Resolution of Vectors, Unit Vectors and the Analytical Method

Why Resolve at All? Because a Ruler Is Not a Formula

Section 2 handed you three constructions — the triangle law, the parallelogram law, the polygon law — and every one of them is a drawing. You pick a scale, you sharpen a pencil, you lay one arrow's tail on another's head, and then you measure the answer off the page.

That works. It is also the problem.

Key Point: The graphical method is exact in principle and approximate in practice. The accuracy of your resultant is the accuracy of your ruler and your protractor — nothing more. Put plainly: the graphical method is sometimes tedious and has limited accuracy.

Think about what an exam actually asks. Two forces of 17.3 N and 25.0 N act at $47°$ to each other; find the resultant to three significant figures. You cannot draw that. Half a degree of protractor error moves the answer by more than the tolerance you are being marked against. And in three dimensions you cannot draw it at all.

So we need a way to turn the geometry into arithmetic — into numbers you add, square and square-root, with no drawing anywhere in sight. That method is called resolution, and the whole of this section is built on one simple reversal of direction:

• Addition takes two vectors and produces one.
• Resolution takes one vector and produces two.

They are inverse operations, and the second one is the useful one, because the two vectors you break a vector into can be chosen to be as convenient as you like.

Resolving along any two directions

Here is the general statement.

Let $\vec{a}$ and $\vec{b}$ be any two non-zero vectors in a plane that are not parallel to each other, and let $\vec{A}$ be another vector in that same plane. Then $\vec{A}$ can always be written as a sum of a multiple of $\vec{a}$ and a multiple of $\vec{b}$.

The construction is a two-line piece of geometry:

1. Let O be the tail and P the head of $\vec{A}$.
2. Through O draw a straight line parallel to $\vec{a}$; through P draw a straight line parallel to $\vec{b}$. Because $\vec{a}$ and $\vec{b}$ are not parallel, these two lines must meet — call the meeting point Q.
3. By the triangle law, $\vec{A} = \vec{OP} = \vec{OQ} + \vec{QP}$.
4. But $\vec{OQ}$ lies along $\vec{a}$ and $\vec{QP}$ lies along $\vec{b}$, so each is just a real multiple of its direction: $\vec{OQ} = \lambda\vec{a}$ and $\vec{QP} = \mu\vec{b}$.

Key Point: Any vector $\vec{A}$ in a plane can be resolved along any two non-zero, non-parallel vectors $\vec{a}$ and $\vec{b}$ in that plane: $$\vec{A} = \lambda\vec{a} + \mu\vec{b}$$ where $\lambda$ and $\mu$ are real numbers. The vectors $\lambda\vec{a}$ and $\mu\vec{b}$ are called the component vectors of $\vec{A}$ along $\vec{a}$ and $\vec{b}$.

Notice that this uses nothing you have not already met: Section 2's multiplication by a real number supplies $\lambda\vec{a}$, and Section 2's triangle law glues the two pieces back together.

Two things students get wrong about this

First: the resolution is unique — once you have chosen $\vec{a}$ and $\vec{b}$. There is exactly one line through O parallel to $\vec{a}$ and exactly one line through P parallel to $\vec{b}$, so Q is pinned down, and with it $\lambda$ and $\mu$. You cannot get two different answers for the same pair of base directions.

Second: change the base pair and the numbers change. Panels (b) and (c) of the figure show the same vector $\vec{A}$ resolved twice, along two different pairs of directions, and the two answers look nothing like each other. Both are correct.

Key Point: The components belong to the axes, not to the vector. $\vec{A}$ itself never changes; only the numbers you describe it with do. This is why a question must always tell you the directions you are resolving along, and why your answer must always say which ones you used.

[JEE Tip] The non-parallel condition is not a technicality. If $\vec{a}$ and $\vec{b}$ were parallel, every combination $\lambda\vec{a} + \mu\vec{b}$ would lie along that one line, and you could never reach a vector pointing anywhere else. Two parallel directions carry only one direction's worth of information.

Why we almost always choose perpendicular directions

Nothing in the construction demands a right angle. But if you pick $\vec{a}$ and $\vec{b}$ perpendicular to each other, three lovely things happen at once:

1. Each component can be read off with a single sine or cosine.
2. The magnitude comes back by Pythagoras, with no cross term.
3. The two components are completely independent — pushing along $x$ has no effect on $y$ whatsoever.

That special case is called rectangular or orthogonal resolution, and it is what the rest of this section — and honestly, the rest of this chapter — actually uses.

Rectangular Components and the Unit Vectors $\hat{i}$, $\hat{j}$, $\hat{k}$

To resolve along perpendicular directions we need a name for "one unit that way". Section 1 introduced the general unit vector $\hat{A} = \vec{A}/|\vec{A}|$. Now we fix three of them permanently, one per coordinate axis.

Key Point: $\hat{i}$, $\hat{j}$ and $\hat{k}$ are the unit vectors along the $x$, $y$ and $z$ axes of a rectangular coordinate system. They satisfy $$|\hat{i}| = |\hat{j}| = |\hat{k}| = 1$$ They are mutually perpendicular, and they are dimensionless and carry no unit — their only job is to point.

That last property is worth a sentence of its own, because it is a one-mark question in disguise. A unit vector is a vector divided by its own magnitude, so the units cancel. If $\vec{v} = 3\hat{i} + 4\hat{j}$ m/s, the "m/s" belongs to the numbers 3 and 4, not to $\hat{i}$ and $\hat{j}$. Writing "$\hat{i}$ metres" is meaningless.

Multiplying a unit vector by a scalar $\lambda$ gives a vector of magnitude $|\lambda|$ along that direction, so any vector splits into "how much" times "which way": $$\vec{A} = |\vec{A}|\,\hat{n}$$ where $\hat{n}$ is the unit vector along $\vec{A}$.

Resolving $\vec{A}$ into $A_x$ and $A_y$

Take a vector $\vec{A}$ lying in the $x$-$y$ plane, with its tail at the origin, making an angle $\theta$ with the positive $x$-axis. Drop a perpendicular from its head onto each axis. You get two vectors $\vec{A_1}$ and $\vec{A_2}$ along the axes with $$\vec{A_1} + \vec{A_2} = \vec{A}$$ and since $\vec{A_1}$ is parallel to $\hat{i}$ and $\vec{A_2}$ is parallel to $\hat{j}$, we can write $\vec{A_1} = A_x\hat{i}$ and $\vec{A_2} = A_y\hat{j}$.

Key Point: $$\vec{A} = A_x\hat{i} + A_y\hat{j}, \qquad A_x = A\cos\theta, \qquad A_y = A\sin\theta$$ $A_x$ and $A_y$ are the $x$- and $y$-components of $\vec{A}$. They are ordinary real numbers — scalars, which may be positive, negative or zero.

[Board Important] The wording here matters and examiners follow it closely: $A_x$ is not a vector, but $A_x\hat{i}$ is. The number is called the component; the arrow $A_x\hat{i}$ is called the component vector. Say $A_x = -6$ and you have said something perfectly sensible — a component is allowed to be negative. Say "the magnitude is $-6$" and you have said something impossible.

Going backwards: from components to magnitude and direction

The forward direction is trigonometry. The reverse direction is Pythagoras.

$$A_x^2 + A_y^2 = A^2\cos^2\theta + A^2\sin^2\theta = A^2(\cos^2\theta + \sin^2\theta) = A^2$$

Key Point: $$A = \sqrt{A_x^2 + A_y^2}, \qquad \tan\theta = \frac{A_y}{A_x}, \qquad \theta = \tan^{-1}\!\left(\frac{A_y}{A_x}\right)$$

So a vector in a plane can be specified in exactly two equivalent ways, and you must be fluent in translating between them:

Specification	Data	Convert with
Polar form	magnitude $A$ and angle $\theta$	$A_x = A\cos\theta$, $A_y = A\sin\theta$
Component form	the pair $A_x$, $A_y$	$A = \sqrt{A_x^2 + A_y^2}$, $\tan\theta = A_y/A_x$

The angles you should be able to write down instantly

$\theta$	$\cos\theta$	$\sin\theta$	$A_x$	$A_y$
$0°$	$1$	$0$	$A$	$0$
$30°$	$0.866$	$0.5$	$0.866A$	$0.5A$
$37°$	$0.8$	$0.6$	$0.8A$	$0.6A$
$45°$	$0.707$	$0.707$	$0.707A$	$0.707A$
$53°$	$0.6$	$0.8$	$0.6A$	$0.8A$
$60°$	$0.5$	$0.866$	$0.5A$	$0.866A$
$90°$	$0$	$1$	$0$	$A$

[NEET Important] The $37°$ and $53°$ rows are not exact — the exact angles are $36.87°$ and $53.13°$ — but they come from the 3-4-5 triangle and problem setters use them constantly precisely because they give clean answers. Recognising "3-4-5" on sight saves fifteen seconds every time.

The one trap in $A\cos\theta$

The formula $A_x = A\cos\theta$ is not a law of nature that says "$x$ gets the cosine". It says something better:

Key Point: The cosine goes with the axis you measured the angle FROM; the sine goes with the other one.

Measure the angle from the $x$-axis and $A_x = A\cos\theta$. Measure the very same vector's angle from the $y$-axis — call it $\phi$, with $\phi = 90° - \theta$ — and now $A_y = A\cos\phi$ while $A_x = A\sin\phi$. Both descriptions are right, and they agree, because $\cos(90° - \theta) = \sin\theta$.

This matters far more than it looks. A projectile launched at "$30°$ above the horizontal" and a projectile launched at "$30°$ to the vertical" are two completely different problems, and thousands of marks a year are lost by students who reach for cosine on autopilot. Draw the angle before you write the formula.

The Quadrant Trap: Why $\tan^{-1}$ Alone Is Not an Answer

Here is a question that has cost more marks than almost anything else in this chapter.

A vector has components $A_x = -3$ and $A_y = 4$. Find its direction.

You do the obvious thing: $\tan\theta = A_y/A_x = 4/(-3) = -1.333$, so $\theta = \tan^{-1}(-1.333) = -53.13°$. And that is wrong. A vector with a negative $x$-component and a positive $y$-component points up and to the left — it is in the second quadrant, somewhere between $90°$ and $180°$. It cannot possibly be at $-53.13°$, which points down and to the right.

The calculator is not broken. It is doing exactly what it is designed to do, and the design has a limitation you have to know about.

Key Point: The $\tan^{-1}$ key on a calculator always returns an angle in the range $-90°$ to $+90°$ — the first and fourth quadrants only. It sees only the ratio $A_y/A_x$, and one ratio corresponds to two opposite directions, $180°$ apart. It cannot tell them apart, so you must.

The sign table — learn this, it is free marks

Quadrant	$A_x$	$A_y$	True $\theta$ lies in	Calculator gives	Your fix
First	$+$	$+$	$0°$ to $90°$	the correct angle	none
Second	$-$	$+$	$90°$ to $180°$	a negative angle	add $180°$
Third	$-$	$-$	$180°$ to $270°$	a positive angle	add $180°$
Fourth	$+$	$-$	$270°$ to $360°$	the correct negative angle	none (or add $360°$)

The pattern is easy to remember once you see where it comes from: whenever $A_x$ is negative, add $180°$. That is the whole rule. A negative $x$-component means the vector points into the left half-plane, and the calculator has quietly given you the arrow pointing the opposite way.

The same drill on all four quadrants

Take the four vectors in the figure. Every one has magnitude $\sqrt{3^2 + 4^2} = 5$ and every one has $|\tan\theta| = 4/3$, so a calculator gives $\pm 53.13°$ for all four. Only the signs separate them:

Components	Quadrant	Calculator	True direction
$(3, 4)$	First	$+53.13°$	$53.13°$
$(-3, 4)$	Second	$-53.13°$	$53.13° + 180° = 126.87°$
$(-3, -4)$	Third	$+53.13°$	$53.13° + 180° = 233.13°$
$(3, -4)$	Fourth	$-53.13°$	$-53.13°$, i.e. $306.87°$

[JEE/NEET] In multiple-choice papers this is a favourite distractor: the "wrong" option is exactly the calculator value, sitting there waiting for anyone who did not check the signs. Before you press $\tan^{-1}$, write down the two signs and say out loud which quadrant you are in.

The habit that never fails

There is a cleaner way to think about it that works every single time, without memorising a table:

1. Compute the acute reference angle from magnitudes only: $\theta_{\text{ref}} = \tan^{-1}\left|\dfrac{A_y}{A_x}\right|$.
2. Sketch the vector using the signs — a five-second sketch, no scale needed.
3. Read the true angle off the sketch: first quadrant $\theta_{\text{ref}}$; second $180° - \theta_{\text{ref}}$; third $180° + \theta_{\text{ref}}$; fourth $360° - \theta_{\text{ref}}$.

[Board Important] In board answers you are usually allowed — often preferred — to describe the direction in words rather than as an angle from the $x$-axis: "$5$ units directed $53°$ north of west" is completely unambiguous and cannot be marked wrong for a quadrant error. Use that phrasing whenever the problem is set in compass directions.

One last reassurance: components carry signs, magnitudes never do. $A_x = -3$ is fine. $A = -5$ is not, ever.

Stepping Out of the Plane: Three Dimensions and Direction Cosines

Everything so far has lived in a plane, because that is what this chapter is about. But the machinery does not care how many axes you have, and JEE does ask, so here is the extension in one short block.

Resolve $\vec{A}$ along all three axes and you get three components instead of two:

Key Point: $$\vec{A} = A_x\hat{i} + A_y\hat{j} + A_z\hat{k}, \qquad A = \sqrt{A_x^2 + A_y^2 + A_z^2}$$

The magnitude formula is just Pythagoras applied twice — once in the $x$-$y$ plane to get $\sqrt{A_x^2 + A_y^2}$, and once more with $A_z$ perpendicular to that.

A position vector in space is written the same way: $$\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$$

Direction cosines

In a plane, one angle fixes a direction. In space you need more, and the tidiest way to give it is with the angles $\alpha$, $\beta$ and $\gamma$ that $\vec{A}$ makes with the $x$, $y$ and $z$ axes.

Key Point: $$A_x = A\cos\alpha, \qquad A_y = A\cos\beta, \qquad A_z = A\cos\gamma$$ The three quantities $\cos\alpha$, $\cos\beta$, $\cos\gamma$ are called the direction cosines of $\vec{A}$.

Notice that all three use a cosine. That is the same rule as before — each angle is measured from the axis whose component you are computing, so each axis gets the cosine.

Substituting into $A^2 = A_x^2 + A_y^2 + A_z^2$ and dividing through by $A^2$ gives an identity that examiners love:

Key Point: $$\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$$ Equivalently, in terms of the angles themselves, $\sin^2\alpha + \sin^2\beta + \sin^2\gamma = 2$.

[JEE Tip] This identity is a ready-made filter. If a question offers you three direction cosines and asks whether they are possible, square them and add: unless you get exactly 1, no such direction exists. And if two are given, the third follows immediately up to a sign.

The direction cosines are also just the components of the unit vector along $\vec{A}$, since $$\hat{A} = \frac{\vec{A}}{A} = \cos\alpha\,\hat{i} + \cos\beta\,\hat{j} + \cos\gamma\,\hat{k}$$ and a unit vector must have magnitude 1 — which is the identity above, wearing different clothes.

One subtlety most students skip: $\alpha$, $\beta$ and $\gamma$ are angles in space, between pairs of lines that are generally not coplanar. They are not three angles you can draw in one flat picture and add up.

The Analytical Method: Add Components, Never Magnitudes

Now we cash in. This is the payoff for all the resolving.

Take two vectors in the $x$-$y$ plane, already written in component form: $$\vec{A} = A_x\hat{i} + A_y\hat{j}, \qquad \vec{B} = B_x\hat{i} + B_y\hat{j}$$

Their sum is $$\vec{R} = \vec{A} + \vec{B} = (A_x\hat{i} + A_y\hat{j}) + (B_x\hat{i} + B_y\hat{j})$$

Vector addition is commutative and associative (Section 2 established both), so we are free to shuffle and regroup the four pieces, collecting the $\hat{i}$ terms together and the $\hat{j}$ terms together: $$\vec{R} = (A_x + B_x)\hat{i} + (A_y + B_y)\hat{j}$$

Comparing with $\vec{R} = R_x\hat{i} + R_y\hat{j}$ gives the result the whole section has been driving at.

Key Point — the analytical method: $$R_x = A_x + B_x, \qquad R_y = A_y + B_y$$ Each component of the resultant is the sum of the corresponding components of the vectors being added. No drawing, no protractor — just ordinary arithmetic, done twice.

Why this is legal, in one sentence

Because $\hat{i}$ and $\hat{j}$ are perpendicular, motion along $x$ can never contribute anything along $y$. The two axes are watertight compartments. That is exactly why we chose perpendicular directions back in the first block, and it is the same idea that will later let us treat a projectile as two independent one-dimensional motions.

The rule to shout at yourself

Key Point: Add components, never magnitudes. $|\vec{A} + \vec{B}|$ is almost never $|\vec{A}| + |\vec{B}|$ — that happens only when the two vectors are parallel.

In the figure, $\vec{A} = 3\hat{i} + \hat{j}$ and $\vec{B} = \hat{i} + 3\hat{j}$ each have magnitude $\sqrt{10} = 3.16$. Add the magnitudes and you would predict 6.32. The true resultant is $4\hat{i} + 4\hat{j}$, of magnitude $\sqrt{32} = 5.66$. Section 2's bound was telling you this all along: $5.66 \leq 6.32$, and the gap is the price of the vectors not being parallel.

More than two vectors, and subtraction

Nothing changes. Add a third vector and you get a third term in each bracket; the components just keep piling up.

Key Point — for any number of vectors: $$R_x = A_x + B_x + C_x + \dots, \qquad R_y = A_y + B_y + C_y + \dots$$ and in three dimensions the same again with $R_z = A_z + B_z + C_z + \dots$
Subtraction is just addition with a minus sign in front of the components: for $\vec{T} = \vec{a} + \vec{b} - \vec{c}$, $$T_x = a_x + b_x - c_x, \qquad T_y = a_y + b_y - c_y, \qquad T_z = a_z + b_z - c_z$$

This is the moment the polygon law becomes obsolete for calculation. Five vectors, ten vectors, any number — the work grows by one line of addition per vector, not by one more careful drawing.

The four-step recipe

Use this every single time, in this order:

1. Choose axes and draw a rough sketch. Take east as $+x$ and north as $+y$ unless the problem suggests something better. The sketch is for signs, not for measurement.
2. Resolve every vector into $A\cos\theta$ and $A\sin\theta$, being careful about which axis the angle was measured from, and attach the correct signs from the sketch.
3. Add all the $x$-components; add all the $y$-components. Keep the two columns physically separate on the page — mixing them is the single most common slip.
4. Rebuild: $R = \sqrt{R_x^2 + R_y^2}$ and $\theta = \tan^{-1}(R_y/R_x)$, then fix the quadrant using the signs of $R_x$ and $R_y$.

[JEE Tip] Step 3 is where a table beats prose. Rule two columns headed $x$ and $y$, write one row per vector, and sum each column. It takes ten seconds longer and eliminates the whole family of "lost a sign" errors.

A quick sanity check before you move on: whatever you get for $R$ must satisfy Section 2's bounds. If you have added a 5 and a 3 and produced 9.4, something is wrong — no arrangement of those two can beat 8.

The Resultant of Two Vectors at an Angle $\theta$

Section 2 could tell you that the resultant of two vectors lies between $|A - B|$ and $A + B$, and it could tell you the answer exactly when the vectors were parallel, antiparallel, perpendicular or equal in magnitude. What it could not do was give you the general answer. Components can. This derivation is a standard one, and it is examinable in its own right.

Setting it up

Let $\vec{A}$ and $\vec{B}$ have magnitudes $A$ and $B$ with an angle $\theta$ between them. We are free to choose our axes, so choose them to make life easy: put the $x$-axis along $\vec{A}$. Then

$$\vec{A} = A\hat{i} + 0\hat{j}, \qquad \vec{B} = (B\cos\theta)\hat{i} + (B\sin\theta)\hat{j}$$

Now just add the components: $$R_x = A + B\cos\theta, \qquad R_y = B\sin\theta$$

The magnitude

$$R^2 = R_x^2 + R_y^2 = (A + B\cos\theta)^2 + (B\sin\theta)^2$$ $$R^2 = A^2 + 2AB\cos\theta + B^2\cos^2\theta + B^2\sin^2\theta$$

The last two terms collapse, since $B^2(\cos^2\theta + \sin^2\theta) = B^2$:

Key Point — the law of cosines for vectors: $$R = \sqrt{A^2 + B^2 + 2AB\cos\theta}$$ where $\theta$ is the angle between $\vec{A}$ and $\vec{B}$ when they are drawn from a common tail.

The direction

The resultant makes an angle $\alpha$ with $\vec{A}$, and since we put the $x$-axis along $\vec{A}$, that angle is just the direction of $\vec{R}$ in our chosen frame:

Key Point: $$\tan\alpha = \frac{R_y}{R_x} = \frac{B\sin\theta}{A + B\cos\theta}$$ $\alpha$ is measured from $\vec{A}$, towards $\vec{B}$. The angle from $\vec{B}$ is then $\theta - \alpha$.

The same two results can be reached by pure geometry, dropping a perpendicular SN from the tip of the resultant onto the line of $\vec{A}$ — and it is worth seeing that the two routes give identical formulas, because $ON = A + B\cos\theta$ and $SN = B\sin\theta$ are precisely our $R_x$ and $R_y$. The component method just gets there without any construction.

The same figure also gives the sine rule for the triangle, which is occasionally quicker: $$\frac{R}{\sin(180° - \theta)} = \frac{A}{\sin\beta} = \frac{B}{\sin\alpha}$$

Does it agree with Section 2? Check the corners

A formula you cannot check is a formula you will misremember. Feed the special cases in:

$\theta$	$\cos\theta$	$R$ becomes	Section 2 said
$0°$	$1$	$\sqrt{A^2+B^2+2AB} = \sqrt{(A+B)^2} = A+B$	the maximum
$60°$	$0.5$	$\sqrt{A^2+B^2+AB}$	between the bounds
$90°$	$0$	$\sqrt{A^2+B^2}$	the Pythagoras case
$120°$	$-0.5$	$\sqrt{A^2+B^2-AB}$	between the bounds
$180°$	$-1$	$\sqrt{(A-B)^2} = \lvert A-B \rvert$	the minimum

Every row matches. The directions check out too: at $\theta = 0°$, $\tan\alpha = 0$, so the resultant lies along $\vec{A}$ — of course it does, the vectors are parallel. At $\theta = 90°$, $\tan\alpha = B/A$, the familiar right-triangle answer.

And the equal-magnitude shortcut from Section 2 falls straight out. Put $B = A$: $$R = \sqrt{2A^2(1 + \cos\theta)} = \sqrt{2A^2 \cdot 2\cos^2(\theta/2)} = 2A\cos\frac{\theta}{2}$$ using $1 + \cos\theta = 2\cos^2(\theta/2)$. One formula, and Section 2's whole table is contained in it.

Subtraction

$\vec{A} - \vec{B}$ is $\vec{A} + (-\vec{B})$, and reversing $\vec{B}$ turns the angle between them from $\theta$ into $180° - \theta$. Since $\cos(180° - \theta) = -\cos\theta$, the middle sign simply flips:

Key Point: $$|\vec{A} - \vec{B}| = \sqrt{A^2 + B^2 - 2AB\cos\theta}$$ Same $\theta$ — still the angle between the original $\vec{A}$ and $\vec{B}$ — just a minus sign.

Two formulas, one character apart. Read the question twice before you choose. A useful memory hook: the plus sign goes with the sum, and a plus sign makes things bigger; the minus goes with the difference.

[JEE/NEET] Combining the two gives a favourite identity: $$|\vec{A}+\vec{B}|^2 + |\vec{A}-\vec{B}|^2 = 2(A^2 + B^2)$$ The cross terms cancel. This is the parallelogram law of geometry — the sum of the squares of the diagonals equals the sum of the squares of the four sides — and it turns several nasty-looking problems into one line.

Mistake checklist

Mistake	Fix
Using the angle one vector makes with the horizontal as $\theta$	$\theta$ is the angle between the two vectors, tails together
Writing $R = A + B$	Only when $\theta = 0°$
Quoting $\tan^{-1}$ straight off the calculator	Check the signs of $R_x$, $R_y$ and fix the quadrant
$A_x = A\cos\theta$ when the angle was given from the $y$-axis	Cosine belongs to the axis the angle was measured from
Calling $A_x$ a vector	$A_x$ is a scalar; $A_x\hat{i}$ is the component vector
Writing a negative magnitude	Components may be negative; magnitudes never
Attaching a unit to $\hat{i}$	Unit vectors are dimensionless
Forgetting $\alpha$ is measured from $\vec{A}$	Swap $A$ and $B$ in the formula if you want the angle from $\vec{B}$
Adding a 5 and a 3 to get 9	Check against the bound $2 \leq R \leq 8$

Where this goes next

You now have a complete algebra of vectors: scale them, add them, subtract them, resolve them, and get exact numbers out at the end. Section 4 stops treating vectors as static arrows and lets them change with time — position, velocity and acceleration become vectors, and the fact that $x$ and $y$ components add independently becomes the fact that $x$ and $y$ motions run independently. That single idea is the whole of projectile motion.

Solved Examples

Example 1: Resolving a force

A force of 50 N acts at $30°$ above the horizontal. (a) Find its horizontal and vertical components. (b) Write the force in $\hat{i}$, $\hat{j}$ notation. (c) Verify your components by rebuilding the magnitude and direction.

Solution:

1. (a) Take $x$ horizontal and $y$ vertical, with $\theta = 30°$ measured from the $x$-axis: $$F_x = F\cos\theta = 50\cos 30° = 50 \times 0.8660 = 43.30\ \text{N}$$ $$F_y = F\sin\theta = 50\sin 30° = 50 \times 0.5 = 25.00\ \text{N}$$
2. (b) $$\vec{F} = 43.30\hat{i} + 25.00\hat{j}\ \text{N}$$
3. (c) Rebuild the magnitude: $$F = \sqrt{43.30^2 + 25.00^2} = \sqrt{1875 + 625} = \sqrt{2500} = 50.0\ \text{N} \ \checkmark$$ Rebuild the direction: $\tan\theta = 25.00/43.30 = 0.5774$, so $\theta = \tan^{-1}(0.5774) = 30.0°$. Both components are positive, so we are in the first quadrant and no correction is needed. $\checkmark$

Final Answer: (a) 43.3 N horizontal, 25.0 N vertical; (b) $\vec{F} = 43.3\hat{i} + 25.0\hat{j}$ N.

Takeaway: Always do part (c). Squaring and adding the components must give back the original magnitude — it is a ten-second check that catches a swapped sine and cosine every time.

Example 2: From components back to magnitude and direction

A displacement vector is $\vec{d} = 5\hat{i} - 12\hat{j}$ metres. Find (a) its magnitude, (b) its direction, (c) the unit vector along it, and (d) the vector of magnitude 26 m pointing the same way.

Solution:

1. (a) $$d = \sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13\ \text{m}$$ Note the square kills the minus sign — the magnitude is $+13$, never $-13$.
2. (b) $\tan\theta = d_y/d_x = -12/5 = -2.4$, and the calculator returns $$\tan^{-1}(-2.4) = -67.38°$$ Check the signs: $d_x = +5$ and $d_y = -12$, so the vector points right and down — the fourth quadrant, where angles run from $270°$ to $360°$. The calculator's negative answer is already a fourth-quadrant direction, so it is correct as it stands; expressed as a positive angle it is $-67.38° + 360° = 292.62°$.
3. (c) $$\hat{d} = \frac{\vec{d}}{d} = \frac{5\hat{i} - 12\hat{j}}{13} = 0.3846\hat{i} - 0.9231\hat{j}$$ Check: $\sqrt{0.3846^2 + 0.9231^2} = \sqrt{0.1479 + 0.8521} = 1.000 \ \checkmark$, and it carries no unit — the metres cancelled.
4. (d) Multiply the unit vector by the required magnitude: $$\vec{D} = 26\,\hat{d} = 26(0.3846\hat{i} - 0.9231\hat{j}) = 10\hat{i} - 24\hat{j}\ \text{m}$$ Check: $\sqrt{10^2 + 24^2} = \sqrt{676} = 26$ m $\checkmark$

Final Answer: (a) 13 m; (b) $67.38°$ below the $+x$ axis, i.e. $292.62°$; (c) $0.3846\hat{i} - 0.9231\hat{j}$; (d) $10\hat{i} - 24\hat{j}$ m.

Takeaway: $(5, 12, 13)$ is a Pythagorean triple worth knowing alongside $(3, 4, 5)$. And note how part (d) works: magnitude times unit vector rebuilds any vector you like along a known direction.

Example 3: The four-quadrant drill

Four vectors have components $(3, 4)$, $(-3, 4)$, $(-3, -4)$ and $(3, -4)$. For each, find the magnitude and the true direction measured anticlockwise from the $+x$ axis.

Solution:

1. Magnitudes. Every one of them squares to the same thing: $$A = \sqrt{(\pm 3)^2 + (\pm 4)^2} = \sqrt{9 + 16} = 5\ \text{units}$$ All four have magnitude 5. The signs vanish under the square.
2. Reference angle. Using magnitudes only, $$\theta_{\text{ref}} = \tan^{-1}\left(\frac{4}{3}\right) = 53.13°$$ This same reference angle serves all four.
3. Now place each one by its signs:

Components	Quadrant	True direction
$(3, 4)$	First ($+$, $+$)	$53.13°$
$(-3, 4)$	Second ($-$, $+$)	$180° - 53.13° = 126.87°$
$(-3, -4)$	Third ($-$, $-$)	$180° + 53.13° = 233.13°$
$(3, -4)$	Fourth ($+$, $-$)	$360° - 53.13° = 306.87°$

4. A calculator fed $A_y/A_x$ directly would have returned $+53.13°$ for both the first and the third rows, and $-53.13°$ for both the second and the fourth. Two pairs, indistinguishable by ratio alone.

Final Answer: All four have magnitude 5 units; directions $53.13°$, $126.87°$, $233.13°$ and $306.87°$.

Takeaway: One value of $\tan\theta$ always corresponds to two directions $180°$ apart. The signs of the components — not the ratio — are what pick the right one.

Example 4: Resolving along two non-perpendicular directions

Express the vector $\vec{A} = 6\hat{i} + 2\hat{j}$ in the form $\lambda\vec{a} + \mu\vec{b}$, where $\vec{a} = \hat{i} + \hat{j}$ and $\vec{b} = \hat{i} - \hat{j}$. (a) Find $\lambda$ and $\mu$. (b) Verify your answer. (c) What are the components of the same $\vec{A}$ along $\hat{i}$ and $\hat{j}$, and what does the comparison tell you?

Solution:

1. (a) Write out the requirement and compare coefficients: $$\lambda(\hat{i} + \hat{j}) + \mu(\hat{i} - \hat{j}) = 6\hat{i} + 2\hat{j}$$ $$(\lambda + \mu)\hat{i} + (\lambda - \mu)\hat{j} = 6\hat{i} + 2\hat{j}$$ Because $\hat{i}$ and $\hat{j}$ are independent directions, the coefficients must match separately: $$\lambda + \mu = 6, \qquad \lambda - \mu = 2$$ Adding: $2\lambda = 8$, so $\lambda = 4$. Subtracting: $2\mu = 4$, so $\mu = 2$.
2. (b) Check by substitution: $$4(\hat{i} + \hat{j}) + 2(\hat{i} - \hat{j}) = 4\hat{i} + 4\hat{j} + 2\hat{i} - 2\hat{j} = 6\hat{i} + 2\hat{j} \ \checkmark$$
3. (c) Along $\hat{i}$ and $\hat{j}$ the components are simply $A_x = 6$ and $A_y = 2$. So the same vector is described by the pair $(4, 2)$ on one base pair and by $(6, 2)$ on another. Neither is more correct than the other. The vector's own magnitude is unaffected: $$A = \sqrt{6^2 + 2^2} = \sqrt{40} = 6.32\ \text{units at } \tan^{-1}(2/6) = 18.43°$$
4. Note also that $\vec{a}$ and $\vec{b}$ here happen to be perpendicular to each other (at $45°$ and $-45°$), but they are not unit vectors — each has magnitude $\sqrt{2}$ — which is exactly why the numbers 4 and 2 are not the same as 6 and 2.

Final Answer: (a) $\lambda = 4$, $\mu = 2$; (c) $A_x = 6$, $A_y = 2$ — same vector, different numbers.

Takeaway: "Comparing coefficients" works because a resolution along a fixed pair of non-parallel directions is unique. Change the pair and the numbers change; the arrow does not.

Example 5: Three forces, one resultant

Three forces act at a point: 10 N along the $+x$ direction, 20 N at $60°$ to the $+x$ direction, and 15 N along the $-x$ direction. Find the magnitude and direction of the resultant.

Solution:

1. Resolve each force. Set out a two-column table — this is the habit that prevents lost signs.

Force	$x$-component (N)	$y$-component (N)
10 N at $0°$	$10\cos 0° = 10.00$	$10\sin 0° = 0$
20 N at $60°$	$20\cos 60° = 10.00$	$20\sin 60° = 17.32$
15 N at $180°$	$15\cos 180° = -15.00$	$15\sin 180° = 0$
Sum	$R_x = 5.00$	$R_y = 17.32$

2. Magnitude: $$R = \sqrt{R_x^2 + R_y^2} = \sqrt{5.00^2 + 17.32^2} = \sqrt{25 + 300} = \sqrt{325} = 18.03\ \text{N}$$
3. Direction: $\tan\theta = 17.32/5.00 = 3.464$, so $\theta = \tan^{-1}(3.464) = 73.90°$. Both $R_x$ and $R_y$ are positive, so we are in the first quadrant and the calculator value stands.
4. Sanity check: the three magnitudes are 10, 20 and 15, so the largest conceivable resultant is 45 N and 18.03 N sits comfortably below it.

Final Answer: $R = 18.0$ N at $73.9°$ from the $+x$ direction.

Takeaway: Notice that the 15 N force did not need any special treatment — pointing it along $-x$ was handled automatically by $\cos 180° = -1$. Let the trigonometry supply the signs; do not add them by hand as well, or you will apply them twice.

Example 6: The general resultant of two vectors at an angle

Find the magnitude and direction of the resultant of two vectors $\vec{A}$ and $\vec{B}$ in terms of their magnitudes and the angle $\theta$ between them. Then apply your result to $A = 5$ units, $B = 3$ units, $\theta = 60°$.

Solution:

1. Choose convenient axes. The answer cannot depend on our choice, so take the $x$-axis along $\vec{A}$. Then $$\vec{A} = A\hat{i}, \qquad \vec{B} = (B\cos\theta)\hat{i} + (B\sin\theta)\hat{j}$$
2. Add components: $$R_x = A + B\cos\theta, \qquad R_y = B\sin\theta$$
3. Magnitude: $$R^2 = (A + B\cos\theta)^2 + (B\sin\theta)^2 = A^2 + 2AB\cos\theta + B^2(\cos^2\theta + \sin^2\theta)$$ $$R = \sqrt{A^2 + B^2 + 2AB\cos\theta}$$
4. Direction. With the $x$-axis along $\vec{A}$, the angle $\alpha$ of the resultant from $\vec{A}$ is $$\tan\alpha = \frac{R_y}{R_x} = \frac{B\sin\theta}{A + B\cos\theta}$$
5. Apply the numbers. With $A = 5$, $B = 3$, $\theta = 60°$ ($\cos 60° = 0.5$, $\sin 60° = 0.8660$): $$R = \sqrt{25 + 9 + 2(5)(3)(0.5)} = \sqrt{25 + 9 + 15} = \sqrt{49} = 7.00\ \text{units}$$ $$\tan\alpha = \frac{3 \times 0.8660}{5 + 3 \times 0.5} = \frac{2.598}{6.5} = 0.3997 \quad \Rightarrow \quad \alpha = 21.79°$$
6. Check against the bounds. $|5 - 3| = 2$ and $5 + 3 = 8$, and indeed $2 \leq 7 \leq 8$. Check the direction too: $\alpha = 21.79°$ is less than $\theta = 60°$, so the resultant lies between the two vectors and closer to the longer one, exactly as it must.

Final Answer: $R = \sqrt{A^2+B^2+2AB\cos\theta}$ at $\tan\alpha = B\sin\theta/(A + B\cos\theta)$ from $\vec{A}$; for the numbers given, $R = 7.00$ units at $21.79°$ from $\vec{A}$.

Takeaway: The trick that makes this derivation short is choosing the $x$-axis along $\vec{A}$. You are always allowed to do that, and it kills half the algebra before it starts.

Example 7: The motorboat and the current

A motorboat is racing towards north at 25 km/h and the water current in that region is 10 km/h in the direction $60°$ east of south. Find the resultant velocity of the boat.

Solution:

1. Set up axes. Take $+x$ = east and $+y$ = north.
2. Resolve the boat's velocity. It points due north, so $$\vec{v}_b = 0\hat{i} + 25\hat{j}\ \text{km/h}$$
3. Resolve the current. "$60°$ east of south" means: start pointing south, then swing $60°$ towards east. Its southward part uses the cosine of $60°$ (measured from south) and its eastward part the sine: $$v_{c,x} = 10\sin 60° = 10(0.8660) = +8.66\ \text{(east)}$$ $$v_{c,y} = -10\cos 60° = -10(0.5) = -5.00\ \text{(south, hence negative)}$$ $$\vec{v}_c = 8.66\hat{i} - 5.00\hat{j}\ \text{km/h}$$
4. Add components: $$R_x = 0 + 8.66 = 8.66, \qquad R_y = 25 - 5.00 = 20.00$$
5. Magnitude: $$R = \sqrt{8.66^2 + 20.00^2} = \sqrt{75 + 400} = \sqrt{475} = 21.79 \approx 22\ \text{km/h}$$
6. Direction. Both components are positive, so the resultant points into the north-east quadrant. Measuring from north (the natural reference for a navigation problem): $$\tan\phi = \frac{R_x}{R_y} = \frac{8.66}{20.00} = 0.4330 \quad \Rightarrow \quad \phi = 23.41°$$ so the boat actually travels at about $23.4°$ east of north. Measured anticlockwise from east instead, that is $90° - 23.41° = 66.59°$.
7. Cross-check with the formula of Example 6. The angle between $\vec{v}_b$ (due north) and $\vec{v}_c$ ($60°$ east of south) is $180° - 60° = 120°$: $$R = \sqrt{25^2 + 10^2 + 2(25)(10)\cos 120°} = \sqrt{625 + 100 - 250} = \sqrt{475} = 21.79\ \text{km/h} \ \checkmark$$

Final Answer: About 22 km/h, directed $23.4°$ east of north.

Takeaway: The boat is pointed due north but travels $23.4°$ east of north — the current takes it sideways. Resolving compass directions is the whole difficulty here: read "$60°$ east of south" as "from south, turn $60°$ towards east", and let the sketch tell you that the north component must come out negative.