Vector Operations

In a coordinate system a vector is given by two numbers, its components. With them, adding and subtracting vectors, multiplying a vector by a number, and finding lengths and angles all come down to arithmetic with numbers. On this page vectors are written as bold letters, such as $\mathbf{u},$ $\mathbf{v}$ and $\mathbf{w},$ or named by their endpoints, such as $\overrightarrow{AC}.$

On this page: Component form · Operations on vectors · Unit vectors · Direction angles · Dot product and angle · Forces in equilibrium · Summary of formulas · Practice

Component Form of a Vector

Consider a vector $\mathbf{v}$ whose initial point is the origin of an $xy$-coordinate system and whose terminal point is $(a, b).$ Such a vector is in standard position and is called a position vector. The ordered pair $(a, b)$ determines it completely, so the pair can be used to name the vector. To tell the vector apart from the point $(a, b)$ and from an interval, we write it with angle brackets:

$\mathbf{v}={\left\langle a, b\right\rangle}.$

The number $a$ is the horizontal component of the vector and $b$ is its vertical component. Both are scalars, that is, ordinary numbers and not vectors. The expression ${\left\langle a, b\right\rangle}$ is the component form of $\mathbf{v}.$

Now take a vector $\overrightarrow{AC}$ that does not start at the origin, with ${A=(x_1, y_1)}$ and ${C=(x_2, y_2)}.$ Slide it, without turning it, until its initial point is at the origin $O.$ Its terminal point is then the point $P$ whose coordinates are found by subtracting the coordinates of $A$ from the coordinates of $C\text{:}$

$P=(x_2-x_1,\ y_2-y_1).$

In the figure $\overrightarrow{AC}$ and $\overrightarrow{OP}$ are hypotenuses of two congruent right triangles with horizontal and vertical legs, so they have the same length and the same direction. Such vectors are called equivalent, or equal, and we write ${\overrightarrow{AC}=\overrightarrow{OP}}.$

Example 1. Find the component form of $\overrightarrow{CF}$ if ${C=(-4, -3)}$ and ${F=(1, 5)}.$

Solution.

${\overrightarrow{CF}}={\left\langle 1-(-4),\ 5-(-3)\right\rangle}={{\left\langle 5, 8\right\rangle}.}$

The vector $\overrightarrow{CF}$ is equivalent to the position vector ${\mathbf{v}=\overrightarrow{OP}}$ with ${P=(5, 8)}\text{:}$ going from $C$ to $F$ and going from $O$ to $P$ both mean 5 units to the right and 8 units up.

Length of a Vector

When the components are known, the length of a vector follows from the Pythagorean theorem. The vector ${\mathbf{v}={\left\langle v_1, v_2\right\rangle}}$ in standard position is the hypotenuse of a right triangle with legs $|v_1|$ and $|v_2|,$ so its length $|\mathbf{v}|$ satisfies ${|\mathbf{v}|^2=v_1^2+v_2^2}$ (this also holds when one of the components is 0).

For the vector in Example 1,${\left|\overrightarrow{CF}\right|}={\sqrt{5^2+8^2}}={\sqrt{89}}\approx{9.43}.$

Two vectors are equivalent if they have the same length and the same direction. In components this means that ${{\left\langle u_1, u_2\right\rangle}={\left\langle v_1, v_2\right\rangle}}$ if and only if ${u_1=v_1}$ and ${u_2=v_2}.$

Operations on Vectors

Multiplying a Vector by a Number

To multiply a vector $\mathbf{v}$ by a positive number, multiply its length by that number and keep its direction. For example, $2\mathbf{v}$ is twice as long as $\mathbf{v},$ and $1.6\mathbf{v}$ is 60% longer than $\mathbf{v};$ both point the same way as $\mathbf{v}.$ To multiply $\mathbf{v}$ by a negative number, multiply its length by the absolute value of the number and reverse its direction: ${-2\mathbf{v}}$ is twice as long as $\mathbf{v}$ and points the opposite way.

Numbers act on vectors as scaling factors, which is where the name scalar comes from. The products $k\mathbf{v}$ are called scalar multiples of $\mathbf{v}.$

Example 2. Let ${\mathbf{u}={\left\langle -5, 4\right\rangle}}$ and ${\mathbf{w}={\left\langle 1, -1\right\rangle}}.$ Find ${-7\mathbf{w}},$ $3\mathbf{u}$ and ${(-1)\mathbf{w}}.$

Solution.

${-7\mathbf{w}}={-7{\left\langle 1, -1\right\rangle}}={{\left\langle -7, 7\right\rangle},}$

${3\mathbf{u}}={3{\left\langle -5, 4\right\rangle}}={{\left\langle -15, 12\right\rangle},}$

${(-1)\mathbf{w}}={(-1){\left\langle 1, -1\right\rangle}}={{\left\langle -1, 1\right\rangle}.}$

Adding Vectors

Two vectors given in component form are added by adding their corresponding components.

In the figure ${\mathbf{u}+\mathbf{v}}$ is the diagonal of the parallelogram built on $\mathbf{u}$ and $\mathbf{v}$ (the parallelogram law), and its terminal point is ${(u_1+v_1,\ u_2+v_2)}.$

For example, if ${\mathbf{v}={\left\langle -3, 2\right\rangle}}$ and ${\mathbf{w}={\left\langle 5, -9\right\rangle}},$ then

${\mathbf{v}+\mathbf{w}}={\left\langle -3+5,\ 2+(-9)\right\rangle}={{\left\langle 2, -7\right\rangle}.}$

Vector Subtraction

Before subtracting we need the opposite of a vector. The opposite of ${\mathbf{v}={\left\langle v_1, v_2\right\rangle}}$ is

${-\mathbf{v}}={(-1)\mathbf{v}}={{\left\langle -v_1, -v_2\right\rangle}.}$

It has the same length as $\mathbf{v}$ and the opposite direction.

To subtract $\mathbf{v}$ is to add its opposite, so subtraction is also done component by component:

${\mathbf{u}-\mathbf{v}}={\mathbf{u}+(-\mathbf{v})}={\left\langle u_1, u_2\right\rangle+\left\langle -v_1, -v_2\right\rangle}={{\left\langle u_1-v_1,\ u_2-v_2\right\rangle}.}$

The difference can be drawn with a parallelogram, just like the sum:

1. Draw $\mathbf{u}$ and $\mathbf{v}$ from the same point.
2. Draw the opposite vector ${-\mathbf{v}}.$
3. Add $\mathbf{u}$ and ${-\mathbf{v}}$ by the parallelogram law: the diagonal is ${\mathbf{u}+(-\mathbf{v})=\mathbf{u}-\mathbf{v}}.$
4. The same vector ${\mathbf{u}-\mathbf{v}}$ goes from the terminal point of $\mathbf{v}$ to the terminal point of $\mathbf{u}.$

So the two diagonals of the parallelogram built on $\mathbf{u}$ and $\mathbf{v}$ are the sum ${\mathbf{u}+\mathbf{v}}$ and the difference ${\mathbf{u}-\mathbf{v}}.$

Example 3. Let ${\mathbf{u}={\left\langle 7, 2\right\rangle}}$ and ${\mathbf{v}={\left\langle -3, 5\right\rangle}}.$ Find: a) ${\mathbf{u}+\mathbf{v}};$ b) ${\mathbf{u}-6\mathbf{v}};$ c) ${3\mathbf{u}+4\mathbf{v}};$ d) ${|5\mathbf{v}-2\mathbf{u}|}.$

Solution.

a) ${\mathbf{u}+\mathbf{v}}={\left\langle 7, 2\right\rangle+\left\langle -3, 5\right\rangle}={\left\langle 7+(-3),\ 2+5\right\rangle}={{\left\langle 4, 7\right\rangle};}$

b) ${\mathbf{u}-6\mathbf{v}}={\left\langle 7, 2\right\rangle-6{\left\langle -3, 5\right\rangle}}={\left\langle 7, 2\right\rangle-\left\langle -18, 30\right\rangle}={{\left\langle 25, -28\right\rangle};}$

c) ${3\mathbf{u}+4\mathbf{v}}={3{\left\langle 7, 2\right\rangle}+4{\left\langle -3, 5\right\rangle}}={\left\langle 21, 6\right\rangle+\left\langle -12, 20\right\rangle}={{\left\langle 9, 26\right\rangle};}$

d) ${|5\mathbf{v}-2\mathbf{u}|}={|5{\left\langle -3, 5\right\rangle}-2{\left\langle 7, 2\right\rangle}|}={|{\left\langle -15, 25\right\rangle}-{\left\langle 14, 4\right\rangle}|}={|{\left\langle -29, 21\right\rangle}|}={\sqrt{(-29)^2+21^2}}={\sqrt{1282}}\approx{35.8.}$

Properties of Vector Addition and Scalar Multiplication

One more special vector is needed: the zero vector ${\mathbf{0}={\left\langle 0, 0\right\rangle}},$ whose initial and terminal points coincide. Its length is 0, and it has no direction. When vectors are added, it plays the role of the number 0: ${\mathbf{v}+\mathbf{0}=\mathbf{v}},$ because ${\left\langle v_1, v_2\right\rangle+\left\langle 0, 0\right\rangle}={\left\langle v_1, v_2\right\rangle}.$

Operations on vectors share many properties with operations on real numbers.

Unit Vectors

A vector of length 1 is called a unit vector. For example, ${\mathbf{v}={\left\langle -\frac35,\ \frac45\right\rangle}}$ is a unit vector, because

${|\mathbf{v}|}={\sqrt{\left(-\frac35\right)^2+\left(\frac45\right)^2}}={\sqrt{\frac{9}{25}+\frac{16}{25}}}={{\sqrt{\frac{25}{25}}=1}.}$

Example 4. Find the unit vector that has the same direction as ${\mathbf{w}={\left\langle -3, 5\right\rangle}}.$

Solution. The length of $\mathbf{w}$ is ${|\mathbf{w}|}={\sqrt{(-3)^2+5^2}}={\sqrt{34}}.$ We need a vector with the direction of $\mathbf{w}$ whose length is $\frac{1}{\sqrt{34}}$ of the length of $\mathbf{w}\text{:}$

${\mathbf{u}}={\frac{1}{\sqrt{34}}\mathbf{w}}={\frac{1}{\sqrt{34}}{\left\langle -3, 5\right\rangle}}={{\left\langle -\frac{3}{\sqrt{34}},\ \frac{5}{\sqrt{34}}\right\rangle}.}$

It is indeed a unit vector:

${|\mathbf{u}|}={\sqrt{\left(-\frac{3}{\sqrt{34}}\right)^2+\left(\frac{5}{\sqrt{34}}\right)^2}}={\sqrt{\frac{9}{34}+\frac{25}{34}}}={{\sqrt{\frac{34}{34}}=1}.}$

The Unit Vectors i and j

Unit vectors can point in any direction, but two of them are used all the time: the unit vectors in the directions of the positive $x$-axis and the positive $y$-axis,

${\mathbf{i}}={{\left\langle 1, 0\right\rangle}\quad\text{and}\quad\mathbf{j}}={{\left\langle 0, 1\right\rangle}.}$

Every vector can be written as a linear combination of $\mathbf{i}$ and $\mathbf{j},$ that is, as a sum of scalar multiples of them. Indeed, for ${\mathbf{v}={\left\langle v_1, v_2\right\rangle}},$

${\mathbf{v}}={\left\langle v_1, v_2\right\rangle}={\left\langle v_1, 0\right\rangle+\left\langle 0, v_2\right\rangle}={v_1{\left\langle 1, 0\right\rangle}+v_2{\left\langle 0, 1\right\rangle}}={v_1\mathbf{i}+v_2\mathbf{j}.}$

Example 5. Express the vector ${\mathbf{r}={\left\langle 2, -6\right\rangle}}$ as a linear combination of $\mathbf{i}$ and $\mathbf{j}.$

Solution. ${\mathbf{r}}={\left\langle 2, -6\right\rangle}={2\mathbf{i}+(-6)\mathbf{j}}={2\mathbf{i}-6\mathbf{j}}.$

Example 6. Write the vector ${\mathbf{q}=-\mathbf{i}+7\mathbf{j}}$ in component form.

Solution. ${\mathbf{q}}={-\mathbf{i}+7\mathbf{j}}={(-1)\mathbf{i}+7\mathbf{j}}={\left\langle -1, 7\right\rangle}.$

Vectors written with $\mathbf{i}$ and $\mathbf{j}$ are added, subtracted and multiplied by numbers term by term, like algebraic expressions.

Example 7. If ${\mathbf{a}=5\mathbf{i}-2\mathbf{j}}$ and ${\mathbf{b}=-\mathbf{i}+8\mathbf{j}},$ find ${3\mathbf{a}-\mathbf{b}}.$

Solution.

${3\mathbf{a}-\mathbf{b}}={3(5\mathbf{i}-2\mathbf{j})}-{(-\mathbf{i}+8\mathbf{j})}={15\mathbf{i}-6\mathbf{j}+\mathbf{i}-8\mathbf{j}}={16\mathbf{i}-14\mathbf{j}.}$

Direction Angles

The terminal point $P$ of a unit vector in standard position lies on the unit circle, so its coordinates are ${(\cos\theta, \sin\theta)}$ for some angle $\theta.$ Therefore every unit vector can be written in component form as

$\mathbf{u}={\left\langle \cos\theta, \sin\theta\right\rangle},$

or as a linear combination of $\mathbf{i}$ and $\mathbf{j},$

$\mathbf{u}=(\cos\theta)\mathbf{i}+(\sin\theta)\mathbf{j}.$

The angle $\theta$ is the direction angle of the vector. It is measured counterclockwise from the positive $x$-axis to the vector. As $\theta$ goes from $0$ to $2\pi,$ the point $P$ runs once around the circle ${x^2+y^2=1},$ so every unit vector has exactly one direction angle with ${0\le\theta\lt 2\pi},$ that is, from $0^\circ$ up to, but not including, $360^\circ.$

Example 8. Calculate and sketch the unit vector ${\mathbf{u}=(\cos\theta)\mathbf{i}+(\sin\theta)\mathbf{j}}$ for ${\theta=\frac{2\pi}{3}}.$ Include the unit circle in the sketch.

Solution.

${\mathbf{u}}={\left(\cos\frac{2\pi}{3}\right)\mathbf{i}+\left(\sin\frac{2\pi}{3}\right)\mathbf{j}}={-\frac12\,\mathbf{i}+\frac{\sqrt3}{2}\,\mathbf{j}.}$

The direction angle of any nonzero vector ${\mathbf{v}={\left\langle v_1, v_2\right\rangle}}$ is measured in the same way, and it is found from the components with the tangent function.

The equation ${\tan\theta=\frac{v_2}{v_1}}$ has two solutions between $0^\circ$ and $360^\circ,$ while a calculator gives one value of $\tan^{-1}\frac{v_2}{v_1},$ between ${-90^\circ}$ and $90^\circ.$ Always check the quadrant of the vector: if ${v_1\lt 0},$ add $180^\circ$ to the calculator value; if ${v_1\gt 0}$ and ${v_2\lt 0},$ add $360^\circ.$ If ${v_1=0},$ the vector points straight up (${\theta=90^\circ}$) or straight down (${\theta=270^\circ}$).

Example 9. Determine the direction angle $\theta$ of the vector ${\mathbf{w}=-4\mathbf{i}-3\mathbf{j}}.$

Solution. In component form ${\mathbf{w}={\left\langle -4, -3\right\rangle}},$ so

$\tan\theta=\frac{-3}{-4}=\frac34.$

A calculator gives ${\tan^{-1}\frac34\approx 37^\circ}.$ Both components of $\mathbf{w}$ are negative, so the vector lies in the third quadrant and ${\theta\approx 180^\circ+37^\circ=217^\circ}.$

A Vector from Its Magnitude and Direction

In applications a vector is often given by its magnitude and its direction angle. If $\theta$ is the direction angle of $\mathbf{v},$ then $\frac{\mathbf{v}}{|\mathbf{v}|}$ is the unit vector with the same direction, so

${\frac{\mathbf{v}}{|\mathbf{v}|}}={(\cos\theta)\mathbf{i}+(\sin\theta)\mathbf{j}.}$

Multiplying both sides by $|\mathbf{v}|$ gives the vector itself.

Example 10 (airplane speed and direction). An airplane is headed on a bearing of $100^\circ$ at an airspeed of 190 km/h, while a wind of 48 km/h blows from the bearing $220^\circ.$ Find the ground speed of the airplane and the direction of its track over the ground.

Solution. A bearing is measured clockwise from north. Let $\overrightarrow{OA}$ be the velocity of the airplane relative to the air and $\overrightarrow{OC}$ the velocity of the wind. The velocity of the airplane over the ground is their sum:

$\mathbf{v}=\overrightarrow{OA}+\overrightarrow{OC}.$

Let the positive $x$-axis point east and the positive $y$-axis point north. Then a bearing $\beta$ corresponds to the direction angle ${90^\circ-\beta}$ (add $360^\circ$ if the result is negative).

• The airplane is headed on a bearing of $100^\circ,$ so the direction angle of $\overrightarrow{OA}$ is ${90^\circ-100^\circ+360^\circ=350^\circ}.$
• The wind blows from $220^\circ,$ that is, toward the bearing ${220^\circ-180^\circ=40^\circ},$ so the direction angle of $\overrightarrow{OC}$ is ${90^\circ-40^\circ=50^\circ}.$

The magnitudes of $\overrightarrow{OA}$ and $\overrightarrow{OC}$ are 190 and 48, so

${\overrightarrow{OA}}={190(\cos 350^\circ)\mathbf{i}}+{190(\sin 350^\circ)\mathbf{j},}$

${\overrightarrow{OC}}={48(\cos 50^\circ)\mathbf{i}}+{48(\sin 50^\circ)\mathbf{j}.}$

Adding the corresponding components, we get

${\mathbf{v}}={(190\cos 350^\circ+48\cos 50^\circ)\mathbf{i}}+{(190\sin 350^\circ+48\sin 50^\circ)\mathbf{j}}\approx{217.97\,\mathbf{i}+3.78\,\mathbf{j}.}$

The ground speed is the length of $\mathbf{v}\text{:}$

${|\mathbf{v}|}\approx{\sqrt{217.97^2+3.78^2}}\approx{218\text{ km/h}.}$

Let $\alpha$ be the direction angle of $\mathbf{v}.$ Then ${\tan\alpha\approx\frac{3.78}{217.97}},$ and since $\mathbf{v}$ lies in the first quadrant, ${\alpha\approx\tan^{-1}\frac{3.78}{217.97}\approx 1^\circ}.$ So the airplane moves over the ground at about 218 km/h on a bearing of ${90^\circ-1^\circ=89^\circ}.$

The Dot Product and the Angle Between Vectors

A scalar multiple of a vector is a vector, and the sum of two vectors is a vector. The dot product of two vectors is different: it is a number. It is used to find the angle between two vectors and to tell whether two vectors are perpendicular.

Example 11. Let ${\mathbf{u}={\left\langle 2, -5\right\rangle}},$ ${\mathbf{v}={\left\langle 0, 4\right\rangle}}$ and ${\mathbf{w}={\left\langle -3, 1\right\rangle}}.$ Find: a) ${\mathbf{u}\cdot\mathbf{w}};$ b) ${\mathbf{w}\cdot\mathbf{v}}.$

Solution.

a) ${\mathbf{u}\cdot\mathbf{w}}={2\cdot(-3)+(-5)\cdot 1}={{-6-5=-11};}$

b) ${\mathbf{w}\cdot\mathbf{v}}={(-3)\cdot 0+1\cdot 4}={{0+4=4}.}$

The dot product shares several properties with the product of numbers:${\mathbf{u}\cdot\mathbf{v}=\mathbf{v}\cdot\mathbf{u}},$ ${\mathbf{u}\cdot(\mathbf{v}+\mathbf{w})}={\mathbf{u}\cdot\mathbf{v}+\mathbf{u}\cdot\mathbf{w}}$ and ${(k\mathbf{u})\cdot\mathbf{v}=k(\mathbf{u}\cdot\mathbf{v})}.$ The dot product of a vector with itself is the square of its length: ${\mathbf{v}\cdot\mathbf{v}}={v_1^2+v_2^2}={|\mathbf{v}|^2}.$

The angle between two nonzero vectors is the smaller of the two angles they form when they are drawn from the same point. The angle $\theta$ between $\mathbf{u}$ and $\mathbf{v}$ is the same as the angle between $\mathbf{v}$ and $\mathbf{u},$ and ${0\le\theta\le\pi}$ (from $0^\circ$ to $180^\circ$).

Why the formula is true

Draw $\mathbf{u}$ and $\mathbf{v}$ from the same point. Together with ${\mathbf{u}-\mathbf{v}}$ they form a triangle with sides $|\mathbf{u}|,$ $|\mathbf{v}|$ and ${|\mathbf{u}-\mathbf{v}|},$ in which $\theta$ is the angle opposite the third side. By the law of cosines,

${|\mathbf{u}-\mathbf{v}|^2}={|\mathbf{u}|^2+|\mathbf{v}|^2}-{2\,|\mathbf{u}|\,|\mathbf{v}|\cos\theta.}$

On the other hand, by the properties of the dot product,

${|\mathbf{u}-\mathbf{v}|^2}={(\mathbf{u}-\mathbf{v})\cdot(\mathbf{u}-\mathbf{v})}={|\mathbf{u}|^2}-{2\,\mathbf{u}\cdot\mathbf{v}+|\mathbf{v}|^2.}$

Comparing the two right-hand sides gives ${\mathbf{u}\cdot\mathbf{v}=|\mathbf{u}|\,|\mathbf{v}|\cos\theta}.$ (When the vectors are parallel there is no triangle, but the equality still holds.)

Example 12. Find the angle between ${\mathbf{u}={\left\langle 3, 7\right\rangle}}$ and ${\mathbf{v}={\left\langle -4, 2\right\rangle}}.$

Solution. First find ${\mathbf{u}\cdot\mathbf{v}},$ $|\mathbf{u}|$ and $|\mathbf{v}|\text{:}$

${\mathbf{u}\cdot\mathbf{v}}={{3\cdot(-4)+7\cdot 2=2},}$

$|\mathbf{u}|=\sqrt{3^2+7^2}=\sqrt{58},$

${|\mathbf{v}|}={\sqrt{(-4)^2+2^2}}={\sqrt{20}.}$

Then

${\cos\theta}={\frac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{u}|\,|\mathbf{v}|}}={\frac{2}{\sqrt{58}\cdot\sqrt{20}},}$

${\theta}={\cos^{-1}\frac{2}{\sqrt{58}\cdot\sqrt{20}}}\approx{86.6^\circ.}$

Perpendicular Vectors

Since ${\cos 90^\circ=0},$ the angle between two nonzero vectors is a right angle exactly when their dot product is zero.

Two nonzero vectors $\mathbf{u}$ and $\mathbf{v}$ are perpendicular if and only if ${\mathbf{u}\cdot\mathbf{v}=0}.$

For example, ${\left\langle 2, -5\right\rangle}$ and ${\left\langle 5, 2\right\rangle}$ are perpendicular, because ${2\cdot 5+(-5)\cdot 2=0}.$ The sign of the dot product also shows what kind of angle the vectors form:${\theta\lt 90^\circ}$ if ${\mathbf{u}\cdot\mathbf{v}\gt 0},$ and ${\theta\gt 90^\circ}$ if ${\mathbf{u}\cdot\mathbf{v}\lt 0}.$

Forces in Equilibrium

When several forces act at the same point of an object, the object is in equilibrium (it stays at rest or moves in a straight line at constant speed) exactly when the vector sum of the forces is $\mathbf{0}.$ Many applied problems about forces are solved with this fact.

Example 13 (suspended block). A 350-lb block hangs from two cables, as shown in the figure. Three forces act at the point $A\text{:}$ the weight $\mathbf{W}$ of the block, pulling down, and the tensions $\mathbf{R}$ and $\mathbf{S}$ in the left and right cables, pulling upward and outward. Find the tension in each cable.

Solution. Draw the three forces from the origin. The left cable makes $55^\circ$ with the horizontal, so the direction angle of $\mathbf{R}$ is ${180^\circ-55^\circ=125^\circ}.$ The right cable makes $37^\circ$ with the horizontal, so the direction angle of $\mathbf{S}$ is $37^\circ.$ The weight $\mathbf{W}$ points straight down, so its direction angle is $270^\circ.$

The block is at rest, so

$\mathbf{R}+\mathbf{S}+\mathbf{W}=\mathbf{0}.$

Write each force using its magnitude and its direction angle:

${\mathbf{R}}={|\mathbf{R}|(\cos 125^\circ)\mathbf{i}}+{|\mathbf{R}|(\sin 125^\circ)\mathbf{j},}$

${\mathbf{S}}={|\mathbf{S}|(\cos 37^\circ)\mathbf{i}}+{|\mathbf{S}|(\sin 37^\circ)\mathbf{j},}$

${\mathbf{W}}={350(\cos 270^\circ)\mathbf{i}}+{350(\sin 270^\circ)\mathbf{j}}={-350\,\mathbf{j},}$

because ${\cos 270^\circ=0}$ and ${\sin 270^\circ=-1}.$ In the sum ${\mathbf{R}+\mathbf{S}+\mathbf{W}}$ the coefficient of $\mathbf{i}$ and the coefficient of $\mathbf{j}$ must both be zero. This gives a system of two equations:

${|\mathbf{R}|\cos 125^\circ}+{|\mathbf{S}|\cos 37^\circ}={0,}$

${|\mathbf{R}|\sin 125^\circ}+{|\mathbf{S}|\sin 37^\circ-350}={0.}$

From the first equation ${|\mathbf{S}|}={-\frac{\cos 125^\circ}{\cos 37^\circ}\,|\mathbf{R}|}\approx{0.718\,|\mathbf{R}|}.$ Substituting this into the second equation gives ${|\mathbf{R}|\,(\sin 125^\circ+0.718\sin 37^\circ)}\approx{350},$ so

${|\mathbf{R}|}\approx{280\quad\text{and}\quad|\mathbf{S}|}\approx{201.}$

The tension is about 280 lb in the left cable and about 201 lb in the right cable.

Summary of Formulas

Quantity	Formula
Vector from ${A=(x_1, y_1)}$ to ${C=(x_2, y_2)}$	${\overrightarrow{AC}={\left\langle x_2-x_1,\ y_2-y_1\right\rangle}}$
Length of ${\mathbf{v}={\left\langle v_1, v_2\right\rangle}}$	${|\mathbf{v}|=\sqrt{v_1^2+v_2^2}}$
Scalar multiple	${k\mathbf{v}={\left\langle kv_1, kv_2\right\rangle}}$
Sum and difference	${\mathbf{u}\pm\mathbf{v}}={\left\langle u_1\pm v_1,\ u_2\pm v_2\right\rangle}$
Unit vector in the direction of $\mathbf{v}$	$\frac{\mathbf{v}}{|\mathbf{v}|}$
Linear combination of $\mathbf{i}$ and $\mathbf{j}$	${{\left\langle v_1, v_2\right\rangle}=v_1\mathbf{i}+v_2\mathbf{j}}$
Direction angle $\theta$	${\tan\theta=\frac{v_2}{v_1}},$ ${\mathbf{v}}={|\mathbf{v}|(\cos\theta)\mathbf{i}}+{|\mathbf{v}|(\sin\theta)\mathbf{j}}$
Dot product	${\mathbf{u}\cdot\mathbf{v}=u_1v_1+u_2v_2}$
Angle between $\mathbf{u}$ and $\mathbf{v}$	${\cos\theta=\frac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{u}|\,|\mathbf{v}|}}$
Perpendicular vectors	${\mathbf{u}\cdot\mathbf{v}=0}$

Practice Problems

Try them yourself, then open the answer.

1. Find the component form and the length of $\overrightarrow{AB}$ if ${A=(2, -1)}$ and ${B=(-3, 4)}.$

${\overrightarrow{AB}}={\left\langle -3-2,\ 4-(-1)\right\rangle}={\left\langle -5, 5\right\rangle},$ ${\left|\overrightarrow{AB}\right|}={\sqrt{25+25}}={5\sqrt2}\approx{7.07}.$

2. Let ${\mathbf{u}={\left\langle 4, -3\right\rangle}}$ and ${\mathbf{v}={\left\langle -2, 6\right\rangle}}.$ Find ${\mathbf{u}+\mathbf{v}},$ ${\mathbf{u}-\mathbf{v}},$ ${2\mathbf{u}-3\mathbf{v}}$ and ${|\mathbf{u}+\mathbf{v}|}.$

${\mathbf{u}+\mathbf{v}={\left\langle 2, 3\right\rangle}},$ ${\mathbf{u}-\mathbf{v}={\left\langle 6, -9\right\rangle}},$ ${2\mathbf{u}-3\mathbf{v}}={\left\langle 8, -6\right\rangle-\left\langle -6, 18\right\rangle}={\left\langle 14, -24\right\rangle},$ ${|\mathbf{u}+\mathbf{v}|}={\sqrt{2^2+3^2}}={\sqrt{13}}\approx{3.61}.$

3. Find the unit vector that has the same direction as ${\mathbf{v}={\left\langle -8, 15\right\rangle}}.$

${|\mathbf{v}|=\sqrt{64+225}=17},$ so the unit vector is ${\frac{\mathbf{v}}{17}={\left\langle -\frac{8}{17},\ \frac{15}{17}\right\rangle}}.$

4. If ${\mathbf{a}=2\mathbf{i}-\mathbf{j}}$ and ${\mathbf{b}=-\mathbf{i}+3\mathbf{j}},$ find ${4\mathbf{a}-3\mathbf{b}}.$

${4\mathbf{a}-3\mathbf{b}}={8\mathbf{i}-4\mathbf{j}+3\mathbf{i}-9\mathbf{j}}={11\mathbf{i}-13\mathbf{j}}.$

5. Find the direction angle of: a) ${\mathbf{v}={\left\langle -5, 5\right\rangle}};$ b) ${\mathbf{w}=3\mathbf{i}-3\sqrt3\,\mathbf{j}}.$

a) ${\tan\theta=-1}$ and the vector is in the second quadrant, so ${\theta=135^\circ}.$

b) ${\tan\theta=-\sqrt3}$ and the vector is in the fourth quadrant, so ${\theta=300^\circ}.$

6. A vector has magnitude 10 and direction angle $150^\circ.$ Write it in component form.

${\mathbf{v}}={\left\langle 10\cos 150^\circ,\ 10\sin 150^\circ\right\rangle}={\left\langle -5\sqrt3,\ 5\right\rangle}\approx{\left\langle -8.66,\ 5\right\rangle}.$

7. Find the angle between ${\mathbf{u}={\left\langle 1, 2\right\rangle}}$ and ${\mathbf{v}={\left\langle 3, 1\right\rangle}}.$

${\mathbf{u}\cdot\mathbf{v}=3+2=5},$ ${|\mathbf{u}|=\sqrt5},$ ${|\mathbf{v}|=\sqrt{10}},$ so ${\cos\theta=\frac{5}{\sqrt5\cdot\sqrt{10}}=\frac{1}{\sqrt2}}$ and ${\theta=45^\circ}.$

8. For which value of $k$ are the vectors ${\left\langle k, 3\right\rangle}$ and ${\left\langle 2, -4\right\rangle}$ perpendicular?

The dot product must be zero: ${2k-12=0},$ so ${k=6}.$

9. A boat heads due east at 12 km/h relative to the water, and the current flows due north at 5 km/h. Find the speed of the boat over the ground and the direction in which it moves.

With the $x$-axis pointing east and the $y$-axis pointing north, the velocity is ${\mathbf{v}=12\mathbf{i}+5\mathbf{j}}.$ The speed is ${|\mathbf{v}|=\sqrt{12^2+5^2}=13}$ km/h. The direction angle is ${\tan^{-1}\frac{5}{12}\approx 22.6^\circ},$ so the boat moves about $22.6^\circ$ north of east, on a bearing of about $67.4^\circ.$

10. A 100-N weight hangs from two cables that make angles of $30^\circ$ and $60^\circ$ with the horizontal ceiling. Find the tension in each cable.

Let $T_1$ be the tension in the $30^\circ$ cable and $T_2$ the tension in the $60^\circ$ cable. The horizontal components cancel: ${T_1\cos 30^\circ=T_2\cos 60^\circ},$ so ${T_2=\sqrt3\,T_1}.$ The vertical components carry the weight: ${T_1\sin 30^\circ+T_2\sin 60^\circ=100},$ that is, ${\frac12T_1+\frac32T_1=100}.$ So ${T_1=50}$ N and ${T_2=50\sqrt3\approx 86.6}$ N.