Vectors

A vector is a quantity that has a magnitude and a direction. In geometry a vector is drawn as an arrow, a directed segment: a segment in which one endpoint is chosen as the initial point and the other as the terminal point.

The vector with initial point $A$ and terminal point $B$ is written $\overrightarrow{AB}$ (read "vector $AB$"). A vector is also named by one small letter with an arrow over it, such as $\vec{a}.$ The magnitude of a vector is its length, written $|\overrightarrow{AB}|$ or $|\vec{a}|;$ the length of $\overrightarrow{AB}$ is the length of the segment $AB.$

On this page vectors are arrows: we add them, multiply them by numbers and find the angles between them using drawings and theorems of geometry, without coordinates. Vectors in component form, such as ${\langle a, b\rangle},$ are covered on the page Vector Operations, and forces and velocities as vectors on a separate page.

On this page: Basic terms · Adding vectors · Subtracting vectors · Multiplying by a number · Resolving a vector · Dot product · Problems

Basic Terms

• The zero vector $\vec{0}$ has the same initial and terminal point, for example $\overrightarrow{AA}.$ Its length is 0 and it has no particular direction.
• Two nonzero vectors are collinear (parallel) if they lie on one line or on parallel lines; the zero vector is considered collinear with every vector. Two nonzero collinear vectors have either the same direction or opposite directions.
• Two vectors are equal if they have the same direction and the same length. All zero vectors are equal. It does not matter where a vector starts: it may be moved parallel to itself, and from every point exactly one vector equal to a given vector can be drawn.
• Two vectors are opposite if they have the same length and opposite directions. The vector opposite to $\vec{a}$ is written $-\vec{a};$ for example, ${\overrightarrow{BA}=-\overrightarrow{AB}}.$

All four vectors in the figure are collinear. The vectors $\vec{a},$ $\vec{b}$ and $\vec{c}$ have the same direction, and $\vec{d}$ has the opposite direction. The vectors $\vec{a}$ and $\vec{b}$ also have the same length, so they are equal, ${\vec{a}=\vec{b},}$ although they lie on different lines. The vector $\vec{d}$ is as long as $\vec{a}$ and points the other way, so ${\vec{d}=-\vec{a}.}$

Vectors can be equal or unequal, but one vector is never "greater" than another; only their lengths can be compared, as in ${|\vec{c}|\gt|\vec{a}|}.$

Equal vectors and parallelograms. If $ABCD$ is a parallelogram, then ${\overrightarrow{AB}=\overrightarrow{DC}}$ and ${\overrightarrow{AD}=\overrightarrow{BC}}$: the opposite sides are parallel and equal, and the arrows point the same way. Conversely, if ${\overrightarrow{AB}=\overrightarrow{DC}}$ and the points $A,$ $B,$ $C,$ $D$ do not lie on one line, then $ABCD$ is a parallelogram, because its sides $AB$ and $DC$ are parallel and equal. Mind the order of the letters: $\overrightarrow{AB}$ equals $\overrightarrow{DC},$ while $\overrightarrow{CD}$ is its opposite.

Adding Vectors

Two vectors are added by laying them head to tail.

The triangle rule holds for any three points, even when they lie on one line: going from $A$ to $B$ and then from $B$ to $C$ takes you from $A$ to $C.$

The two rules give the same vector. In the parallelogram $OACB$ we have ${\overrightarrow{AC}=\overrightarrow{OB}},$ so by the triangle rule ${\overrightarrow{OA}+\overrightarrow{OB}}={\overrightarrow{OA}+\overrightarrow{AC}=\overrightarrow{OC}.}$ The parallelogram rule is handy when both vectors start at the same point, as two forces acting on one body do.

Example 1. Two forces of 3 N and 4 N act on a body at a right angle to each other. Find the magnitude of their sum, the resultant force.

Solution. The parallelogram built on the two forces is a rectangle with sides 3 and 4. The resultant is its diagonal, so by the Pythagorean theorem its magnitude is ${\sqrt{3^2+4^2}}={\sqrt{25}=5}$ N.

Properties of Addition

• ${\vec{a}+\vec{b}=\vec{b}+\vec{a}}$ (commutative law). In the figure of the parallelogram rule, ${\overrightarrow{OA}+\overrightarrow{AC}=\vec{a}+\vec{b}}$ and ${\overrightarrow{OB}+\overrightarrow{BC}=\vec{b}+\vec{a}},$ and both sums are $\overrightarrow{OC}.$
• ${(\vec{a}+\vec{b})+\vec{c}}={\vec{a}+(\vec{b}+\vec{c})}$ (associative law), so a sum of several vectors is written without brackets and the vectors can be added in any order.
• ${\vec{a}+\vec{0}=\vec{a}}$ and ${\vec{a}+(-\vec{a})=\vec{0}}.$

Polygon rule. Several vectors are added by laying each one at the end of the previous one. The sum goes from the initial point of the first vector to the terminal point of the last one:

$\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}+\overrightarrow{DE}=\overrightarrow{AE}.$

If the chain of vectors ends where it began, the sum is the zero vector; for example, ${\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}=\vec{0}}$ for every triangle $ABC.$

Example 2. Simplify ${\overrightarrow{AB}+\overrightarrow{CD}+\overrightarrow{BC}}.$

Solution. Change the order so that each vector starts where the previous one ends: ${\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}=\overrightarrow{AD}.}$

The length of a sum is in general not the sum of the lengths. Always ${|\vec{a}+\vec{b}|\le|\vec{a}|+|\vec{b}|},$ with equality only when $\vec{a}$ and $\vec{b}$ have the same direction or one of them is $\vec{0}.$ This is the triangle inequality; Problem 13 below proves it with vectors for the sides of a triangle.

Subtracting Vectors

The difference ${\vec{a}-\vec{b}}$ is the vector that gives $\vec{a}$ when it is added to $\vec{b}$: ${\vec{b}+(\vec{a}-\vec{b})=\vec{a}}.$ It is equal to ${\vec{a}+(-\vec{b})},$ so subtracting a vector means adding its opposite.

Check: ${\overrightarrow{OB}+\overrightarrow{BA}=\overrightarrow{OA}}$ by the triangle rule. Read from right to left, the rule expresses any vector in terms of two vectors drawn from a chosen point $O$: ${\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}}$ ("terminal point minus initial point").

Example 3. $ABCD$ is a parallelogram, ${\vec{a}=\overrightarrow{AB}}$ and ${\vec{b}=\overrightarrow{AD}}.$ Express the vectors $\overrightarrow{AC}$ and $\overrightarrow{DB}$ in terms of $\vec{a}$ and $\vec{b}.$

Solution. ${\overrightarrow{BC}=\overrightarrow{AD}=\vec{b}},$ so ${\overrightarrow{AC}}={\overrightarrow{AB}+\overrightarrow{BC}=\vec{a}+\vec{b}}.$ The vectors $\overrightarrow{AB}$ and $\overrightarrow{AD}$ start at $A,$ so ${\overrightarrow{DB}}={\overrightarrow{AB}-\overrightarrow{AD}=\vec{a}-\vec{b}}.$ One diagonal of a parallelogram is the sum and the other the difference of the vectors along two sides from one vertex.

Multiplying a Vector by a Number

Multiplying by $k$ stretches or shrinks the arrow and, for a negative $k,$ turns it around. For a nonzero vector $\vec{a}$:

• if ${k\gt 1},$ then $k\vec{a}$ is longer than $\vec{a}$ and has the same direction;
• if ${0\lt k\lt 1},$ then $k\vec{a}$ is shorter than $\vec{a}$ and has the same direction;
• if ${k\lt 0},$ then $k\vec{a}$ has the opposite direction; it is longer than $\vec{a}$ if ${k\lt -1}$ and shorter if ${-1\lt k\lt 0}.$ In particular, ${(-1)\vec{a}=-\vec{a}}.$

The usual rules of algebra hold: ${k(\vec{a}+\vec{b})}={k\vec{a}+k\vec{b}},$ ${(k+l)\vec{a}}={k\vec{a}+l\vec{a}}$ and ${k(l\vec{a})}={(kl)\vec{a}}.$ For example, ${2(\vec{a}-3\vec{b})+5\vec{b}}={2\vec{a}-6\vec{b}+5\vec{b}}={2\vec{a}-\vec{b}.}$

So three points $A,$ $B,$ $C$ with ${A\ne B}$ lie on one line exactly when ${\overrightarrow{AC}=k\overrightarrow{AB}}$ for some number $k.$

Midpoint of a Segment

Proof. By the triangle rule ${\overrightarrow{OM}=\overrightarrow{OA}+\overrightarrow{AM}}$ and ${\overrightarrow{OM}=\overrightarrow{OB}+\overrightarrow{BM}}.$ Because $M$ is the midpoint, $\overrightarrow{AM}$ and $\overrightarrow{BM}$ are opposite vectors, so ${\overrightarrow{AM}+\overrightarrow{BM}=\vec{0}}.$ Adding the two equalities gives ${2\overrightarrow{OM}=\overrightarrow{OA}+\overrightarrow{OB}}.$

Example 4. $M$ and $N$ are the midpoints of the sides $AB$ and $AC$ of the triangle $ABC.$ Prove that the segment $MN$ is parallel to $BC$ and half as long.

Solution. ${\overrightarrow{AM}=\frac{1}{2}\overrightarrow{AB}}$ and ${\overrightarrow{AN}=\frac{1}{2}\overrightarrow{AC}},$ so

${\overrightarrow{MN}}={\overrightarrow{AN}-\overrightarrow{AM}}={\frac{1}{2}\overrightarrow{AC}-\frac{1}{2}\overrightarrow{AB}}={\frac{1}{2}\left(\overrightarrow{AC}-\overrightarrow{AB}\right)}={\frac{1}{2}\overrightarrow{BC}.}$

The vector $\overrightarrow{MN}$ is half of $\overrightarrow{BC},$ so ${MN\parallel BC}$ and ${MN=\frac{1}{2}BC}.$ This is the midsegment theorem.

Resolving a Vector Along Two Directions

To resolve a vector along two directions means to write it as a sum of two vectors along these directions. In a plane this can always be done, and in exactly one way, along any two non-parallel directions.

Construction. Draw $\vec{a},$ $\vec{b}$ and $\vec{c}$ from one point $O$; let ${\overrightarrow{OC}=\vec{c}}.$ The line through $C$ parallel to $\vec{b}$ meets the line of $\vec{a}$ at $A_1,$ and the line through $C$ parallel to $\vec{a}$ meets the line of $\vec{b}$ at $B_1.$ The quadrilateral $OA_1CB_1$ is a parallelogram, so by the parallelogram rule ${\vec{c}=\overrightarrow{OA_1}+\overrightarrow{OB_1}}.$ The vector $\overrightarrow{OA_1}$ is collinear with $\vec{a},$ so ${\overrightarrow{OA_1}=x\vec{a}};$ in the same way ${\overrightarrow{OB_1}=y\vec{b}}.$ (If $\vec{c}$ lies on the line of $\vec{a}$ or of $\vec{b},$ one of the numbers is 0.)

The numbers are unique: if ${x\vec{a}+y\vec{b}}={x'\vec{a}+y'\vec{b}}$ and ${x\ne x'},$ then ${\vec{a}=\frac{y'-y}{x-x'}\vec{b}},$ and $\vec{a}$ would be collinear with $\vec{b}.$ So ${x=x'},$ and then ${y=y'}$ as well.

Example 5. $AM$ is a median of the triangle $ABC$ and $G$ is the centroid, the point where the medians meet. Express $\overrightarrow{AM}$ and $\overrightarrow{AG}$ in terms of $\overrightarrow{AB}$ and $\overrightarrow{AC},$ and prove that ${\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\vec{0}}.$

Solution. $M$ is the midpoint of $BC,$ so by the midpoint rule with ${O=A}$

$\overrightarrow{AM}=\frac{1}{2}\overrightarrow{AB}+\frac{1}{2}\overrightarrow{AC}.$

The centroid divides each median in the ratio ${2:1}$ from the vertex, so ${\overrightarrow{AG}=\frac{2}{3}\overrightarrow{AM}}={\frac{1}{3}\overrightarrow{AB}+\frac{1}{3}\overrightarrow{AC}}.$ Now express $\overrightarrow{GB}$ and $\overrightarrow{GC}$ in terms of $\overrightarrow{GA}$: ${\overrightarrow{GB}=\overrightarrow{GA}+\overrightarrow{AB}}$ and ${\overrightarrow{GC}=\overrightarrow{GA}+\overrightarrow{AC}}.$ Since ${\overrightarrow{AB}+\overrightarrow{AC}=3\overrightarrow{AG}}$ and ${\overrightarrow{GA}=-\overrightarrow{AG}},$ we get

${\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}}={3\overrightarrow{GA}+\overrightarrow{AB}+\overrightarrow{AC}}={-3\overrightarrow{AG}+3\overrightarrow{AG}=\vec{0}.}$

The Dot Product (Scalar Product)

The angle between two nonzero vectors $\vec{a}$ and $\vec{b}$ is the angle $\varphi$ that they form when they are drawn from a common point. It is between $0^\circ$ and $180^\circ$: ${\varphi=0^\circ}$ when the vectors have the same direction, ${\varphi=180^\circ}$ when they have opposite directions, and the vectors are perpendicular when ${\varphi=90^\circ}.$

Other notations for the dot product are $\vec{a}\,\vec{b}$ and ${(\vec{a},\vec{b})}.$ The result of the dot product of two vectors is always a real number, not a vector.

Example 6. ${|\vec{a}|=3}$ and ${|\vec{b}|=4}.$ Find ${\vec{a}\cdot\vec{b}}$ if the angle between the vectors is $60^\circ,$ $90^\circ$ or $120^\circ.$

Solution. For $60^\circ$: ${\vec{a}\cdot\vec{b}}={3\cdot 4\cdot\cos 60^\circ}={12\cdot\frac{1}{2}=6.}$ For $90^\circ$: ${\vec{a}\cdot\vec{b}}={12\cdot 0=0.}$ For $120^\circ$: ${\vec{a}\cdot\vec{b}}={12\cdot\left(-\frac{1}{2}\right)=-6.}$

Example 7. $ABC$ is an equilateral triangle with side 2. Find ${\overrightarrow{AB}\cdot\overrightarrow{AC}}$ and ${\overrightarrow{AB}\cdot\overrightarrow{BC}}.$

Solution. The vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ start at $A,$ and the angle between them is the angle of the triangle, $60^\circ$: ${\overrightarrow{AB}\cdot\overrightarrow{AC}}={2\cdot 2\cdot\cos 60^\circ=2.}$

The vectors $\overrightarrow{AB}$ and $\overrightarrow{BC}$ do not start at the same point. Draw ${\overrightarrow{BD}=\overrightarrow{AB}}$ from $B$; the angle between the vectors is ${\angle DBC}={180^\circ-60^\circ=120^\circ},$ not the angle $B$ of the triangle. So ${\overrightarrow{AB}\cdot\overrightarrow{BC}}={2\cdot 2\cdot\cos 120^\circ=-2.}$

Dot Product Properties

• ${\vec{a}\cdot\vec{b}=\vec{b}\cdot\vec{a}};$
• ${k(\vec{a}\cdot\vec{b})}={(k\vec{a})\cdot\vec{b}}={\vec{a}\cdot(k\vec{b})},$ where $k$ is a number;
• ${\vec{a}\cdot(\vec{b}+\vec{c})}={\vec{a}\cdot\vec{b}+\vec{a}\cdot\vec{c}};$
• ${\vec{a}\cdot\vec{a}=|\vec{a}|^2},$ because the angle between a vector and itself is $0^\circ$ and ${\cos 0^\circ=1}.$

The sign of the dot product shows whether the angle between two nonzero vectors is acute or obtuse: if ${\vec{a}\cdot\vec{b}\gt 0},$ the angle is acute or $0^\circ,$ and if ${\vec{a}\cdot\vec{b}\lt 0},$ it is obtuse or $180^\circ.$ The angle itself is found from

$\cos\varphi=\frac{\vec{a}\cdot\vec{b}}{|\vec{a}|\cdot|\vec{b}|}.$

These properties let us multiply out brackets as in algebra. Since ${|\vec{a}+\vec{b}|^2}={(\vec{a}+\vec{b})\cdot(\vec{a}+\vec{b})},$ the length of a sum or a difference is found from

${|\vec{a}+\vec{b}|^2}={|\vec{a}|^2+2\,\vec{a}\cdot\vec{b}+|\vec{b}|^2,}$

${|\vec{a}-\vec{b}|^2}={|\vec{a}|^2-2\,\vec{a}\cdot\vec{b}+|\vec{b}|^2.}$

Example 8. ${|\vec{a}|=3},$ ${|\vec{b}|=5}$ and the angle between the vectors is $60^\circ.$ Find ${|\vec{a}+\vec{b}|}.$

Solution. ${\vec{a}\cdot\vec{b}}={3\cdot 5\cdot\frac{1}{2}=7.5},$ so ${|\vec{a}+\vec{b}|^2}={9+2\cdot 7.5+25=49}$ and ${|\vec{a}+\vec{b}|=7.}$

Example 9. ${|\vec{a}|=2},$ ${|\vec{b}|=3}$ and ${|\vec{a}-\vec{b}|=\sqrt{7}}.$ Find the angle between $\vec{a}$ and $\vec{b}.$

Solution. By the formula for ${|\vec{a}-\vec{b}|^2},$ we get ${7=4-2\,\vec{a}\cdot\vec{b}+9},$ so ${\vec{a}\cdot\vec{b}=3}$ and ${\cos\varphi}={\frac{3}{2\cdot 3}=\frac{1}{2}}.$ The angle is ${\varphi=60^\circ}.$

Proofs with the Dot Product

Example 10. The law of cosines. In the triangle $ABC$ let ${a=BC},$ ${b=CA},$ ${c=AB}$ and let $\gamma$ be the angle at $C.$ Prove that ${c^2=a^2+b^2-2ab\cos\gamma}.$

Solution. The vectors $\overrightarrow{CB}$ and $\overrightarrow{CA}$ start at $C,$ so the angle between them is $\gamma$ and ${\overrightarrow{CB}\cdot\overrightarrow{CA}=ab\cos\gamma}.$ By the rule for a difference ${\overrightarrow{AB}=\overrightarrow{CB}-\overrightarrow{CA}},$ so

${c^2}={|\overrightarrow{CB}-\overrightarrow{CA}|^2}={|\overrightarrow{CB}|^2}-{2\,\overrightarrow{CB}\cdot\overrightarrow{CA}}+{|\overrightarrow{CA}|^2}={a^2-2ab\cos\gamma+b^2.}$

For ${\gamma=90^\circ}$ this is the Pythagorean theorem ${c^2=a^2+b^2}.$

Example 11. Prove that the diagonals of a rhombus are perpendicular.

Solution. Let ${\vec{a}=\overrightarrow{AB}}$ and ${\vec{b}=\overrightarrow{AD}}$ in the rhombus $ABCD.$ As in Example 3, the diagonals are ${\overrightarrow{AC}=\vec{a}+\vec{b}}$ and ${\overrightarrow{DB}=\vec{a}-\vec{b}},$ and the sides are equal, ${|\vec{a}|=|\vec{b}|}.$ Then

${\overrightarrow{AC}\cdot\overrightarrow{DB}}={(\vec{a}+\vec{b})\cdot(\vec{a}-\vec{b})}={|\vec{a}|^2-|\vec{b}|^2=0,}$

so the diagonals are perpendicular. Read backwards, the same equality shows that a parallelogram with perpendicular diagonals is a rhombus.

Vector Problems

Click a problem to see the solution.

1. If ${\vec{a}=-1\cdot\vec{b}},$ what can we say about these two vectors?

${\vec{a}=-\vec{b}}.$ If ${\vec{b}\ne\vec{0}},$ the vectors are opposite: they are collinear (parallel), have the same length and point in opposite directions. If ${\vec{b}=\vec{0}},$ then ${\vec{a}=\vec{0}}$ too.

2. Simplify ${\overrightarrow{AB}-\overrightarrow{CB}+\overrightarrow{CD}}.$

${-\overrightarrow{CB}=\overrightarrow{BC}},$ so the sum is ${\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}=\overrightarrow{AD}.}$

3. $B$ is the midpoint of the segment $AC.$ Find the numbers $k$ and $l$ for which ${\overrightarrow{AC}=k\overrightarrow{AB}}$ and ${\overrightarrow{CA}=l\overrightarrow{AB}}.$

$\overrightarrow{AC}$ is twice as long as $\overrightarrow{AB}$ and has the same direction, so ${k=2}.$ $\overrightarrow{CA}$ is twice as long and has the opposite direction, so ${l=-2}.$

4. $O$ is the point where the diagonals of the parallelogram $ABCD$ meet, ${\vec{a}=\overrightarrow{AB}}$ and ${\vec{b}=\overrightarrow{AD}}.$ Express $\overrightarrow{AO}$ and $\overrightarrow{BO}$ in terms of $\vec{a}$ and $\vec{b}.$

The diagonals of a parallelogram bisect each other. ${\overrightarrow{AC}=\vec{a}+\vec{b}},$ so ${\overrightarrow{AO}=\frac{1}{2}(\vec{a}+\vec{b})}.$ ${\overrightarrow{BD}}={\overrightarrow{AD}-\overrightarrow{AB}=\vec{b}-\vec{a}},$ so ${\overrightarrow{BO}=\frac{1}{2}(\vec{b}-\vec{a})}.$

5. $M$ and $N$ are the midpoints of the sides $AB$ and $CD$ of the quadrilateral $ABCD.$ Prove that ${\overrightarrow{MN}=\frac{1}{2}\left(\overrightarrow{AD}+\overrightarrow{BC}\right)}.$

Go from $M$ to $N$ in two ways: ${\overrightarrow{MN}=\overrightarrow{MA}+\overrightarrow{AD}+\overrightarrow{DN}}$ and ${\overrightarrow{MN}=\overrightarrow{MB}+\overrightarrow{BC}+\overrightarrow{CN}}.$ Since $M$ and $N$ are midpoints, ${\overrightarrow{MA}+\overrightarrow{MB}=\vec{0}}$ and ${\overrightarrow{DN}+\overrightarrow{CN}=\vec{0}}.$ Adding the two equalities gives ${2\overrightarrow{MN}=\overrightarrow{AD}+\overrightarrow{BC}}.$

6. What is the dot product ${\vec{a}\cdot\vec{b}}$ if ${|\vec{a}|=5},$ ${|\vec{b}|=7}$ and the angle between the two vectors is $30^\circ$?

${\vec{a}\cdot\vec{b}}={5\cdot 7\cdot\cos 30^\circ}={35\cdot\frac{\sqrt{3}}{2}}={\frac{35\sqrt{3}}{2}\approx 30.31.}$

7. ${|\vec{a}|=4},$ ${|\vec{b}|=3}$ and the angle between the vectors is $120^\circ.$ Find ${\vec{a}\cdot\vec{b}},$ ${|\vec{a}+\vec{b}|}$ and ${|\vec{a}-\vec{b}|}.$

${\vec{a}\cdot\vec{b}}={4\cdot 3\cdot\cos 120^\circ=-6}.$ ${|\vec{a}+\vec{b}|^2}={16+2\cdot(-6)+9=13},$ so ${|\vec{a}+\vec{b}|=\sqrt{13}}.$ ${|\vec{a}-\vec{b}|^2}={16-2\cdot(-6)+9=37},$ so ${|\vec{a}-\vec{b}|=\sqrt{37}}.$

8. ${|\vec{a}|=|\vec{b}|=1}$ and ${|\vec{a}+\vec{b}|=\sqrt{3}}.$ Find the angle between $\vec{a}$ and $\vec{b}.$

${3=1+2\,\vec{a}\cdot\vec{b}+1},$ so ${\vec{a}\cdot\vec{b}=\frac{1}{2}}$ and ${\cos\varphi}={\frac{1/2}{1\cdot 1}=\frac{1}{2}}.$ The angle is $60^\circ.$

9. ${|\vec{a}|=3},$ ${|\vec{b}|=2}$ and the vectors are not collinear. For which numbers $k$ are the vectors ${\vec{a}+k\vec{b}}$ and ${\vec{a}-k\vec{b}}$ perpendicular?

${(\vec{a}+k\vec{b})\cdot(\vec{a}-k\vec{b})}={|\vec{a}|^2-k^2|\vec{b}|^2}={9-4k^2}.$ It is 0 for ${k=\frac{3}{2}}$ and ${k=-\frac{3}{2}}.$ Both vectors are nonzero because $\vec{a}$ and $\vec{b}$ are not collinear, so they are perpendicular exactly for ${k=\pm\frac{3}{2}}.$

10. Prove that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

For the parallelogram $ABCD$ let ${\vec{a}=\overrightarrow{AB}}$ and ${\vec{b}=\overrightarrow{AD}}.$ The diagonals are ${\vec{a}+\vec{b}}$ and ${\vec{a}-\vec{b}},$ and adding the formulas for their squared lengths gives

${AC^2+BD^2}={|\vec{a}+\vec{b}|^2+|\vec{a}-\vec{b}|^2}={2|\vec{a}|^2+2|\vec{b}|^2}={AB^2+BC^2+CD^2+DA^2.}$

11. Prove that the angle $C$ of the triangle $ABC$ is obtuse if ${\overrightarrow{CA}\cdot\overrightarrow{CB}\lt 0}.$

Both vectors start at $C,$ so the angle between them is the angle $C$ of the triangle. ${\overrightarrow{CA}\cdot\overrightarrow{CB}}={CA\cdot CB\cdot\cos C}$ is negative only when ${\cos C\lt 0},$ that is, when ${C\gt 90^\circ}.$

12. $AB$ is a diameter of a circle with centre $O,$ and $C$ is a point of the circle other than $A$ and $B.$ Prove with vectors that ${\angle ACB=90^\circ}.$

${\overrightarrow{OB}=-\overrightarrow{OA}},$ and all three radii have the same length $r.$ Then ${\overrightarrow{CA}=\overrightarrow{OA}-\overrightarrow{OC}}$ and ${\overrightarrow{CB}=-\overrightarrow{OA}-\overrightarrow{OC}},$ so

${\overrightarrow{CA}\cdot\overrightarrow{CB}}={-(\overrightarrow{OA}-\overrightarrow{OC})}\cdot{(\overrightarrow{OA}+\overrightarrow{OC})}={-\left(|\overrightarrow{OA}|^2-|\overrightarrow{OC}|^2\right)}={-(r^2-r^2)=0.}$

The vectors $\overrightarrow{CA}$ and $\overrightarrow{CB}$ are nonzero, so they are perpendicular and ${\angle ACB=90^\circ}.$

13. Prove with vectors that in every triangle each side is shorter than the sum of the other two sides.

In the triangle $ABC$ let ${\vec{a}=\overrightarrow{AB}}$ and ${\vec{b}=\overrightarrow{BC}}.$ Then ${\overrightarrow{AC}=\vec{a}+\vec{b}}$ and

${|\vec{a}+\vec{b}|^2}={|\vec{a}|^2+2\,\vec{a}\cdot\vec{b}+|\vec{b}|^2.}$

The points $A,$ $B$ and $C$ do not lie on one line, so $\vec{a}$ and $\vec{b}$ are not collinear and the angle $\varphi$ between them is not $0^\circ.$ Then ${\cos\varphi\lt 1}$ and ${\vec{a}\cdot\vec{b}\lt|\vec{a}|\cdot|\vec{b}|},$ so ${|\vec{a}+\vec{b}|^2}\lt{(|\vec{a}|+|\vec{b}|)^2},$ that is, ${AC\lt AB+BC}.$ The same holds for the other two sides.