Section 12.2: Vectors

The term vector is used by scientists to indicate a quantity (such as displacement or velocity or force) that has both magnitude and direction.

A vector is often represented by an arrow or a directed line segment. The length of the arrow represents the magnitude of the vector and the arrow points in the direction of the vector. We denote a vector by printing a letter in boldface (v) or by putting an arrow above the letter (\(\vec{v}\)).

For instance, suppose a particle moves along a line segment from point A to point B. The corresponding displacement vector v, shown in Figure 1, has initial point A (the tail) and terminal point B (the tip) and we indicate this by writing \(\mathbf{v} = \vec{AB}\). Notice that the vector \(\mathbf{u} = \vec{CD}\) has the same length and the same direction as v even though it is in a different position. We say that u and v are equivalent (or equal) and we write \(\mathbf{u} = \mathbf{v}\). The zero vector, denoted by 0, has length 0. It is the only vector with no specific direction.

Suppose a particle moves from A to B, so its displacement vector is \(\vec{AB}\). Then the particle changes direction and moves from B to C, with displacement vector \(\vec{BC}\) as in Figure 2.

The combined effect of these displacements is that the particle has moved from A to C. The resulting displacement vector \(\vec{AC}\) is called the sum of \(\vec{AB}\) and \(\vec{BC}\) and we write \[ \vec{AC} = \vec{AB} + \vec{BC} \] In general, if we start with vectors u and v, we first move v so that its tail coincides with the tip of u and define the sum of u and v as follows.

Definition of Vector Addition If u and v are vectors positioned so the initial point of v is at the terminal point of u, then the sum \(\mathbf{u} + \mathbf{v}\) is the vector from the initial point of u to the terminal point of v.

The definition of vector addition is illustrated in Figure 3. You can see why this definition is sometimes called the Triangle Law.

In Figure 4 we start with the same vectors u and v as in Figure 3 and draw another copy of v with the same initial point as u. Completing the parallelogram, we see that \(\mathbf{u} + \mathbf{v} = \mathbf{v} + \mathbf{u}\). This also gives another way to construct the sum: if we place u and v so they start at the same point, then \(\mathbf{u} + \mathbf{v}\) lies along the diagonal of the parallelogram with u and v as sides. (This is called the Parallelogram Law.)

EXAMPLE 1 Draw the sum of the vectors a and b shown in Figure 5.

It is possible to multiply a vector by a real number \(c\). (In this context we call the real number \(c\) a scalar to distinguish it from a vector.) For instance, we want \(2\mathbf{v}\) to be the same vector as \(\mathbf{v} + \mathbf{v}\), which has the same direction as v but is twice as long. In general, we multiply a vector by a scalar as follows.

Definition of Scalar Multiplication If \(c\) is a scalar and v is a vector, then the scalar multiple \(c\mathbf{v}\) is the vector whose length is \(|c|\) times the length of v and whose direction is the same as v if \(c > 0\) and is opposite to v if \(c < 0\). If \(c = 0\) or \(\mathbf{v} = \mathbf{0}\), then \(c\mathbf{v} = \mathbf{0}\).

This definition is illustrated in Figure 7. We see that real numbers work like scaling factors here; that’s why we call them scalars. Notice that two nonzero vectors are parallel if they are scalar multiples of one another. In particular, the vector \(-\mathbf{v} = (-1)\mathbf{v}\) has the same length as v but points in the opposite direction. We call it the negative of v.

Recall, the vector \(-\mathbf{v} = (-1)\mathbf{v}\) has the same length as v but points in the opposite direction. We call it the negative of v.

By the difference \(\mathbf{u} - \mathbf{v}\) of two vectors we mean \[ \mathbf{u} - \mathbf{v} = \mathbf{u} + (-\mathbf{v}) \]

So we can construct \(\mathbf{u} - \mathbf{v}\) by first drawing the negative of v, \(-\mathbf{v}\), and then adding it to u by the Parallelogram Law as in Figure 8(a). Alternatively, since \(\mathbf{v} + (\mathbf{u} - \mathbf{v}) = \mathbf{u}\), the vector \(\mathbf{u} - \mathbf{v}\), when added to v, gives u. So we could construct \(\mathbf{u} - \mathbf{v}\) as in Figure 8(b) by means of the Triangle Law. Notice that if u and v both start from the same initial point, then \(\mathbf{u} - \mathbf{v}\) connects the tip of v to the tip of u.

EXAMPLE 2 If a and b are the vectors shown in Figure 9, draw \(\mathbf{a} - 2\mathbf{b}\).

For some purposes it’s best to introduce a coordinate system and treat vectors algebraically. If we place the initial point of a vector a at the origin of a rectangular coordinate system, then the terminal point of a has coordinates of the form \((a_1, a_2)\) or \((a_1, a_2, a_3)\), depending on whether our coordinate system is two- or three-dimensional (see Figure 11).

These coordinates are called the components of a and we write \[ \mathbf{a} = \langle a_1, a_2 \rangle \quad \text{or} \quad \mathbf{a} = \langle a_1, a_2, a_3 \rangle \]

We use the notation \(\langle a_1, a_2 \rangle\) for the ordered pair that refers to a vector so as not to confuse it with the ordered pair \((a_1, a_2)\) that refers to a point in the plane.

For instance, the vectors shown in Figure 12 are all equivalent to the vector \(\vec{OP} = \langle 3, 2 \rangle\) whose terminal point is \(P(3, 2)\). What they have in common is that the terminal point is reached from the initial point by a displacement of three units to the right and two upward. We can think of all these geometric vectors as representations of the algebraic vector \(\mathbf{a} = \langle 3, 2 \rangle\). The particular representation \(\vec{OP}\) from the origin to the point \(P(3, 2)\) is called the position vector of the point \(P\).

In three dimensions, the vector \(\mathbf{a} = \vec{OP} = \langle a_1, a_2, a_3 \rangle\) is the position vector of the point \(P(a_1, a_2, a_3)\). (See Figure 13.)

Let’s consider any other representation \(\vec{AB}\) of a, where the initial point is \(A(x_1, y_1, z_1)\) and the terminal point is \(B(x_2, y_2, z_2)\). Then we must have \(x_1 + a_1 = x_2, y_1 + a_2 = y_2,\) and \(z_1 + a_3 = z_2\) and so \(a_1 = x_2 - x_1, a_2 = y_2 - y_1,\) and \(a_3 = z_2 - z_1\). Thus we have the following result.

Given the points \(A(x_1, y_1, z_1)\) and \(B(x_2, y_2, z_2)\), the vector a with representation \(\vec{AB}\) is \[ \mathbf{a} = \langle x_2 - x_1, y_2 - y_1, z_2 - z_1 \rangle \tag{1} \]

EXAMPLE 3 Find the vector represented by the directed line segment with initial point \(A(2, -3, 4)\) and terminal point \(B(-2, 1, 1)\).

The magnitude or length of the vector v is the length of any of its representations and is denoted by the symbol \(|\mathbf{v}|\) or \(||\mathbf{v}||\). By using the distance formula to compute the length of a segment OP, we obtain the following formulas.

The length of the two-dimensional vector \(\mathbf{a} = \langle a_1, a_2 \rangle\) is \[ |\mathbf{a}| = \sqrt{a_1^2 + a_2^2} \] The length of the three-dimensional vector \(\mathbf{a} = \langle a_1, a_2, a_3 \rangle\) is \[ |\mathbf{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2} \]

Figure 14 shows that if \(\mathbf{a} = \langle a_1, a_2 \rangle\) and \(\mathbf{b} = \langle b_1, b_2 \rangle\), then the sum is \(\mathbf{a} + \mathbf{b} = \langle a_1 + b_1, a_2 + b_2 \rangle\), at least for the case where the components are positive. In other words, to add algebraic vectors we add corresponding components. Similarly, to subtract vectors we subtract corresponding components. From the similar triangles in Figure 15 we see that the components of \(c\mathbf{a}\) are \(ca_1\) and \(ca_2\). So to multiply a vector by a scalar we multiply each component by that scalar.

If \(\mathbf{a} = \langle a_1, a_2 \rangle\) and \(\mathbf{b} = \langle b_1, b_2 \rangle\), then \[ \mathbf{a} + \mathbf{b} = \langle a_1 + b_1, a_2 + b_2 \rangle \quad \mathbf{a} - \mathbf{b} = \langle a_1 - b_1, a_2 - b_2 \rangle \] \[ c\mathbf{a} = \langle ca_1, ca_2 \rangle \] Similarly, for three-dimensional vectors, \[ \langle a_1, a_2, a_3 \rangle + \langle b_1, b_2, b_3 \rangle = \langle a_1 + b_1, a_2 + b_2, a_3 + b_3 \rangle \] \[ \langle a_1, a_2, a_3\rangle - \langle b_1, b_2, b_3 \rangle = \langle a_1 - b_1, a_2 - b_2, a_3 - b_3 \rangle \] \[ c\langle a_1, a_2, a_3 \rangle = \langle ca_1, ca_2, ca_3 \rangle \]

EXAMPLE 4 If \(\mathbf{a} = \langle 4, 0, 3 \rangle\) and \(\mathbf{b} = \langle -2, 1, 5 \rangle\), find \(|\mathbf{a}|\) and the vectors \(\mathbf{a} + \mathbf{b}, \mathbf{a} - \mathbf{b}, 3\mathbf{b},\) and \(2\mathbf{a} + 5\mathbf{b}\).

We denote by \(V_2\) the set of all two-dimensional vectors and by \(V_3\) the set of all three-dimensional vectors. More generally, we will later need to consider the set \(V_n\) of all n-dimensional vectors. An n-dimensional vector is an ordered n-tuple: \[ \mathbf{a} = \langle a_1, a_2, ..., a_n \rangle \] where \(a_1, a_2, ..., a_n\) are real numbers that are called the components of a. Addition and scalar multiplication are defined in terms of components just as for the cases \(n = 2\) and \(n = 3\).

Properties of Vectors If a, b, and c are vectors in \(V_n\) and \(c\) and \(d\) are scalars, then 1. \(\mathbf{a} + \mathbf{b} = \mathbf{b} + \mathbf{a}\) 2. \(\mathbf{a} + (\mathbf{b} + \mathbf{c}) = (\mathbf{a} + \mathbf{b}) + \mathbf{c}\) 3. \(\mathbf{a} + \mathbf{0} = \mathbf{a}\) 4. \(\mathbf{a} + (-\mathbf{a}) = \mathbf{0}\) 5. \(c(\mathbf{a} + \mathbf{b}) = c\mathbf{a} + c\mathbf{b}\) 6. \((c + d)\mathbf{a} = c\mathbf{a} + d\mathbf{a}\) 7. \((cd)\mathbf{a} = c(d\mathbf{a})\) 8. \(1\mathbf{a} = \mathbf{a}\) —

These eight properties of vectors can be readily verified either geometrically or algebraically. For instance, Property 1 can be seen from Figure 4 (it’s equivalent to the Parallelogram Law) or as follows for the case \(n = 2\): \[ \mathbf{a} + \mathbf{b} = \langle a_1, a_2 \rangle + \langle b_1, b_2 \rangle = \langle a_1 + b_1, a_2 + b_2 \rangle \] \[ = \langle b_1 + a_1, b_2 + a_2 \rangle = \langle b_1, b_2 \rangle + \langle a_1, a_2 \rangle = \mathbf{b} + \mathbf{a} \] We can see why Property 2 (the associative law) is true by looking at Figure 16 and applying the Triangle Law several times: the vector \(\vec{PQ}\) is obtained either by first constructing \(\mathbf{a} + \mathbf{b}\) and then adding c or by adding a to the vector \(\mathbf{b} + \mathbf{c}\).

Three vectors in \(V_3\) play a special role. Let \[ \mathbf{i} = \langle 1, 0, 0 \rangle \quad \mathbf{j} = \langle 0, 1, 0 \rangle \quad \mathbf{k} = \langle 0, 0, 1 \rangle \]

These vectors i, j, and k are called the standard basis vectors. They have length 1 and point in the directions of the positive x-, y-, and z-axes. Similarly, in two dimensions we define \(\mathbf{i} = \langle 1, 0 \rangle\) and \(\mathbf{j} = \langle 0, 1 \rangle\). (See Figure 17.)

If \(\mathbf{a} = \langle a_1, a_2, a_3 \rangle\), then we can write \[ \mathbf{a} = \langle a_1, a_2, a_3 \rangle = \langle a_1, 0, 0 \rangle + \langle 0, a_2, 0 \rangle + \langle 0, 0, a_3 \rangle \] \[ = a_1\langle 1, 0, 0 \rangle + a_2\langle 0, 1, 0 \rangle + a_3\langle 0, 0, 1 \rangle \] \[ \mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k} \tag{2} \] Thus any vector in \(V_3\) can be expressed in terms of i, j, and k. For instance, \[ \langle 1, -2, 6 \rangle = \mathbf{i} - 2\mathbf{j} + 6\mathbf{k} \] Similarly, in two dimensions, we can write \[ \mathbf{a} = \langle a_1, a_2 \rangle = a_1\mathbf{i} + a_2\mathbf{j} \tag{3} \] See Figure 18 for the geometric interpretation of Equations 3 and 2 and compare with Figure 17.

EXAMPLE 5 If \(\mathbf{a} = \mathbf{i} + 2\mathbf{j} - 3\mathbf{k}\) and \(\mathbf{b} = 4\mathbf{i} + 7\mathbf{k}\), express the vector \(2\mathbf{a} + 3\mathbf{b}\) in terms of i, j, and k.

A unit vector is a vector whose length is 1. For instance, i, j, and k are all unit vectors. In general, if \(\mathbf{a} \ne \mathbf{0}\), then the unit vector that has the same direction as a is \[ \mathbf{u} = \frac{1}{|\mathbf{a}|}\mathbf{a} = \frac{\mathbf{a}}{|\mathbf{a}|} \tag{4} \] In order to verify this, we let \(c = 1/|\mathbf{a}|\). Then \(\mathbf{u} = c\mathbf{a}\) and \(c\) is a positive scalar, so u has the same direction as a. Also \[ |\mathbf{u}| = |c\mathbf{a}| = |c||\mathbf{a}| = \frac{1}{|\mathbf{a}|}|\mathbf{a}| = 1 \]

EXAMPLE 6 Find the unit vector in the direction of the vector \(2\mathbf{i} - \mathbf{j} - 2\mathbf{k}\).

Vectors are useful in many aspects of physics and engineering. In Chapter 13 we will see how they describe the velocity and acceleration of objects moving in space. Here we look at forces.

A force is represented by a vector because it has both a magnitude (measured in pounds or newtons) and a direction. If several forces are acting on an object, the resultant force experienced by the object is the vector sum of these forces.

EXAMPLE 7 A 100-lb weight hangs from two wires as shown in Figure 19. Find the tensions (forces) \(\mathbf{T}_1\) and \(\mathbf{T}_2\) in both wires and the magnitudes of the tensions.

What is the relationship between the point (4, 7) and the vector \(\langle 4, 7 \rangle\)? Illustrate with a sketch.

Write each combination of vectors as a single vector. (a) \(\vec{AB} + \vec{BC}\) (b) \(\vec{CD} + \vec{DB}\) (c) \(\vec{DB} - \vec{AB}\) (d) \(\vec{DC} + \vec{CA} + \vec{AB}\)

Copy the vectors in the figure and use them to draw the following vectors. (a) \(\mathbf{u} + \mathbf{v}\) (b) \(\mathbf{u} + \mathbf{w}\) (c) \(\mathbf{v} + \mathbf{w}\) (d) \(\mathbf{u} - \mathbf{v}\) (e) \(\mathbf{v} + \mathbf{u} + \mathbf{w}\) (f) \(\mathbf{u} - \mathbf{w} - \mathbf{v}\)

Copy the vectors in the figure and use them to draw the following vectors. (a) \(\mathbf{a} + \mathbf{b}\) (b) \(\mathbf{a} - \mathbf{b}\) (c) \(\frac{1}{2}\mathbf{a}\) (d) \(-3\mathbf{b}\) (e) \(\mathbf{a} + 2\mathbf{b}\) (f) \(2\mathbf{b} - \mathbf{a}\)

If the vectors in the figure satisfy \(|\mathbf{u}| = |\mathbf{v}| = 1\) and \(\mathbf{u} + \mathbf{v} + \mathbf{w} = \mathbf{0}\), what is \(|\mathbf{w}|\)?

Find a vector a with representation given by the directed line segment \(\vec{AB}\). Draw \(\vec{AB}\) and the equivalent representation starting at the origin. \(A(-2, 1), B(1, 2)\)

Find a vector a with representation given by the directed line segment \(\vec{AB}\). Draw \(\vec{AB}\) and the equivalent representation starting at the origin. \(A(-5, -1), B(-3, 3)\)

Find a vector a with representation given by the directed line segment \(\vec{AB}\). Draw \(\vec{AB}\) and the equivalent representation starting at the origin. \(A(3, -1), B(2, 3)\)

Find a vector a with representation given by the directed line segment \(\vec{AB}\). Draw \(\vec{AB}\) and the equivalent representation starting at the origin. \(A(3, 2), B(1, 0)\)

Find a vector a with representation given by the directed line segment \(\vec{AB}\). Draw \(\vec{AB}\) and the equivalent representation starting at the origin. \(A(0, 3, 1), B(2, 3, -1)\)

Find a vector a with representation given by the directed line segment \(\vec{AB}\). Draw \(\vec{AB}\) and the equivalent representation starting at the origin. \(A(0, 6, -1), B(3, 4, 4)\)

Find the sum of the given vectors and illustrate geometrically. \(\langle -1, 4 \rangle, \langle 6, -2 \rangle\)

Find the sum of the given vectors and illustrate geometrically. \(\langle 3, -1 \rangle, \langle -1, 5 \rangle\)

Find the sum of the given vectors and illustrate geometrically. \(\langle 3, 0, 1 \rangle, \langle 0, 8, 0 \rangle\)

Find the sum of the given vectors and illustrate geometrically. \(\langle 1, 3, -2 \rangle, \langle 0, 0, 6 \rangle\)

Find \(\mathbf{a} + \mathbf{b}, 4\mathbf{a} + 2\mathbf{b}, |\mathbf{a}|,\) and \(|\mathbf{a} - \mathbf{b}|\). \(\mathbf{a} = \langle -3, 4 \rangle, \mathbf{b} = \langle 9, -1 \rangle\)

Find \(\mathbf{a} + \mathbf{b}, 4\mathbf{a} + 2\mathbf{b}, |\mathbf{a}|,\) and \(|\mathbf{a} - \mathbf{b}|\). \(\mathbf{a} = 5\mathbf{i} + 3\mathbf{j}, \mathbf{b} = -\mathbf{i} - 2\mathbf{j}\)

Find \(\mathbf{a} + \mathbf{b}, 4\mathbf{a} + 2\mathbf{b}, |\mathbf{a}|,\) and \(|\mathbf{a} - \mathbf{b}|\). \(\mathbf{a} = 4\mathbf{i} - 3\mathbf{j} + 2\mathbf{k}, \mathbf{b} = 2\mathbf{i} - 4\mathbf{k}\)

Find \(\mathbf{a} + \mathbf{b}, 4\mathbf{a} + 2\mathbf{b}, |\mathbf{a}|,\) and \(|\mathbf{a} - \mathbf{b}|\). \(\mathbf{a} = \langle 8, 1, -4 \rangle, \mathbf{b} = \langle 5, -2, 1 \rangle\)

Find a unit vector that has the same direction as the given vector. \(\langle 6, -2 \rangle\)

Find a unit vector that has the same direction as the given vector. \(-5\mathbf{i} + 3\mathbf{j} - \mathbf{k}\)

Find a unit vector that has the same direction as the given vector. \(8\mathbf{i} - \mathbf{j} + 4\mathbf{k}\)

Find the vector that has the same direction as \(\langle 6, 2, -3 \rangle\) but has length 4.

What is the angle between the given vector and the positive direction of the x-axis? \(\mathbf{i} + \sqrt{3}\mathbf{j}\)

What is the angle between the given vector and the positive direction of the x-axis? \(8\mathbf{i} + 6\mathbf{j}\)

If v lies in the first quadrant and makes an angle \(\pi/3\) with the positive x-axis and \(|\mathbf{v}| = 4\), find v in component form.

If a child pulls a sled through the snow on a level path with a force of 50 N exerted at an angle of \(38^\circ\) above the horizontal, find the horizontal and vertical components of the force.

A quarterback throws a football with angle of elevation \(40^\circ\) and speed 60 ft/s. Find the horizontal and vertical components of the velocity vector.

A crane suspends a 500-lb steel beam horizontally by support cables (with negligible weight) attached from a hook to each end of the beam. The support cables each make an angle of \(60^\circ\) with the beam. Find the tension vector in each support cable and the magnitude of each tension.

A block-and-tackle pulley hoist is suspended in a warehouse by ropes of lengths 2 m and 3 m. The hoist weighs 350 N. The ropes, fastened at different heights, make angles of \(50^\circ\) and \(38^\circ\) with the horizontal. Find the tension in each rope and the magnitude of each tension.

Three forces act on an object. Two of the forces are at an angle of \(100^\circ\) to each other and have magnitudes 25 N and 12 N. The third is perpendicular to the plane of these two forces and has magnitude 4 N. Calculate the magnitude of the force that would exactly counterbalance these three forces.

Find the unit vectors that are parallel to the tangent line to the parabola \(y = x^2\) at the point (2, 4).

1. Find the unit vectors that are parallel to the tangent line to the curve \(y = 2 \sin x\) at the point \((\pi/6, 1)\).
2. Find the unit vectors that are perpendicular to the tangent line.
3. Sketch the curve \(y = 2 \sin x\) and the vectors in parts (a) and (b), all starting at \((\pi/6, 1)\).

If A, B, and C are the vertices of a triangle, find \(\vec{AB} + \vec{BC} + \vec{CA}\).

Let C be the point on the line segment AB that is twice as far from B as it is from A. If \(\mathbf{a} = \vec{OA}, \mathbf{b} = \vec{OB},\) and \(\mathbf{c} = \vec{OC}\), show that \(\mathbf{c} = \frac{2}{3}\mathbf{a} + \frac{1}{3}\mathbf{b}\).

1. Draw the vectors \(\mathbf{a} = \langle 3, 2 \rangle, \mathbf{b} = \langle 2, -1 \rangle,\) and \(\mathbf{c} = \langle 7, 1 \rangle\).
2. Show, by means of a sketch, that there are scalars s and t such that \(\mathbf{c} = s\mathbf{a} + t\mathbf{b}\).
3. Use the sketch to estimate the values of s and t.
4. Find the exact values of s and t.

Suppose that a and b are nonzero vectors that are not parallel and c is any vector in the plane determined by a and b. Give a geometric argument to show that c can be written as \(\mathbf{c} = s\mathbf{a} + t\mathbf{b}\) for suitable scalars s and t. Then give an argument using components.

If \(\mathbf{r} = \langle x, y, z \rangle\) and \(\mathbf{r}_0 = \langle x_0, y_0, z_0 \rangle\), describe the set of all points \((x, y, z)\) such that \(|\mathbf{r} - \mathbf{r}_0| = 1\).

If \(\mathbf{r} = \langle x, y \rangle, \mathbf{r}_1 = \langle x_1, y_1 \rangle,\) and \(\mathbf{r}_2 = \langle x_2, y_2 \rangle\), describe the set of all points \((x, y)\) such that \(|\mathbf{r} - \mathbf{r}_1| + |\mathbf{r} - \mathbf{r}_2| = k\), where \(k > |\mathbf{r}_1 - \mathbf{r}_2|\).

Suppose the three coordinate planes are all mirrored and a light ray given by the vector \(\mathbf{a} = \langle a_1, a_2, a_3 \rangle\) first strikes the xz-plane, as shown in the figure. Use the fact that the angle of incidence equals the angle of reflection to show that the direction of the reflected ray is given by \(\mathbf{b} = \langle a_1, -a_2, a_3 \rangle\). Deduce that, after being reflected by all three mutually perpendicular mirrors, the resulting ray is parallel to the initial ray.