}=\overrightarrow{OC}=\overrightarrow{ED}=\overrightarrow{AB}$.

4 Reverse each arrow: $\overrightarrow{BA}$, $\overrightarrow{OF}$, $\overrightarrow{CO}$, $\overrightarrow{DE}$.

5 (a) False — they may differ in length or sense. (b) False — the directions may differ. (c) True. (d) True, since $\vec 0$ may be given any direction. (e) True (definition). (f) False — e.g. $\hat i$ and $\hat j$ have equal magnitudes but are not collinear.

6 An arrow $4$ cm long, drawn at $30^\circ$ from the north line towards the east, with the arrowhead at the far end.

Position vectors, magnitudes and direction cosines.

1 Find $|\overrightarrow{OP}|$ for $P(1,-2,2)$.
2 Find the magnitude and direction cosines of $\overrightarrow{OP}$ for $P(3,4,12)$.
3 Find the magnitude and direction cosines of $\overrightarrow{OP}$ for $P(-6,2,3)$.
4 Find the direction cosines of $\overrightarrow{OP}$ for $P(1,1,1)$ and the angle it makes with each axis.
5 Find the direction cosines of a vector with direction ratios $2,3,-6$.
6 Find the direction cosines of a vector with direction ratios $1,-1,0$, and the angles it makes with the axes (take the first sign).
7 Can $\tfrac12,\tfrac12,\tfrac12$ be the direction cosines of a vector? Explain.
8 A vector makes angles of $60^\circ$ with $OX$ and $45^\circ$ with $OY$. Find the angle it makes with $OZ$.
9 Find the direction cosines of a vector that is equally inclined to the three axes.
10 Write down the direction cosines of (a) the positive $z$-axis; (b) the negative $x$-axis.

💡 Show solutions▶

1 $\sqrt{1+4+4}=3$.

2 $\sqrt{9+16+144}=\sqrt{169}=13$; direction cosines $\tfrac3{13},\tfrac4{13},\tfrac{12}{13}$.

3 $\sqrt{36+4+9}=7$; direction cosines $-\tfrac67,\tfrac27,\tfrac37$.

4 $|\overrightarrow{OP}|=\sqrt3$, so $l=m=n=\tfrac1{\sqrt3}$, and each angle is $\cos^{-1}\tfrac1{\sqrt3}\approx54.74^\circ$.

5 $\sqrt{4+9+36}=7$: $\ \pm\left(\tfrac27,\tfrac37,-\tfrac67\right)$.

6 $\sqrt{1+1+0}=\sqrt2$: $\ \tfrac1{\sqrt2},-\tfrac1{\sqrt2},0$ (or all signs reversed). Angles $45^\circ$, $135^\circ$, $90^\circ$ — the vector lies in the $xy$-plane.

7 No: $\tfrac14+\tfrac14+\tfrac14=\tfrac34\ne1$. (They are direction ratios of the line making equal angles with the axes.)

8 $\cos^2\gamma=1-\tfrac14-\tfrac12=\tfrac14$, so $\cos\gamma=\pm\tfrac12$: $\gamma=60^\circ$ or $120^\circ$.

9 $l=m=n$ and $3l^2=1$: $\ l=m=n=\tfrac1{\sqrt3}$ or $l=m=n=-\tfrac1{\sqrt3}$.

10 (a) $0,0,1$ (angles $90^\circ,90^\circ,0^\circ$). (b) $-1,0,0$ (angles $180^\circ,90^\circ,90^\circ$).

III

Components and unit vectors.

1 Find $|\vec a|$ and $\hat a$ for $\vec a=3\hat i-4\hat j$.
2 Find $|\vec b|$ and $\hat b$ for $\vec b=\hat i+2\hat j-2\hat k$.
3 Find the unit vector in the direction of $\vec c=4\hat i+4\hat j-7\hat k$.
4 Find the vector of magnitude $10$ in the direction of $3\hat i+4\hat k$.
5 Find $x,y,z$ if $(x-2)\hat i+(y+3)\hat j+2z\hat k=5\hat i-\hat j+8\hat k$.
6 Are $\vec a=2\hat i+3\hat j$ and $\vec b=3\hat i+2\hat j$ equal? Compare their magnitudes.
7 Write down the scalar components and the vector components of $\vec a=-2\hat i+5\hat k$.
8 For which values of $p$ is $p\hat i+p\hat j+p\hat k$ a unit vector?
9 Find the vector of magnitude $6$ that makes equal acute angles with the three axes.

Check: magnitude, direction cosines and unit vector

$\vec a$ = $\hat i$ + $\hat j$ + $\hat k$

—

Whole-number components give an exact surd answer; decimals are accepted too.

💡 Show solutions▶

1 $|\vec a|=\sqrt{9+16}=5$; $\ \hat a=\tfrac35\hat i-\tfrac45\hat j$.

2 $|\vec b|=\sqrt{1+4+4}=3$; $\ \hat b=\tfrac13\hat i+\tfrac23\hat j-\tfrac23\hat k$.

3 $|\vec c|=\sqrt{16+16+49}=9$; $\ \hat c=\tfrac49\hat i+\tfrac49\hat j-\tfrac79\hat k$.

4 $|3\hat i+4\hat k|=5$, unit vector $\tfrac35\hat i+\tfrac45\hat k$; times $10$: $\ 6\hat i+8\hat k$.

5 $x-2=5$, $y+3=-1$, $2z=8$: $\ x=7$, $y=-4$, $z=4$.

6 Not equal — the $\hat i$-components $2$ and $3$ differ (Theorem 10.2(b)) — although $|\vec a|=|\vec b|=\sqrt{13}$.

7 Scalar components $-2,\ 0,\ 5$; vector components $-2\hat i$, $\ 0\hat j=\vec 0$, $\ 5\hat k$.

8 $\sqrt{3p^2}=|p|\sqrt3=1$, so $p=\pm\tfrac1{\sqrt3}$.

9 The direction cosines are $\tfrac1{\sqrt3}$ each (acute angles), so the vector is $\tfrac6{\sqrt3}(\hat i+\hat j+\hat k)=2\sqrt3\,\hat i+2\sqrt3\,\hat j+2\sqrt3\,\hat k$. Check: $\sqrt{3\cdot12}=6$ ✓.

Reasoning and applications.

1 Drone. With $x$ east, $y$ north and $z$ up (in metres), a drone is at $P(30,40,120)$ relative to its launch pad $O$. Find its distance from $O$ and the angle between $\overrightarrow{OP}$ and the vertical.
2 Show that no vector can make angles of $30^\circ$ with $OX$ and $45^\circ$ with $OY$.
3 If $\alpha,\beta,\gamma$ are the direction angles of a vector, show that $\sin^2\alpha+\sin^2\beta+\sin^2\gamma=2$ and $\cos2\alpha+\cos2\beta+\cos2\gamma=-1$.
4 A point $P$ has $|\overrightarrow{OP}|=6$, and the direction ratios of $\overrightarrow{OP}$ are $1,2,2$ with all angles acute. Find the coordinates of $P$.
5 Phone sensor (illustrative reading). A phone lying tilted on a desk reports the vector $(0,\ 5.88,\ 7.84)$ m/s² for gravity along its own $x$-, $y$- and $z$-axes. Find the magnitude of this vector and the angle between gravity and the phone's $z$-axis.
6 Find $k$ if $\tfrac13,\ k,\ \tfrac23$ are the direction cosines of a vector.

💡 Show solutions▶

1 $|\overrightarrow{OP}|=\sqrt{900+1600+14400}$ $=\sqrt{16900}=130$ m. The angle with the vertical ($OZ$) is $\gamma$ with $\cos\gamma=\tfrac{120}{130}=\tfrac{12}{13}$: $\ \gamma=\cos^{-1}\tfrac{12}{13}\approx22.62^\circ$.

2 $\cos^2 30^\circ+\cos^2 45^\circ=\tfrac34+\tfrac12=\tfrac54\gt1$, so $\cos^2\gamma$ would be $-\tfrac14$, which is impossible.

3 $\sin^2\alpha+\sin^2\beta+\sin^2\gamma$ $=3-(\cos^2\alpha+\cos^2\beta+\cos^2\gamma)$ $=3-1=2$. Also $\cos2\alpha=2\cos^2\alpha-1$, so the sum is $2(1)-3=-1$.

4 $\sqrt{1+4+4}=3$, so $l,m,n=\tfrac13,\tfrac23,\tfrac23$ and $P=(6l,6m,6n)=(2,4,4)$.

5 $\sqrt{0+34.5744+61.4656}=\sqrt{96.04}=9.8$ m/s². $\cos\gamma=\tfrac{7.84}{9.8}=0.8$, so $\gamma=\cos^{-1}0.8\approx36.87^\circ$: the phone is tilted about $37^\circ$ from lying flat.

6 $k^2=1-\tfrac19-\tfrac49=\tfrac49$, so $k=\pm\tfrac23$.

🎯 Exam corner Board · 1 markBoard · 2 marksJEE MainJEE Advanced

How this is examined: Board papers ask for a unit vector, a vector of given magnitude, or direction cosines (1–2 marks), and use the identity $l^2+m^2+n^2=1$ in MCQs and Assertion–Reason. JEE combines the identity with angle conditions and sign choices.

E1. Board · 1 mark If a vector makes angles $\alpha,\beta,\gamma$ with the coordinate axes, then $\sin^2\alpha+\sin^2\beta+\sin^2\gamma$ equals (A) $1$ (B) $2$ (C) $3$ (D) $0$

💡 Show solution▶

Idea $\sin^2=1-\cos^2$ for each angle, so the sum is $3-(l^2+m^2+n^2)=3-1=2$.

Answer (B). Trap (A) is the value for the cosines squared.

E2. Board · 1 mark Assertion–Reason. Assertion (A): $\tfrac1{\sqrt2},\tfrac12,\tfrac12$ can be the direction cosines of a vector. Reason (R): the direction cosines $l,m,n$ of any vector satisfy $l^2+m^2+n^2=1$. (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

💡 Show solution▶

Idea R is Theorem 10.1. Here $\tfrac12+\tfrac14+\tfrac14=1$, so the numbers satisfy the identity; and any three numbers with squares adding to $1$ are the direction cosines of the position vector of the point $(l,m,n)$, which is a unit vector. So A is true, and R is what makes it checkable.

Answer (A). (They are the direction cosines of the vector at $45^\circ$, $60^\circ$, $60^\circ$ to the axes.)

E3. Board · 2 marks Find a vector of magnitude $9$ units in the direction of $\vec a=2\hat i-\hat j+2\hat k$.

💡 Show solution▶

Unit vector $|\vec a|=\sqrt{4+1+4}=3$, so $\hat a=\tfrac23\hat i-\tfrac13\hat j+\tfrac23\hat k$. (1 mark)

Answer $9\hat a=6\hat i-3\hat j+6\hat k$. Check: $\sqrt{36+9+36}=9$ ✓. (1 mark)

E4. Board · 2 marks Find $x$ and $y$ so that $\vec a=2x\hat i+3\hat j$ and $\vec b=8\hat i+(y-1)\hat j$ are equal. Then find $|\vec a|$ and the direction cosines of $\vec a$.

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Equality $2x=8$ and $3=y-1$: $\ x=4$, $y=4$. (1 mark)

Answer $\vec a=8\hat i+3\hat j$, $|\vec a|=\sqrt{64+9}=\sqrt{73}$; direction cosines $\tfrac8{\sqrt{73}},\tfrac3{\sqrt{73}},0$. (1 mark) The third is $0$ because $\vec a$ has no $\hat k$-component — it is perpendicular to $OZ$.

E5. JEE Main A vector $\vec r$ with $|\vec r|=6$ makes angles $\tfrac\pi4$ with $OX$ and $\tfrac\pi3$ with $OY$, and an obtuse angle with $OZ$. Then $\vec r$ equals (A) $3\sqrt2\,\hat i+3\hat j-3\hat k$ (B) $3\sqrt2\,\hat i+3\hat j+3\hat k$ (C) $3\hat i+3\sqrt2\,\hat j-3\hat k$ (D) $3\sqrt2\,\hat i-3\hat j+3\hat k$

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Method $l=\tfrac1{\sqrt2}$, $m=\tfrac12$, so $n^2=1-\tfrac12-\tfrac14=\tfrac14$ and, the angle being obtuse, $n=-\tfrac12$. Then $\vec r=6(l\hat i+m\hat j+n\hat k)=3\sqrt2\,\hat i+3\hat j-3\hat k$.

Answer (A). Trap (B) ignores “obtuse”; (C) swaps the $x$- and $y$-angles; (D) puts the minus sign on the wrong component.

⏱ Fast route Only (A) and (B) have the right $x$- and $y$-components; obtuse with $OZ$ means a negative $\hat k$-component.

E6. JEE Advanced (One or more correct options.) A non-zero vector makes an angle $\theta$ with $OX$, the same angle $\theta$ with $OY$, and the angle $2\theta$ with $OZ$, where $0\lt\theta\le\tfrac\pi2$. Then $\theta$ can be (A) $\tfrac\pi4$ (B) $\tfrac\pi2$ (C) $\tfrac\pi3$ (D) $\tfrac\pi6$

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Method $2\cos^2\theta+\cos^2 2\theta=1$. Since $2\cos^2\theta=1+\cos2\theta$, this is $\cos2\theta+\cos^2 2\theta=0$, i.e. $\cos2\theta\,(1+\cos2\theta)=0$.

Solve $\cos2\theta=0$ gives $\theta=\tfrac\pi4$ (direction $\tfrac1{\sqrt2},\tfrac1{\sqrt2},0$); $\cos2\theta=-1$ gives $\theta=\tfrac\pi2$, $2\theta=\pi$ (the direction of $-\hat k$). Both angles $2\theta$ lie in $[0,\pi]$ as required.

Answer (A) and (B). Check (C): $2\cdot\tfrac14+\tfrac14=\tfrac34\ne1$; (D): $2\cdot\tfrac34+\tfrac14=\tfrac74\ne1$.

🔬 Research–Teaching NexusMini project · about 45 minutes

Research question. How does your phone know which way is “down” — and can you use direction cosines to measure the tilt of a ramp, a roof or a staircase?

Why it matters. Every smartphone has an accelerometer that reports a 3-D vector along the phone's own $x$-, $y$- and $z$-axes. At rest that vector is (very nearly) gravity, so its direction cosines tell you exactly how the phone is tilted. Screen rotation, spirit-level apps, step counters and fall detectors in fitness bands all use this.

What to do.

Install a free sensor app that shows raw accelerometer readings (for example phyphox). Lay the phone flat and note the three components; check that the magnitude is close to $9.8$ m/s².
Place the phone on five different slopes (a book propped up, a ramp, a staircase rail…). For each, record $(a_x,a_y,a_z)$, compute the magnitude and the direction cosines, and find the tilt angle $\gamma=\cos^{-1}\dfrac{a_z}{|\vec a|}$.
Measure each slope with a protractor or by rise/run and compare with your vector answer.
Compute $l^2+m^2+n^2$ for every reading, first dividing by the magnitude you calculated and then dividing by $9.8$ instead. Why is the first sum always exactly $1$ while the second is not?

Tools: Phone with a sensor app; spreadsheet (SQRT, ACOS, DEGREES); protractor or measuring tape.

Hand in: A table of readings, magnitudes, direction cosines and angles, a comparison with your direct measurements, and a short paragraph on the sources of error.

Going further. Hold the phone still in your hand for 20 seconds and record many readings. How much do the direction cosines wobble? How would a fitness band decide that a sudden change means a fall?

Instructor note: the computed magnitude typically lies between $9.7$ and $9.9$ m/s² because of sensor noise and calibration offsets. Dividing by the computed $|\vec a|$ makes $l^2+m^2+n^2=1$ exactly (Theorem 10.1), whereas dividing by $9.8$ gives sums such as $0.98$–$1.02$ — a good discussion of why direction cosines must use the vector's own magnitude. Tilt from the $z$-component agrees with protractor readings to within $1^\circ$–$2^\circ$ on a stable surface.

🧪 Self-check quiz

0/5 correct

Q1. Which of these is a vector quantity?

Q2. $|2\hat i-3\hat j+6\hat k|$ equals

Q3. The direction cosines of $\hat i+\hat j+\hat k$ are

Q4. A vector makes angles $90^\circ$ with $OX$ and $60^\circ$ with $OY$. Then $\cos\gamma$ (for the angle with $OZ$) is

Q5. Which pair of vectors is collinear but not equal?

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