Skip to content
Topos Uranos
English

Vectors in the plane and parametric equations of a line

Summary

This lesson introduces the vectors of the plane, which represent displacements, as ordered pairs of real numbers, and defines their sum, their multiplication by a number, their norm and their dot product, together with their properties. From the dot product it derives the condition for perpendicularity, the Cauchy-Schwarz inequality, the triangle inequality, the orthogonal projection and the classification of angles into acute, right and obtuse. With vectors, the line is written in vector and parametric form, its normal vector and the formula for the distance from a point to a line are obtained, lines given by parameters are intersected, line segments are divided in a given ratio and uniform motion is described; the problems apply these tools to geometric proofs, to reflection in a mirror and to the closest approach of two moving bodies.

Learning objectives

  1. Operate with vectors of the plane, by components and geometrically, and express a vector as a linear combination of two others.
  2. Compute norms and dot products, and use them to recognize perpendicular vectors, classify angles and project one vector onto another.
  3. Write a line in vector, parametric and general form, convert from one form to another, and compute with vectors intersections, distances and points that divide a line segment.
  4. Prove geometric properties by means of vectors and model uniform motion with parametric lines.

Vectors in the plane

In the lesson on the equation of a line, each point of the plane was identified with the pair of its coordinates. Many quantities, however, are described not by a point but by a displacement: moving three units to the right and two upwards is the same instruction whatever the starting point. The same holds for velocity or for force, which have a magnitude, a direction and a sense, and which are studied in the course Foundations of Mechanics. A displacement is drawn as an arrow, and it is determined by two numbers: the horizontal shift and the vertical shift. If this is so, the natural definition is the following.

A vector of the plane is an ordered pair of real numbers, u⃗=(u1,u2)\vec{u} = \left( u_1, u_2 \right); the numbers u1u_1 and u2u_2 are its components. The vector from the point P=(x1,y1)P = (x_1, y_1) to the point Q=(x2,y2)Q = (x_2, y_2) is the pair of the differences of their coordinates, and it is written PQ⃗\vec{PQ}; the arrow with initial point PP and tip QQ represents it.
PQ⃗=(x2−x1,y2−y1)\vec{PQ} = \left( x_2 - x_1, y_2 - y_1 \right)

One and the same vector has infinitely many representations: two arrows represent the same vector when they produce the same horizontal shift and the same vertical shift, that is, when one is obtained from the other by a translation. Among all of them there is exactly one that starts at the origin OO, the arrow that ends at the point (u1,u2)(u_1, u_2). It follows that OP⃗\vec{OP}, called the position vector of PP, has as its components the coordinates of PP. From now on, P+u⃗P + \vec{u} denotes the point reached from PP by the displacement u⃗\vec{u}, that is, the tip of the arrow that represents u⃗\vec{u} with initial point PP. The zero vector 0⃗=(0,0)\vec{0} = \left( 0, 0 \right) corresponds to not moving, and the opposite of u⃗\vec{u} is −u⃗=(−u1,−u2)-\vec{u} = \left( -u_1, -u_2 \right), the reverse displacement; thus QP⃗=−PQ⃗\vec{QP} = -\vec{PQ}.

LemmaEqual vectors and parallelograms

Let PP, QQ, RR and SS be points of the plane. The equalities PQ⃗=RS⃗\vec{PQ} = \vec{RS} and PR⃗=QS⃗\vec{PR} = \vec{QS} are equivalent, and both are equivalent to the line segments PSPS and QRQR having the same midpoint. If the four points are not collinear, then PQ⃗=RS⃗\vec{PQ} = \vec{RS} if and only if the quadrilateral with vertices PP, QQ, SS and RR, in that order, is a parallelogram.

The three equivalences reduce to rearranging an equality between numbers, coordinate by coordinate; the last one relies on the theorem on the intersection of two lines from the lesson on the line. We call p1p_1, q1q_1, r1r_1 and s1s_1 the abscissas of the four points.

Proof

  1. q1−p1=s1−r1↔r1−p1=s1−q1\dato{a}{q_1 - p_1 = s_1 - r_1} \leftrightarrow \resaltar{r_1 - p_1 = s_1 - q_1}

    The equality of the first components of PQ⃗\vec{PQ} and of RS⃗\vec{RS} is equivalent, by adding r1−q1r_1 - q_1 to both sides, to the equality of the first components of PR⃗\vec{PR} and of QS⃗\vec{QS}. In the figure, P=(1,0)P = (1, 0), Q=(5,1)Q = (5, 1), R=(2,3)R = (2, 3) and S=(6,4)S = (6, 4).

  2. q1−p1=s1−r1↔p1+s12=q1+r12q_1 - p_1 = s_1 - r_1 \leftrightarrow \resaltar{\frac{p_1 + s_1}{2} = \frac{q_1 + r_1}{2}}

    The same equality is equivalent, by adding p1p_1 and r1r_1 to both sides and dividing by 22, to the midpoints of PSPS and of QRQR having the same abscissa. The ordinates are handled in the same way; therefore the three statements are equivalent.

  3. q2−p2q1−p1=s2−r2s1−r1\dato{p}{\frac{q_2 - p_2}{q_1 - p_1} = \frac{s_2 - r_2}{s_1 - r_1}}

    Suppose now that the four points are not collinear and that PQ⃗=RS⃗\vec{PQ} = \vec{RS}. The lines PQPQ and RSRS have the same direction: the same slope, the quotient of the components of that vector, or both are vertical if its first component is zero. They are distinct, because if they coincided the four points would be collinear; therefore they are parallel: if they are not vertical, by the theorem on the intersection of two nonvertical lines, and if they are, because two distinct vertical lines do not meet.

  4. PR⃗=QS⃗→r2−p2r1−p1=s2−q2s1−q1\vec{PR} = \vec{QS} \rightarrow \resaltar{\frac{r_2 - p_2}{r_1 - p_1} = \frac{s_2 - q_2}{s_1 - q_1}}

    By the first equivalence, also PR⃗=QS⃗\vec{PR} = \vec{QS}, and the same argument shows that the sides PRPR and QSQS are parallel. The opposite sides of the quadrilateral PQSRPQSR are parallel in pairs: it is a parallelogram.

  5. RS1⃗=PQ⃗,QS1⃗=PR⃗\vec{RS_1} = \vec{PQ}, \quad \vec{QS_1} = \vec{PR}

    Conversely, let PQSRPQSR be a parallelogram and let S1=R+PQ⃗S_1 = R + \vec{PQ}. By what has already been proved, applied to the points PP, QQ, RR and S1S_1, the point S1S_1 lies on the parallel to PQPQ through RR and on the parallel to PRPR through QQ.

  6. S1=S→RS⃗=PQ⃗S_1 = S \rightarrow \resaltar{\vec{RS} = \vec{PQ}}

    These two parallels are not parallel to each other, because PP, QQ and RR are not collinear, and they meet in a single point. Since the vertex SS of the parallelogram also lies on both, it coincides with S1S_1.

ExampleThe vector from one point to another

Compute the vector from P=(1,1)P = (1, 1) to Q=(4,3)Q = (4, 3), translate it to the point R=(−2,2)R = (-2, 2) and to the origin, and compare it with the vector from QQ to PP.

Proof

  1. PQ⃗=(4−1,3−1)=(3,2)\vec{PQ} = \left( \resaltar{4 - 1}, \resaltar{3 - 1} \right) = \dato{v}{\left( 3, 2 \right)}

    We subtract the coordinates of the initial point of the arrow from those of its tip: the displacement is 33 units to the right and 22 upwards.

  2. S=R+PQ⃗=(−2+3,2+2)=(1,4)S = R + \vec{PQ} = \left( -2 + \resaltar{3}, 2 + \resaltar{2} \right) = \left( 1, 4 \right)

    With initial point RR, the same displacement leads to the point SS.

  3. PR⃗=(−2−1,2−1)=(−3,1)=QS⃗\vec{PR} = \left( \resaltar{-2 - 1}, \resaltar{2 - 1} \right) = \left( -3, 1 \right) = \vec{QS}

    By the lemma, PR⃗\vec{PR} and QS⃗\vec{QS} are also equal, and the points PP, QQ, SS and RR are the vertices of a parallelogram.

  4. OA⃗=(3,2),A=(3,2)\vec{OA} = \resaltar{\left( 3, 2 \right)}, \quad A = \left( 3, 2 \right)

    With initial point OO, the arrow ends at the point AA whose coordinates are the components of the vector: PQ⃗\vec{PQ} is the position vector of AA.

  5. QP⃗=(1−4,1−3)=(−3,−2)=−PQ⃗\vec{QP} = \left( \resaltar{1 - 4}, \resaltar{1 - 3} \right) = \left( -3, -2 \right) = -\vec{PQ}

    The arrow from QQ to PP runs along the same line segment in the opposite sense: it represents the opposite vector.

Vector addition and multiplication by a number

The sum of the vectors u⃗=(u1,u2)\vec{u} = \left( u_1, u_2 \right) and v⃗=(v1,v2)\vec{v} = \left( v_1, v_2 \right) is u⃗+v⃗=(u1+v1,u2+v2)\vec{u} + \vec{v} = \left( u_1 + v_1, u_2 + v_2 \right), and the product of the real number tt and u⃗\vec{u} is tu⃗=(tu1,tu2)t \vec{u} = \left( t u_1, t u_2 \right). The difference of u⃗\vec{u} and v⃗\vec{v} is u⃗−v⃗=u⃗+(−v⃗)\vec{u} - \vec{v} = \vec{u} + \left( -\vec{v} \right).

The sum corresponds to performing displacements one after another: if u⃗\vec{u} is travelled and then v⃗\vec{v}, the total horizontal shift is u1+v1u_1 + v_1 and the vertical one u2+v2u_2 + v_2. In terms of points, for any three points AA, BB and CC the Chasles relation holds, because the differences of coordinates follow on from one another: (b1−a1)+(c1−b1)=c1−a1(b_1 - a_1) + (c_1 - b_1) = c_1 - a_1, and likewise with the ordinates.

AB⃗+BC⃗=AC⃗\vec{AB} + \vec{BC} = \vec{AC}

Two constructions of the sum follow: the triangle rule, which places the initial point of v⃗\vec{v} at the tip of u⃗\vec{u}, and the parallelogram rule, which draws both vectors from the same initial point and takes the diagonal of the parallelogram they determine. In particular, PQ⃗=OQ⃗−OP⃗\vec{PQ} = \vec{OQ} - \vec{OP}: every vector between two points is the difference of their position vectors. The product tu⃗t \vec{u}, for its part, multiplies both shifts by tt: if t>0t > 0, it keeps the direction and the sense of u⃗\vec{u}; if t<0t < 0, it keeps the direction and reverses the sense; and, as will be seen with the norm, it multiplies the length by ∣t∣\abs{t}.

ExampleThe sum, the difference and the parallelogram rule

Let u⃗=(4,1)\vec{u} = \left( 4, 1 \right) and v⃗=(1,3)\vec{v} = \left( 1, 3 \right). Construct u⃗+v⃗\vec{u} + \vec{v} by the triangle rule and by the parallelogram rule, and construct u⃗−v⃗\vec{u} - \vec{v}.

Proof

  1. u⃗=(4,1),v⃗=(1,3)\dato{u}{\vec{u} = \left( 4, 1 \right)}, \quad \dato{v}{\vec{v} = \left( 1, 3 \right)}

    We draw both vectors with initial point OO.

  2. u⃗+v⃗=(4+1,1+3)=(5,4)\vec{u} + \vec{v} = \left( \resaltar{4 + 1}, \resaltar{1 + 3} \right) = \dato{s}{\left( 5, 4 \right)}

    Triangle rule: with initial point at the tip of u⃗\vec{u}, v⃗\vec{v} leads to the point (5,4)(5, 4); the sum is computed component by component.

  3. v⃗+u⃗=(5,4)=u⃗+v⃗\vec{v} + \vec{u} = \resaltar{\left( 5, 4 \right)} = \vec{u} + \vec{v}

    Parallelogram rule: if one travels first v⃗\vec{v} and then u⃗\vec{u}, one reaches the same point. The sum is the diagonal of the parallelogram with sides u⃗\vec{u} and v⃗\vec{v} that starts at OO.

  4. −v⃗=(−1,−3)=(−1)v⃗-\vec{v} = \left( \resaltar{-1}, \resaltar{-3} \right) = \left( -1 \right) \vec{v}

    The opposite of v⃗\vec{v} is its product by −1-1: it has the same length and the opposite sense.

  5. u⃗−v⃗=(4−1,1−3)=(3,−2)\vec{u} - \vec{v} = \left( \resaltar{4 - 1}, \resaltar{1 - 3} \right) = \left( 3, -2 \right)

    The difference is the sum of u⃗\vec{u} and −v⃗-\vec{v}. Translated, it is the arrow from the tip of v⃗\vec{v} to the tip of u⃗\vec{u}: the other diagonal of the parallelogram.

The two operations inherit from the real numbers the following properties, valid for all vectors u⃗\vec{u}, v⃗\vec{v} and w⃗\vec{w} and all real numbers ss and tt:

  1. Commutativity of addition: u⃗+v⃗=v⃗+u⃗\vec{u} + \vec{v} = \vec{v} + \vec{u}.
  2. Associativity of addition: (u⃗+v⃗)+w⃗=u⃗+(v⃗+w⃗)\left( \vec{u} + \vec{v} \right) + \vec{w} = \vec{u} + \left( \vec{v} + \vec{w} \right).
  3. Neutral element: u⃗+0⃗=u⃗\vec{u} + \vec{0} = \vec{u}.
  4. Opposite element: u⃗+(−u⃗)=0⃗\vec{u} + \left( -\vec{u} \right) = \vec{0}.
  5. Associativity of the product: s(tu⃗)=(st)u⃗s \left( t \vec{u} \right) = \left( s t \right) \vec{u}.
  6. Identity: 1u⃗=u⃗1 \vec{u} = \vec{u}.
  7. Distributivity over the sum of vectors: t(u⃗+v⃗)=tu⃗+tv⃗t \left( \vec{u} + \vec{v} \right) = t \vec{u} + t \vec{v}.
  8. Distributivity over the sum of numbers: (s+t)u⃗=su⃗+tu⃗\left( s + t \right) \vec{u} = s \vec{u} + t \vec{u}.
TheoremProperties of the operations on vectors

For all vectors u⃗\vec{u}, v⃗\vec{v} and w⃗\vec{w} of the plane and all real numbers ss and tt, the eight properties of the preceding list hold.

The two operations act component by component, so each property reduces to the corresponding property of the real numbers, applied in each component. It is enough to write the first component, because the second is identical with the subscript 22.

Proof

  1. u1+v1=v1+u1,(u1+v1)+w1=u1+(v1+w1)\resaltar{u_1 + v_1} = \resaltar{v_1 + u_1}, \qquad \resaltar{\left( u_1 + v_1 \right) + w_1} = \resaltar{u_1 + \left( v_1 + w_1 \right)}

    Commutativity and associativity: they are inherited from those of the addition of real numbers.

  2. u1+0=u1,u1+(−u1)=0\resaltar{u_1 + 0} = \resaltar{u_1}, \qquad \resaltar{u_1 + \left( -u_1 \right)} = \resaltar{0}

    Neutral and opposite: 00 is the neutral element of addition and −u1-u_1 is the opposite of u1u_1.

  3. s(tu1)=(st)u1,1⋅u1=u1\resaltar{s \left( t u_1 \right)} = \resaltar{\left( s t \right) u_1}, \qquad \resaltar{1 \cdot u_1} = \resaltar{u_1}

    Associativity of the product and identity: the product of real numbers is associative and has 11 as its neutral element.

  4. t(u1+v1)=tu1+tv1,(s+t)u1=su1+tu1\resaltar{t \left( u_1 + v_1 \right)} = \resaltar{t u_1 + t v_1}, \qquad \resaltar{\left( s + t \right) u_1} = \resaltar{s u_1 + t u_1}

    The two distributive laws: the product of real numbers is distributive over addition.

  5. u⃗+v⃗=(u1+v1,u2+v2)=(v1+u1,v2+u2)=v⃗+u⃗\vec{u} + \vec{v} = \left( \resaltar{u_1 + v_1}, \resaltar{u_2 + v_2} \right) = \left( \resaltar{v_1 + u_1}, \resaltar{v_2 + u_2} \right) = \vec{v} + \vec{u}

    Two vectors are equal if and only if their two components are; therefore the eight equalities hold between vectors. For example, the first one.

These eight properties are the vector space axioms: every set whose elements can be added to one another and multiplied by real numbers satisfying those eight rules is called a vector space, and everything deduced from them without using components will hold in any of them. The polynomials of the lesson on the algebra of polynomials, with their sum and their product by a number, form another vector space.

Linear combinations and parallel vectors

A linear combination of the vectors u⃗\vec{u} and v⃗\vec{v} is a vector of the form su⃗+tv⃗s \vec{u} + t \vec{v}, with ss and tt real. The vectors e⃗1=(1,0)\vec{e}_1 = \left( 1, 0 \right) and e⃗2=(0,1)\vec{e}_2 = \left( 0, 1 \right), which advance one unit in the direction of each axis, make it possible to write every vector as a combination of them, and in only one way, because ae⃗1+be⃗2=(a,b)a \vec{e}_1 + b \vec{e}_2 = \left( a, b \right) equals u⃗\vec{u} only if a=u1a = u_1 and b=u2b = u_2. That is why they are called the standard basis of the plane:

u⃗=(u1,u2)=u1e⃗1+u2e⃗2\vec{u} = \left( u_1, u_2 \right) = u_1 \vec{e}_1 + u_2 \vec{e}_2

Two nonzero vectors u⃗\vec{u} and v⃗\vec{v} are parallel when one is a multiple of the other, that is, when v⃗=tu⃗\vec{v} = t \vec{u} for some real number tt, which cannot be zero; they have the same sense if t>0t > 0 and opposite senses if t<0t < 0. The arrows that represent them with a common initial point then lie on one and the same line.

LemmaCriterion for parallelism

Two nonzero vectors u⃗\vec{u} and v⃗\vec{v} are parallel if and only if u1v2−u2v1=0u_1 v_2 - u_2 v_1 = 0.

One implication is a computation; for the other, the number tt with v⃗=tu⃗\vec{v} = t \vec{u} has to be constructed, by dividing by a nonzero component of u⃗\vec{u}.

Proof

  1. u1(tu2)−u2(tu1)=0u_1 \resaltar{\left( t u_2 \right)} - u_2 \resaltar{\left( t u_1 \right)} = 0

    If v⃗=tu⃗\vec{v} = t \vec{u}, the components of v⃗\vec{v} are tu1t u_1 and tu2t u_2, and the expression vanishes. In the figure, u⃗=(2,1)\vec{u} = \left( 2, 1 \right) and v⃗=(6,3)=3u⃗\vec{v} = \left( 6, 3 \right) = 3 \vec{u}.

  2. u1v2=u2v1,t=v1u1→v1=tu1\dato{h}{u_1 v_2 = u_2 v_1}, \quad t = \frac{v_1}{u_1} \rightarrow \resaltar{v_1 = t u_1}

    Conversely, suppose that the expression vanishes, and suppose first that u1≠0u_1 \neq 0. We define tt as the quotient of the first components, so that the first component of v⃗\vec{v} is tu1t u_1. In the figure, t=62=3t = \frac{6}{2} = 3.

  3. v2=u2v1u1=u2tu1u1=tu2v_2 = \frac{u_2 \resaltar{v_1}}{u_1} = \frac{u_2 \resaltar{t u_1}}{u_1} = t u_2

    We solve the hypothesis for v2v_2: the second component is also tu2t u_2, and therefore v⃗=tu⃗\vec{v} = t \vec{u}. In the figure, 3=3⋅13 = 3 \cdot 1.

  4. u1=0→v1=0,t=v2u2u_1 = 0 \rightarrow \resaltar{v_1 = 0}, \quad t = \frac{v_2}{u_2}

    If u1=0u_1 = 0, then u2≠0u_2 \neq 0, because u⃗\vec{u} is not zero; the hypothesis gives u2v1=0u_2 v_1 = 0, whence v1=0v_1 = 0, and with the quotient of the second components we get v⃗=tu⃗\vec{v} = t \vec{u}. In the figure, u⃗=(0,2)\vec{u} = \left( 0, 2 \right) and v⃗=(0,−3)\vec{v} = \left( 0, -3 \right), with t=−32t = -\frac{3}{2}.

The number u1v2−u2v1u_1 v_2 - u_2 v_1 is the determinant from the lesson on linear systems, and the following lemma explains why.

LemmaCombinations of two nonparallel vectors

If u⃗\vec{u} and v⃗\vec{v} are nonzero, nonparallel vectors, every w⃗\vec{w} is a linear combination of them in a unique way: there exist numbers ss and tt, and only one such pair, such that w⃗=su⃗+tv⃗\vec{w} = s \vec{u} + t \vec{v}.

The required equality, written by components, is a system of two linear equations in the unknowns ss and tt, and it is enough to apply Cramer’s rule to it.

Proof

  1. {u1s+v1t=w1u2s+v2t=w2\begin{cases} u_1 s + v_1 t = w_1 \\ u_2 s + v_2 t = w_2 \end{cases}

    We write the equality su⃗+tv⃗=w⃗s \vec{u} + t \vec{v} = \vec{w} component by component. In the figure, u⃗=(2,1)\vec{u} = \left( 2, 1 \right), v⃗=(1,3)\vec{v} = \left( 1, 3 \right) and w⃗=(4,7)\vec{w} = \left( 4, 7 \right).

  2. D=u1v2−u2v1≠0\dato{D}{D = u_1 v_2 - u_2 v_1 \neq 0}

    The determinant of the system is not zero, by the criterion for parallelism, because the vectors are not parallel. In the figure, 2⋅3−1⋅1=52 \cdot 3 - 1 \cdot 1 = 5.

  3. s=w1v2−v1w2D,t=u1w2−w1u2Ds = \frac{w_1 v_2 - v_1 w_2}{\resaltar{D}}, \quad t = \frac{u_1 w_2 - w_1 u_2}{\resaltar{D}}

    By Cramer’s rule, the system has exactly one solution, given by the following quotients. In the figure, s=12−75=1s = \frac{12 - 7}{5} = 1 and t=14−45=2t = \frac{14 - 4}{5} = 2, and indeed w⃗=u⃗+2v⃗\vec{w} = \vec{u} + 2 \vec{v}.

  4. su⃗+tv⃗=s1u⃗+t1v⃗→s=s1∧t=t1s \vec{u} + t \vec{v} = s_1 \vec{u} + t_1 \vec{v} \rightarrow \resaltar{s = s_1 \wedge t = t_1}

    In particular, if two combinations of u⃗\vec{u} and v⃗\vec{v} are equal, both are the solution of the system with the same w⃗\vec{w}, and their coefficients coincide. This comparison of coefficients is the tool of several proofs in this lesson.