Chapter 3 ยท Motion in a Plane

3.2 Resolution of Vectors and Analytical Addition

Every vector can be split into perpendicular components along the axes. Components turn vector addition into ordinary arithmetic and give the formula for the resultant of any two vectors.

Pull a suitcase along a station platform by its handle, which makes $30^\circ$ with the ground. Part of your pull drags the suitcase forward; part of it lifts the suitcase, taking some of its weight off the wheels. A phone's satnav does the same kind of splitting all the time: it does not store your velocity as "$5.0\ \text{m/s}$ in some direction", but as two numbers โ€” how fast you are going east and how fast you are going north.

Splitting a vector into perpendicular parts is called resolving it, and the parts are its components. Section 3.1 added vectors by drawing, which is slow and only as accurate as your protractor. With components, adding vectors becomes adding numbers: add all the $x$-components, add all the $y$-components, and you are done. In this section you will learn to resolve vectors, rebuild a vector from its components, add any number of vectors exactly, and derive the formula for the resultant of two vectors at any angle.

Unit vectors

A vector is a magnitude times a direction. It helps to have an arrow that carries only the direction.

๐Ÿ“˜ Definition 3.5 โ€” Unit vector

A unit vector is a vector of magnitude exactly $1$, with no unit and no dimensions; it only specifies a direction. The unit vector along a non-zero vector $\mathbf{A}$ is

$$\hat{\mathbf{A}}=\frac{\mathbf{A}}{A},\qquad\text{so that}\qquad \mathbf{A}=A\,\hat{\mathbf{A}} .$$

The unit vectors along the positive $x$, $y$ and $z$ axes of a right-handed coordinate system are written $\hat{\mathbf{i}}$, $\hat{\mathbf{j}}$ and $\hat{\mathbf{k}}$. They are mutually perpendicular.

In $\mathbf{A}=A\,\hat{\mathbf{A}}$ the magnitude $A$ carries the number and the unit ("$5.0\ \text{m}$"), and $\hat{\mathbf{A}}$ carries the direction. This is Result 3.1 with $\lambda=A$.

Resolving a vector into components

Draw a vector $\mathbf{A}$ in the $xy$-plane from the origin, making an angle $\theta$ with the $+x$ axis, measured anticlockwise. Drop perpendiculars from its head to the two axes. By the triangle law, $\mathbf{A}$ is the sum of an arrow along the $x$-axis and an arrow along the $y$-axis, and these are multiples of $\hat{\mathbf{i}}$ and $\hat{\mathbf{j}}$.

๐Ÿ“˜ Definition 3.6 โ€” Rectangular components

Any vector $\mathbf{A}$ in the $xy$-plane can be written in exactly one way as

$$\mathbf{A}=A_x\,\hat{\mathbf{i}}+A_y\,\hat{\mathbf{j}},\qquad A_x=A\cos\theta,\quad A_y=A\sin\theta ,$$

where $\theta$ is the angle from the $+x$ axis to $\mathbf{A}$, measured anticlockwise. The numbers $A_x$ and $A_y$ are the components of $\mathbf{A}$; the vectors $A_x\hat{\mathbf{i}}$ and $A_y\hat{\mathbf{j}}$ are its component vectors. Components are scalars with signs: $A_x\lt0$ means the $x$-part points along $-x$. Resolving into perpendicular parts like this is called resolution of the vector.

๐Ÿ“ Result 3.4 โ€” Magnitude and direction from components

In two dimensions, by Pythagoras,

$$A=\sqrt{A_x^2+A_y^2},\qquad \tan\theta=\frac{A_y}{A_x},$$

where the quadrant of $\theta$ must be chosen from the signs of $A_x$ and $A_y$ (a calculator's $\tan^{-1}$ only returns angles between $-90^\circ$ and $90^\circ$).

In three dimensions, $\mathbf{A}=A_x\hat{\mathbf{i}}+A_y\hat{\mathbf{j}}+A_z\hat{\mathbf{k}}$ with

$$A=\sqrt{A_x^2+A_y^2+A_z^2},$$

and if $\alpha$, $\beta$, $\gamma$ are the angles $\mathbf{A}$ makes with the $+x$, $+y$, $+z$ axes, then $A_x=A\cos\alpha$, $A_y=A\cos\beta$, $A_z=A\cos\gamma$, so $\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1$.

Nothing forces us to use horizontal and vertical axes. On a slope, for example, it is often best to resolve along the slope and perpendicular to it (Exercise I.4). The components change when the axes are turned, but the vector โ€” and its magnitude โ€” do not.

โœ๏ธ Example 1 โ€” Pulling a suitcase
A traveller pulls a suitcase with a force of $20\ \text{N}$ along its handle, at $30^\circ$ above the horizontal. Find the horizontal and vertical components of the pull.
Axes $x$ horizontal in the direction of travel, $y$ vertically up; $\theta=30^\circ$.
Resolve $F_x=20\cos30^\circ=17.3\ \text{N}$ (moves the case forward); $F_y=20\sin30^\circ=10.0\ \text{N}$ (lifts it).
Answer $\mathbf{F}=(17.3\,\hat{\mathbf{i}}+10.0\,\hat{\mathbf{j}})\ \text{N}$.
Check $\sqrt{17.3^2+10.0^2}=20.0\ \text{N}$ โœ“. A lower handle ($\theta\to0$) puts more of the pull into moving the case.
โœ๏ธ Example 2 โ€” From components back to the vector
A displacement is $\mathbf{r}=(3.0\,\hat{\mathbf{i}}-4.0\,\hat{\mathbf{j}})\ \text{m}$. Find (a) its magnitude, (b) its direction, (c) the unit vector along it. (d) A student says $-3.0\,\hat{\mathbf{i}}+4.0\,\hat{\mathbf{j}}$ points the same way "because $\tan\theta$ is the same". Is she right?
(a) $r=\sqrt{3.0^2+(-4.0)^2}=5.0\ \text{m}$.
(b) $\tan\theta=\dfrac{-4.0}{3.0}$, $\theta=-53.1^\circ$. Since $r_x\gt0$ and $r_y\lt0$ the vector is in the fourth quadrant, so this calculator value is right: $53.1^\circ$ below the $+x$ axis.
(c) $\hat{\mathbf{r}}=\dfrac{\mathbf{r}}{r}=0.60\,\hat{\mathbf{i}}-0.80\,\hat{\mathbf{j}}$ (no unit; check $0.60^2+0.80^2=1$ โœ“).
(d) No. $-3.0\,\hat{\mathbf{i}}+4.0\,\hat{\mathbf{j}}$ is in the second quadrant, at $180^\circ-53.1^\circ=126.9^\circ$: it is $-\mathbf{r}$, pointing exactly the opposite way. The ratio $A_y/A_x$ cannot tell opposite directions apart โ€” the signs can.
โœ๏ธ Example 3 โ€” A vector in three dimensions
A drone's displacement is $\mathbf{r}=(2.0\,\hat{\mathbf{i}}+3.0\,\hat{\mathbf{j}}+6.0\,\hat{\mathbf{k}})\ \text{m}$, with $z$ vertical. Find its magnitude and the angles it makes with the three axes.
Magnitude $r=\sqrt{2.0^2+3.0^2+6.0^2}=\sqrt{49}=7.0\ \text{m}$.
Cosines $\cos\alpha=\tfrac27$, $\cos\beta=\tfrac37$, $\cos\gamma=\tfrac67$.
Angles $\alpha=73.4^\circ$, $\beta=64.6^\circ$, $\gamma=31.0^\circ$ โ€” the drone climbed steeply.
Check $\tfrac{4+9+36}{49}=1$ โœ“.
Interactive 1 โ€” Components Round the Circle

Set the magnitude and direction of $\mathbf{A}$ (or drag its head). The red and blue arrows along the axes are its component vectors $A_x\hat{\mathbf{i}}$ and $A_y\hat{\mathbf{j}}$; the graph on the right plots $A_x/A=\cos\theta$ and $A_y/A=\sin\theta$. Press Play to turn $\mathbf{A}$ all the way round and watch the components change sign. Compare the calculator value of $\tan^{-1}(A_y/A_x)$ with the true direction.

4.030ยฐ
$A_x=A\cos\theta$โ€”
$A_y=A\sin\theta$โ€”
$\sqrt{A_x^2+A_y^2}$โ€”
calculator $\tan^{-1}(A_y/A_x)$โ€”

Adding vectors by components

If $\mathbf{A}=A_x\hat{\mathbf{i}}+A_y\hat{\mathbf{j}}$ and $\mathbf{B}=B_x\hat{\mathbf{i}}+B_y\hat{\mathbf{j}}$, the commutative and associative laws let us regroup: $$\mathbf{A}+\mathbf{B}=(A_x+B_x)\,\hat{\mathbf{i}}+(A_y+B_y)\,\hat{\mathbf{j}} .$$

๐Ÿ“ Result 3.5 โ€” Vector algebra in components
  • Addition: if $\mathbf{R}=\mathbf{A}+\mathbf{B}+\mathbf{C}+\dots$, then $R_x=A_x+B_x+C_x+\dots$, $\ R_y=A_y+B_y+C_y+\dots$ (and $R_z$ likewise).
  • Subtraction: $\mathbf{A}-\mathbf{B}=(A_x-B_x)\,\hat{\mathbf{i}}+(A_y-B_y)\,\hat{\mathbf{j}}$.
  • Multiplication by a number: $\lambda\mathbf{A}=\lambda A_x\,\hat{\mathbf{i}}+\lambda A_y\,\hat{\mathbf{j}}$.
  • Equality: $\mathbf{A}=\mathbf{B}$ if and only if $A_x=B_x$ and $A_y=B_y$ (and $A_z=B_z$). One vector equation is two (or three) ordinary equations.
๐Ÿงญ Problem-solving strategy โ€” adding vectors by components
  1. Choose axes (usually east = $+x$, north = $+y$) and draw every vector from the origin.
  2. Resolve each vector: $A_x=A\cos\theta_A$, $A_y=A\sin\theta_A$, with $\theta$ from $+x$; check each sign against your sketch.
  3. Make a table and add the columns: $R_x=\sum A_x$, $R_y=\sum A_y$.
  4. $R=\sqrt{R_x^2+R_y^2}$; find the direction from $\tan\theta_R=R_y/R_x$ and the signs of $R_x$, $R_y$.
  5. Check that the answer is reasonable against a rough head-to-tail sketch.
โœ๏ธ Example 4 โ€” A three-leg walk
A walker goes $2.0\ \text{km}$ due east, then $4.0\ \text{km}$ at $60^\circ$ north of east, then $3.0\ \text{km}$ due west. Find her resultant displacement.
Resolve $\mathbf{A}$: $(2.0,\ 0)$; $\ \mathbf{B}$: $(4.0\cos60^\circ,\ 4.0\sin60^\circ)=(2.0,\ 3.46)$; $\ \mathbf{C}$: $(-3.0,\ 0)$, all in km.
Add $R_x=2.0+2.0-3.0=1.0\ \text{km}$; $\ R_y=0+3.46+0=3.46\ \text{km}$.
Combine $R=\sqrt{1.0^2+3.46^2}=3.6\ \text{km}$; $\ \tan\theta=\tfrac{3.46}{1.0}$, and both components are positive, so $\theta=73.9^\circ$: $3.6\ \text{km}$ at $74^\circ$ north of east.
Check In a sketch the walk ends a little east of north of the start โœ“. Interactive 2 draws this example with its components.

The resultant of two vectors at an angle $\theta$

For just two vectors there is a neat closed formula. Let $\theta$ be the angle between $\mathbf{A}$ and $\mathbf{B}$ drawn tail to tail, and choose the $x$-axis along $\mathbf{A}$. Then $\mathbf{A}=A\,\hat{\mathbf{i}}$ and $\mathbf{B}=B\cos\theta\,\hat{\mathbf{i}}+B\sin\theta\,\hat{\mathbf{j}}$, so

$$R_x=A+B\cos\theta,\qquad R_y=B\sin\theta .$$

Squaring and adding, and using $\cos^2\theta+\sin^2\theta=1$:

$$R^2=A^2+2AB\cos\theta+B^2(\cos^2\theta+\sin^2\theta)$$ $$\quad=A^2+B^2+2AB\cos\theta .$$
๐Ÿ“ Result 3.6 โ€” Resultant of two vectors (law of cosines)

If $\theta$ is the angle between $\mathbf{A}$ and $\mathbf{B}$ when they are drawn tail to tail, then $\mathbf{R}=\mathbf{A}+\mathbf{B}$ has

$$R=\sqrt{A^2+B^2+2AB\cos\theta},$$ $$\tan\alpha=\frac{B\sin\theta}{A+B\cos\theta},$$

where $\alpha$ is the angle between $\mathbf{R}$ and $\mathbf{A}$. For the difference, replace $\theta$ by $180^\circ-\theta$: $\ |\mathbf{A}-\mathbf{B}|=\sqrt{A^2+B^2-2AB\cos\theta}$.

For two vectors of equal magnitude $A$: $\ R=2A\cos\tfrac{\theta}{2}$, along the bisector of the angle between them.

The three special cases of Section 3.1 drop out: $\theta=0^\circ$ gives $R=A+B$, $\theta=180^\circ$ gives $R=|A-B|$, and $\theta=90^\circ$ gives $R=\sqrt{A^2+B^2}$. Since $-1\le\cos\theta\le1$, the formula also proves $|A-B|\le R\le A+B$.

โœ๏ธ Example 5 โ€” The scale drawing, done exactly
In Example 5 of Section 3.1 a walker moved $4.0\ \text{m}$ due east and then $3.0\ \text{m}$ at $60^\circ$ north of east; the scale drawing gave about $6.1\ \text{m}$ at $25^\circ$. Calculate the resultant (a) with Result 3.6, (b) with components.
(a) $\theta=60^\circ$: $R=\sqrt{4.0^2+3.0^2+2(4.0)(3.0)\cos60^\circ}=\sqrt{16+9+12}=\sqrt{37}=6.08\ \text{m}$.
Direction $\tan\alpha=\dfrac{3.0\sin60^\circ}{4.0+3.0\cos60^\circ}=\dfrac{2.60}{5.50}=0.472$, $\alpha=25.3^\circ$ north of east.
(b) $R_x=4.0+3.0\cos60^\circ=5.50\ \text{m}$, $R_y=3.0\sin60^\circ=2.60\ \text{m}$; $R=\sqrt{5.50^2+2.60^2}=6.08\ \text{m}$ โœ“ โ€” the same arithmetic, because the derivation of Result 3.6 is the component method with $x$ along $\mathbf{A}$.
Answer $6.08\ \text{m}$ at $25.3^\circ$ north of east, matching the drawing to its accuracy.
โœ๏ธ Example 6 โ€” Two ropes on a boat
(a) Two ropes pull a boat with forces of $120\ \text{N}$ and $80\ \text{N}$; the angle between the ropes is $50^\circ$. Find the resultant force. (b) If instead both ropes pull with $100\ \text{N}$ at $50^\circ$ to each other, find the resultant.
(a) Size $R^2=120^2+80^2+2(120)(80)\cos50^\circ=14\,400+6400+12\,342=33\,142\ \text{N}^2$, so $R=182\ \text{N}$.
(a) Direction $\tan\alpha=\dfrac{80\sin50^\circ}{120+80\cos50^\circ}=\dfrac{61.3}{171.4}=0.358$, $\alpha=19.7^\circ$ from the $120\ \text{N}$ rope โ€” closer to the stronger pull, as expected.
(b) Equal magnitudes: $R=2(100)\cos25^\circ=181\ \text{N}$ along the bisector, $25^\circ$ from each rope.
Check Both resultants lie between $|A-B|$ and $A+B$ ($40$โ€“$200\ \text{N}$ and $0$โ€“$200\ \text{N}$) โœ“; with ropes at $50^\circ$ about $9\%$ of the $200\ \text{N}$ total effort is "wasted" pulling sideways against the other rope.
โš ๏ธ Common mistakes

1. The wrong quadrant. $\tan^{-1}$ on a calculator gives $-90^\circ$ to $90^\circ$. If $A_x\lt0$, add $180^\circ$ to the calculator answer.

2. Sine and cosine swapped. $A\cos\theta$ is the component adjacent to the angle $\theta$. If the angle is given from the $y$-axis (a bearing, or an angle from the vertical), then $A_y=A\cos\theta$ and $A_x=A\sin\theta$.

3. The wrong angle in Result 3.6. $\theta$ is the angle between the vectors drawn tail to tail, not the angle inside the head-to-tail triangle (which is $180^\circ-\theta$ and flips the sign of the cosine term).

4. Adding magnitudes of components. Signs matter: a component of $-3.0\ \text{km}$ (west) cancels part of $+2.0\ \text{km}$ (east).

5. Units on unit vectors. $\hat{\mathbf{i}}$, $\hat{\mathbf{j}}$, $\hat{\mathbf{k}}$ and $\hat{\mathbf{A}}$ are pure numbers of size $1$; the unit belongs to the components.

Interactive 2 โ€” Adding Three Vectors by Components

Three vectors are laid head to tail; each one is drawn together with its own $x$-step (red, dashed) and $y$-step (blue, dashed). The resultant's components are just the sums of the steps. Choose a set, or set the third vector $\mathbf{C}$ yourself โ€” can you make the resultant zero? Press Play to build the sum one vector at a time.

3.0180ยฐ
$R_x=A_x+B_x+C_x$โ€”
$R_y=A_y+B_y+C_y$โ€”
resultant $R$โ€”
direction of $\mathbf{R}$โ€”

Exercises

I
Resolving vectors and unit vectors. Take $x$ east (or horizontal) and $y$ north (or vertically up).
  • 1 Resolve a force of $50\ \text{N}$ acting at $37^\circ$ above the horizontal.
  • 2 Resolve a velocity of $12\ \text{m/s}$ at $120^\circ$ anticlockwise from $+x$.
  • 3 Resolve a displacement of $8.0\ \text{m}$ at $30^\circ$ south of west.
  • 4 A box of weight $50\ \text{N}$ rests on a slope inclined at $30^\circ$. Resolve its weight into components along the slope and perpendicular to it.
  • 5 A vector has components $A_x=-5.0$ and $A_y=12.0$ (in N). Find its magnitude and direction.
  • 6 Find the unit vector along $6\,\hat{\mathbf{i}}-8\,\hat{\mathbf{j}}$.
  • 7 Find the magnitude of $\hat{\mathbf{i}}+2\hat{\mathbf{j}}+2\hat{\mathbf{k}}$ and the cosines of the angles it makes with the axes.
  • 8 A vector makes $60^\circ$ with the $x$-axis and $45^\circ$ with the $y$-axis. What angle can it make with the $z$-axis?
1 $F_x=50\cos37^\circ=39.9\ \text{N}$, $F_y=50\sin37^\circ=30.1\ \text{N}$.
2 $v_x=12\cos120^\circ=-6.0\ \text{m/s}$, $v_y=12\sin120^\circ=10.4\ \text{m/s}$.
3 Both parts point negative: $x=-8.0\cos30^\circ=-6.9\ \text{m}$, $y=-8.0\sin30^\circ=-4.0\ \text{m}$.
4 The weight makes $30^\circ$ with the normal to the slope. Along the slope (downhill): $50\sin30^\circ=25\ \text{N}$; into the slope: $50\cos30^\circ=43.3\ \text{N}$.
5 $A=\sqrt{25+144}=13.0\ \text{N}$. Calculator: $\tan^{-1}\tfrac{12.0}{-5.0}=-67.4^\circ$; but $A_x\lt0$, $A_y\gt0$ (second quadrant), so $\theta=180^\circ-67.4^\circ=112.6^\circ$ from $+x$.
6 Magnitude $\sqrt{36+64}=10$: unit vector $0.6\,\hat{\mathbf{i}}-0.8\,\hat{\mathbf{j}}$.
7 $\sqrt{1+4+4}=3$; $\cos\alpha=\tfrac13$, $\cos\beta=\tfrac23$, $\cos\gamma=\tfrac23$.
8 $\cos^2\gamma=1-\cos^260^\circ-\cos^245^\circ=1-\tfrac14-\tfrac12=\tfrac14$, so $\cos\gamma=\pm\tfrac12$: $\gamma=60^\circ$ or $120^\circ$.
II
Vector algebra in components. Let $\mathbf{A}=3\hat{\mathbf{i}}+4\hat{\mathbf{j}}$, $\mathbf{B}=-\hat{\mathbf{i}}+2\hat{\mathbf{j}}$ and $\mathbf{C}=2\hat{\mathbf{i}}-5\hat{\mathbf{j}}$ (all in metres).
  • 1 Find $\mathbf{A}+\mathbf{B}$ and its magnitude.
  • 2 Find $\mathbf{A}-\mathbf{B}$ and its magnitude.
  • 3 Find $\mathbf{A}+\mathbf{B}+\mathbf{C}$, its magnitude and its direction.
  • 4 Find $2\mathbf{A}-3\mathbf{B}$ and its magnitude.
  • 5 Find the vector $\mathbf{D}$ for which $\mathbf{A}+\mathbf{B}+\mathbf{C}+\mathbf{D}=\mathbf{0}$.
  • 6 If $(p+1)\,\hat{\mathbf{i}}+4\,\hat{\mathbf{j}}=3\,\hat{\mathbf{i}}+(q-2)\,\hat{\mathbf{j}}$, find $p$ and $q$.
  • 7 Three forces of $10\ \text{N}$ each act at $0^\circ$, $120^\circ$ and $240^\circ$ to the $x$-axis. Show by components that their resultant is zero.
  • 8 A hiker walks $3.0\ \text{km}$ north, then $5.0\ \text{km}$ at $30^\circ$ south of east, then $2.0\ \text{km}$ west. Find her resultant displacement.
1 $2\hat{\mathbf{i}}+6\hat{\mathbf{j}}$ m; $\sqrt{4+36}=6.3\ \text{m}$.
2 $4\hat{\mathbf{i}}+2\hat{\mathbf{j}}$ m; $\sqrt{20}=4.5\ \text{m}$.
3 $4\hat{\mathbf{i}}+\hat{\mathbf{j}}$ m; $\sqrt{17}=4.1\ \text{m}$ at $\tan^{-1}\tfrac14=14.0^\circ$ above $+x$.
4 $(6+3)\hat{\mathbf{i}}+(8-6)\hat{\mathbf{j}}=9\hat{\mathbf{i}}+2\hat{\mathbf{j}}$ m; $\sqrt{85}=9.2\ \text{m}$.
5 $\mathbf{D}=-(\mathbf{A}+\mathbf{B}+\mathbf{C})=-4\hat{\mathbf{i}}-\hat{\mathbf{j}}$ m.
6 Equate components: $p+1=3$, $p=2$; $4=q-2$, $q=6$.
7 $R_x=10(\cos0^\circ+\cos120^\circ+\cos240^\circ)=10(1-\tfrac12-\tfrac12)=0$; $R_y=10(0+\tfrac{\sqrt3}2-\tfrac{\sqrt3}2)=0$.
8 $R_x=0+5.0\cos30^\circ-2.0=2.33\ \text{km}$; $R_y=3.0-5.0\sin30^\circ+0=0.50\ \text{km}$. $R=\sqrt{2.33^2+0.50^2}=2.4\ \text{km}$ at $\tan^{-1}\tfrac{0.50}{2.33}=12^\circ$ north of east.
III
The resultant of two vectors (Result 3.6).
  • 1 Forces of $3.0\ \text{N}$ and $5.0\ \text{N}$ act at $60^\circ$ to each other. Find the resultant and its angle with the $3.0\ \text{N}$ force.
  • 2 Repeat part 1 for an angle of $120^\circ$ (magnitude only).
  • 3 Two forces of equal magnitude $F$ have a resultant of $\sqrt3\,F$. Find the angle between them.
  • 4 If $|\mathbf{A}+\mathbf{B}|=|\mathbf{A}-\mathbf{B}|$, find the angle between $\mathbf{A}$ and $\mathbf{B}$.
  • 5 Forces $P=5.0\ \text{N}$ and $Q=10\ \text{N}$ have a resultant perpendicular to $P$. Find the angle between them and the resultant.
  • 6 The largest and smallest resultants of two forces are $17\ \text{N}$ and $7\ \text{N}$. Find the forces, and their resultant when they are perpendicular.
  • 7 Vectors of magnitudes $4$ and $6$ units have a resultant of $8$ units. Find the angle between them.
1 $R=\sqrt{9+25+2(3)(5)(0.5)}=\sqrt{49}=7.0\ \text{N}$; $\tan\alpha=\dfrac{5.0\sin60^\circ}{3.0+5.0\cos60^\circ}=\dfrac{4.33}{5.5}$, $\alpha=38.2^\circ$.
2 $R=\sqrt{9+25-15}=\sqrt{19}=4.4\ \text{N}$.
3 $2F\cos\tfrac\theta2=\sqrt3F$, $\cos\tfrac\theta2=\tfrac{\sqrt3}2$, $\tfrac\theta2=30^\circ$: $\theta=60^\circ$.
4 $A^2+B^2+2AB\cos\theta=A^2+B^2-2AB\cos\theta$ gives $\cos\theta=0$: $\theta=90^\circ$ (the parallelogram is a rectangle).
5 $\mathbf{R}\perp\mathbf{P}$ means the component of $\mathbf{R}$ along $\mathbf{P}$ is zero: $P+Q\cos\theta=0$, $\cos\theta=-\tfrac12$, $\theta=120^\circ$. Then $R=Q\sin\theta=10\times0.866=8.7\ \text{N}$ (check: $\sqrt{25+100-50}=\sqrt{75}=8.7$ โœ“).
6 $A+B=17$, $A-B=7$: $A=12\ \text{N}$, $B=5\ \text{N}$; perpendicular: $\sqrt{144+25}=13\ \text{N}$.
7 $64=16+36+48\cos\theta$, $\cos\theta=0.25$, $\theta=75.5^\circ$.
IV
Applications.
  • 1 A child pulls a sledge with a rope at $25^\circ$ above the horizontal, with a tension of $150\ \text{N}$. Find the horizontal and vertical components of the pull.
  • 2 A footballer kicks a ball at $18\ \text{m/s}$ at $40^\circ$ above the horizontal. Find the horizontal and vertical components of its initial velocity. (You will use such components in Section 3.3.)
  • 3 A player runs $12\ \text{m}$ at $30^\circ$ to the $+x$ axis and then $8.0\ \text{m}$ at $120^\circ$. Find the resultant displacement.
  • 4 A $60\ \text{N}$ lamp hangs from the middle of a wire; each half of the wire makes $10^\circ$ with the horizontal. The vertical components of the two equal tensions $T$ together balance the weight. Find $T$, and comment.
  • 5 A GPS unit reports a velocity of $v_\text{east}=-3.0\ \text{m/s}$, $v_\text{north}=4.0\ \text{m/s}$. Find the speed and the direction of motion.
  • 6 $\mathbf{A}=3.0\,\hat{\mathbf{i}}+4.0\,\hat{\mathbf{j}}$. Find its components along new axes $x'$, $y'$ obtained by turning the $x$, $y$ axes $30^\circ$ anticlockwise, and check that the magnitude is unchanged.
1 Horizontal $150\cos25^\circ=136\ \text{N}$; vertical $150\sin25^\circ=63.4\ \text{N}$.
2 $u_x=18\cos40^\circ=13.8\ \text{m/s}$, $u_y=18\sin40^\circ=11.6\ \text{m/s}$.
3 The legs are perpendicular ($120^\circ-30^\circ=90^\circ$): $R=\sqrt{144+64}=14.4\ \text{m}$ at $30^\circ+\tan^{-1}\tfrac{8.0}{12}=30^\circ+33.7^\circ=63.7^\circ$ from $+x$. (Components: $R_x=10.39-4.00=6.39$, $R_y=6.00+6.93=12.93$; same answer.)
4 $2T\sin10^\circ=60$, $T=\dfrac{60}{2\times0.1736}=173\ \text{N}$ โ€” almost three times the weight. A nearly horizontal wire needs a huge tension; it can never be pulled perfectly straight.
5 Speed $\sqrt{9+16}=5.0\ \text{m/s}$; direction $\tan^{-1}\tfrac{3.0}{4.0}=36.9^\circ$ west of north (that is, $126.9^\circ$ anticlockwise from east).
6 $\mathbf{A}$ has $A=5.0$ at $53.1^\circ$ from $x$, i.e. $23.1^\circ$ from $x'$: $A_{x'}=5.0\cos23.1^\circ=4.60$, $A_{y'}=5.0\sin23.1^\circ=1.96$. $\sqrt{4.60^2+1.96^2}=5.0$ โœ“ โ€” the components changed, the vector did not.
๐ŸŽฏ Exam corner BoardJEE MainJEE AdvancedNEET

E1. Board ยท 3 marks Two vectors $\mathbf{A}$ and $\mathbf{B}$ are inclined at an angle $\theta$. Using components, derive expressions for the magnitude of their resultant and for the angle it makes with $\mathbf{A}$.

Set up Take $x$ along $\mathbf{A}$: $\mathbf{A}=A\hat{\mathbf{i}}$, $\mathbf{B}=B\cos\theta\,\hat{\mathbf{i}}+B\sin\theta\,\hat{\mathbf{j}}$. [1]
Magnitude $R_x=A+B\cos\theta$, $R_y=B\sin\theta$; $R^2=R_x^2+R_y^2=A^2+B^2+2AB\cos\theta$. [1]
Direction $\tan\alpha=\dfrac{R_y}{R_x}=\dfrac{B\sin\theta}{A+B\cos\theta}$. [1]
Exam tip Draw the parallelogram with the perpendicular dropped from the head of $\mathbf{R}$ โ€” many marking schemes expect the figure.

E2. NEET If $\mathbf{A}=2\hat{\mathbf{i}}+3\hat{\mathbf{j}}$ and $\mathbf{B}=\hat{\mathbf{i}}-4\hat{\mathbf{j}}$, the magnitude of $2\mathbf{A}-\mathbf{B}$ is (a) $\sqrt{29}$ (b) $\sqrt{109}$ (c) $\sqrt{50}$ (d) $11.3$

Components $2\mathbf{A}-\mathbf{B}=(4-1)\hat{\mathbf{i}}+(6+4)\hat{\mathbf{j}}=3\hat{\mathbf{i}}+10\hat{\mathbf{j}}$.
Answer (b) $\sqrt{9+100}=\sqrt{109}$.
Exam tip (a) is $|2\mathbf{A}+\mathbf{B}|$, (c) is $|\mathbf{A}-\mathbf{B}|$ and (d) is $2A+B$, adding magnitudes. Work in components, never with magnitudes.

E3. JEE Main The resultant of two forces $\mathbf{P}$ and $\mathbf{Q}$ is perpendicular to $\mathbf{P}$ and equal to it in magnitude. The angle between $\mathbf{P}$ and $\mathbf{Q}$, and the ratio $Q/P$, are (a) $135^\circ$, $\sqrt2$ (b) $120^\circ$, $2$ (c) $135^\circ$, $2$ (d) $150^\circ$, $\sqrt2$

Components With $x$ along $\mathbf{P}$: $R_x=P+Q\cos\theta=0$ and $R_y=Q\sin\theta=P$.
Solve $Q\cos\theta=-P$, $Q\sin\theta=P$: $\tan\theta=-1$ with $\sin\theta\gt0$, so $\theta=135^\circ$; $Q^2=P^2+P^2$, $Q=\sqrt2\,P$.
Answer (a).
Exam tip "Resultant perpendicular to $\mathbf{P}$" always means $P+Q\cos\theta=0$ โ€” remember it as a one-line condition.

E4. NEET A vector of magnitude $10$ units makes an angle of $150^\circ$ with the $+x$ axis (anticlockwise). Its components $(A_x,A_y)$ are (a) $(-8.66,\ 5.0)$ (b) $(8.66,\ -5.0)$ (c) $(-5.0,\ 8.66)$ (d) $(5.0,\ 8.66)$

Resolve $A_x=10\cos150^\circ=-8.66$, $A_y=10\sin150^\circ=+5.0$.
Answer (a) โ€” second quadrant: left and up.
Exam tip Sketch first: a $150^\circ$ arrow points mostly left and a little up, which rules out (b), (c) and (d) at a glance.

E5. JEE Advanced (Integer answer) $\mathbf{A}=3\hat{\mathbf{i}}+4\hat{\mathbf{j}}$. A vector $\mathbf{B}$ in the $xy$-plane has magnitude $5$, and $\mathbf{A}+\mathbf{B}$ points along the $+x$ axis. Find $|\mathbf{A}+\mathbf{B}|$.

y-component $\mathbf{A}+\mathbf{B}$ along $+x$ needs $4+B_y=0$, so $B_y=-4$.
x-component $B_x^2+16=25$ gives $B_x=\pm3$. With $B_x=-3$, $\mathbf{A}+\mathbf{B}=\mathbf{0}$, which has no direction, so it does not "point along $+x$". Hence $B_x=+3$.
Answer $\mathbf{A}+\mathbf{B}=6\hat{\mathbf{i}}$: $|\mathbf{A}+\mathbf{B}|=6$.
Exam tip One vector condition = two scalar equations. Discard roots that violate a stated condition โ€” integer-type questions are built around exactly this kind of trap.

E6. Board ยท Case study A ship is towed through a narrow harbour by two tugs. Each tow rope has a tension of $5.0\ \text{kN}$ and makes $25^\circ$ with the ship's direction of motion, one on each side. (a) Resolve each tension along and perpendicular to the ship's direction. [2] (b) Find the resultant pull on the ship. [1] (c) One tug changes its rope angle to $40^\circ$, keeping the tension $5.0\ \text{kN}$. Will the ship still be pulled straight ahead? Explain. [1]

(a) Along: $5.0\cos25^\circ=4.53\ \text{kN}$ each. Perpendicular: $5.0\sin25^\circ=2.11\ \text{kN}$, one to each side. [2]
(b) The sideways parts cancel: $R=2\times4.53=9.1\ \text{kN}$ straight ahead. [1]
(c) No: the sideways parts are now $2.11\ \text{kN}$ and $5.0\sin40^\circ=3.21\ \text{kN}$, leaving a net $1.10\ \text{kN}$ sideways towards the second tug. [1]
Exam tip In case studies, state which components cancel and why (symmetry) โ€” that sentence carries the mark.
๐Ÿ”ฌ Researchโ€“Teaching NexusResearch-tutored ยท optional, about 30 minutes

Research question. What is the "average wind" over a day โ€” and why do weather services insist on averaging it with components?

Why it matters. Wind is a velocity, so it is a vector. Meteorologists, airport controllers and wind-farm engineers store it as an eastward component $u$ and a northward component $v$, and average those. Averaging the directions as plain numbers gives nonsense: a wind from $350^\circ$ and a wind from $10^\circ$ (both nearly northerly) "average" to $180^\circ$, a southerly. Choosing turbine sites, runway directions and pollution forecasts all depend on getting this right.

What to do.

  1. From a weather website or app, record the wind speed and direction every hour for 8โ€“12 hours at one place. (Wind direction is the direction the wind blows from, measured clockwise from north.)
  2. Convert each reading to components of the velocity: $u=-V\sin\phi$ (towards east) and $v=-V\cos\phi$ (towards north), where $\phi$ is the reported direction.
  3. Average $u$ and $v$ separately, then rebuild the mean wind speed and direction (Result 3.4, watching the quadrant).
  4. Compare with the plain average of the directions and the plain average of the speeds. When are they very different?

Tools: A weather website or app, a spreadsheet (use ATAN2, which picks the quadrant for you).

Hand in: Your data table, the vector-mean wind (speed and direction), the plain averages, and a paragraph explaining why the vector-mean speed is never larger than the average speed.

Going further. The ratio (vector-mean speed) รท (average speed) is called the steadiness of the wind. Find its value for your data. Why would a wind-farm engineer care about it?

Instructor note: for two equal winds from $350^\circ$ and $10^\circ$, the vector mean blows from $0^\circ$ (north) with speed $V\cos10^\circ=0.985V$, while the naive direction average is $180^\circ$. Steadiness is $1$ for a perfectly steady wind and near $0$ for winds that swing round; it is a magnitude-of-resultant รท sum-of-magnitudes ratio, which is why it cannot exceed $1$ (the triangle inequality of Section 3.1).

๐Ÿงช Self-check quiz

0/5 correct

Q1. A force of $10\ \text{N}$ acts at $30^\circ$ above the $+x$ axis. Its components are

Q2. The magnitude of $2\hat{\mathbf{i}}-3\hat{\mathbf{j}}+6\hat{\mathbf{k}}$ is

Q3. In which direction (anticlockwise from $+x$) does $-4\hat{\mathbf{i}}-4\hat{\mathbf{j}}$ point?

Q4. Two vectors of equal magnitude $A$ make an angle of $60^\circ$. Their resultant is

Q5. Which statement about the unit vector $\hat{\mathbf{A}}=\mathbf{A}/A$ is correct?

โ† 3.1 Scalars, Vectors and Graphical Vector Addition3.3 Motion in a Plane with Constant Acceleration and Projectile Motion โ†’