Section 12.3: Dot Product

So far we have added two vectors and multiplied a vector by a scalar. The question arises: is it possible to multiply two vectors so that their product is a useful quantity? One such product is the dot product, whose definition follows. Another is the cross product, which is discussed in the next section.

Definition 1 If \(\mathbf{a} = \langle a_1, a_2, a_3 \rangle\) and \(\mathbf{b} = \langle b_1, b_2, b_3 \rangle\), then the dot product of a and b is the number \(\mathbf{a} \cdot \mathbf{b}\) given by \[ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \]

Thus, to find the dot product of a and b, we multiply corresponding components and add. The result is not a vector. It is a real number, that is, a scalar. For this reason, the dot product is sometimes called the scalar product (or inner product). Although Definition 1 is given for three-dimensional vectors, the dot product of two-dimensional vectors is defined in a similar fashion: \[ \langle a_1, a_2 \rangle \cdot \langle b_1, b_2 \rangle = a_1b_1 + a_2b_2 \]

EXAMPLE 1 \[ \langle 2, 4 \rangle \cdot \langle 3, -1 \rangle = 2(3) + 4(-1) = 2 \] \[ \langle -1, 7, 4 \rangle \cdot \langle 6, 2, -\frac{1}{2} \rangle = (-1)(6) + 7(2) + 4(-\frac{1}{2}) = 6 \] \[ (\mathbf{i} + 2\mathbf{j} - 3\mathbf{k}) \cdot (2\mathbf{j} - \mathbf{k}) = 1(0) + 2(2) + (-3)(-1) = 7 \]

The dot product obeys many of the laws that hold for ordinary products of real numbers. These are stated in the following theorem.

Properties of the Dot Product 2 If a, b, and c are vectors in \(V_3\) and \(c\) is a scalar, then 1. \(\mathbf{a} \cdot \mathbf{a} = |\mathbf{a}|^2\) 2. \(\mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a}\) 3. \(\mathbf{a} \cdot (\mathbf{b} + \mathbf{c}) = \mathbf{a} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{c}\) 4. \((c\mathbf{a}) \cdot \mathbf{b} = c(\mathbf{a} \cdot \mathbf{b}) = \mathbf{a} \cdot (c\mathbf{b})\) 5. \(\mathbf{0} \cdot \mathbf{a} = 0\)

These properties are easily proved using Definition 1. For instance, here are the proofs of Properties 1 and 3: 1. \(\mathbf{a} \cdot \mathbf{a} = a_1^2 + a_2^2 + a_3^2 = |\mathbf{a}|^2\) 3. \(\mathbf{a} \cdot (\mathbf{b} + \mathbf{c}) = \langle a_1, a_2, a_3 \rangle \cdot \langle b_1 + c_1, b_2 + c_2, b_3 + c_3 \rangle\) \(= a_1(b_1 + c_1) + a_2(b_2 + c_2) + a_3(b_3 + c_3)\) \(= a_1b_1 + a_1c_1 + a_2b_2 + a_2c_2 + a_3b_3 + a_3c_3\) \(= (a_1b_1 + a_2b_2 + a_3b_3) + (a_1c_1 + a_2c_2 + a_3c_3)\) \(= \mathbf{a} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{c}\)

The proofs of the remaining properties are left as exercises.

The dot product \(\mathbf{a} \cdot \mathbf{b}\) can be given a geometric interpretation in terms of the angle \(\theta\) between a and b, which is defined to be the angle between the representations of a and b that start at the origin, where \(0 \le \theta \le \pi\). In other words, \(\theta\) is the angle between the line segments \(\vec{OA}\) and \(\vec{OB}\) in Figure 1.

Note that if a and b are parallel vectors, then \(\theta = 0\) or \(\theta = \pi\).

The formula in the following theorem is used by physicists as the definition of the dot product.

Theorem 3 If \(\theta\) is the angle between the vectors a and b, then \[ \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta \]

PROOF If we apply the Law of Cosines to triangle OAB in Figure 1, we get \[ |\vec{AB}|^2 = |\vec{OA}|^2 + |\vec{OB}|^2 - 2|\vec{OA}||\vec{OB}|\cos\theta \tag{4} \] (Observe that the Law of Cosines still applies in the limiting cases when \(\theta = 0\) or \(\pi\), or \(\mathbf{a} = \mathbf{0}\) or \(\mathbf{b} = \mathbf{0}\).) But \(|\vec{OA}| = |\mathbf{a}|, |\vec{OB}| = |\mathbf{b}|\), and \(|\vec{AB}| = |\mathbf{a} - \mathbf{b}|\), so Equation 4 becomes \[ |\mathbf{a} - \mathbf{b}|^2 = |\mathbf{a}|^2 + |\mathbf{b}|^2 - 2|\mathbf{a}||\mathbf{b}|\cos\theta \tag{5} \] Using Properties 1, 2, and 3 of the dot product, we can rewrite the left side of this equation as follows: \[ |\mathbf{a} - \mathbf{b}|^2 = (\mathbf{a} - \mathbf{b}) \cdot (\mathbf{a} - \mathbf{b}) = \mathbf{a} \cdot \mathbf{a} - \mathbf{a} \cdot \mathbf{b} - \mathbf{b} \cdot \mathbf{a} + \mathbf{b} \cdot \mathbf{b} = |\mathbf{a}|^2 - 2\mathbf{a} \cdot \mathbf{b} + |\mathbf{b}|^2 \] Therefore Equation 5 gives \[ |\mathbf{a}|^2 - 2\mathbf{a} \cdot \mathbf{b} + |\mathbf{b}|^2 = |\mathbf{a}|^2 + |\mathbf{b}|^2 - 2|\mathbf{a}||\mathbf{b}|\cos\theta \] Thus \[ -2\mathbf{a} \cdot \mathbf{b} = -2|\mathbf{a}||\mathbf{b}|\cos\theta \] or \[ \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta \]

EXAMPLE 2 If the vectors a and b have lengths 4 and 6, and the angle between them is \(\pi/3\), find \(\mathbf{a} \cdot \mathbf{b}\).

EXAMPLE 3 Find the angle between the vectors \(\mathbf{a} = \langle 2, 2, -1 \rangle\) and \(\mathbf{b} = \langle 5, -3, 2 \rangle\).

Two nonzero vectors a and b are called perpendicular or orthogonal if the angle between them is \(\theta = \pi/2\). Then Theorem 3 gives \[ \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos(\pi/2) = 0 \] and conversely if \(\mathbf{a} \cdot \mathbf{b} = 0\), then \(\cos\theta = 0\), so \(\theta = \pi/2\). The zero vector 0 is considered to be perpendicular to all vectors. Therefore we have the following method for determining whether two vectors are orthogonal.

Theorem 7: Two vectors a and b are orthogonal if and only if

\[\mathbf{a} \cdot \mathbf{b} = 0\].

EXAMPLE 4 Show that \(2\mathbf{i} + 2\mathbf{j} - \mathbf{k}\) is perpendicular to \(5\mathbf{i} - 4\mathbf{j} + 2\mathbf{k}\).

Because \(\cos\theta > 0\) if \(0 \le \theta < \pi/2\) and \(\cos\theta < 0\) if \(\pi/2 < \theta \le \pi\), we see that \(\mathbf{a} \cdot \mathbf{b}\) is positive for \(\theta < \pi/2\) and negative for \(\theta > \pi/2\). We can think of \(\mathbf{a} \cdot \mathbf{b}\) as measuring the extent to which a and b point in the same direction. The dot product \(\mathbf{a} \cdot \mathbf{b}\) is positive if a and b point in the same general direction, 0 if they are perpendicular, and negative if they point in generally opposite directions (see Figure 2). In the extreme case where a and b point in exactly the same direction, we have \(\theta = 0\), so \(\cos\theta = 1\) and \[ \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}| \] If a and b point in exactly opposite directions, then we have \(\theta = \pi\) and so \(\cos\theta = -1\) and \(\mathbf{a} \cdot \mathbf{b} = -|\mathbf{a}||\mathbf{b}|\).

Direction Angles and Direction Cosines

The direction angles of a nonzero vector a are the angles \(\alpha, \beta,\) and \(\gamma\) (in the interval \([0, \pi]\)) that a makes with the positive x-, y-, and z-axes, respectively. (See Figure 3.)

The cosines of these direction angles, \(\cos\alpha, \cos\beta,\) and \(\cos\gamma\), are called the direction cosines of the vector a. Using Corollary 6 with b replaced by i, we obtain \[ \cos\alpha = \frac{\mathbf{a} \cdot \mathbf{i}}{|\mathbf{a}||\mathbf{i}|} = \frac{a_1}{|\mathbf{a}|} \tag{8} \] (This can also be seen directly from Figure 3.) Similarly, we also have \[ \cos\beta = \frac{a_2}{|\mathbf{a}|} \quad \cos\gamma = \frac{a_3}{|\mathbf{a}|} \tag{9} \] By squaring the expressions in Equations 8 and 9 and adding, we see that \[ \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1 \tag{10} \] We can also use Equations 8 and 9 to write \[ \mathbf{a} = \langle a_1, a_2, a_3 \rangle = \langle |\mathbf{a}|\cos\alpha, |\mathbf{a}|\cos\beta, |\mathbf{a}|\cos\gamma \rangle = |\mathbf{a}|\langle \cos\alpha, \cos\beta, \cos\gamma \rangle \] Therefore \[ \frac{1}{|\mathbf{a}|}\mathbf{a} = \langle \cos\alpha, \cos\beta, \cos\gamma \rangle \tag{11} \] which says that the direction cosines of a are the components of the unit vector in the direction of a.

EXAMPLE 5 Find the direction angles of the vector \(\mathbf{a} = \langle 1, 2, 3 \rangle\).

Figure 4 shows representations \(\vec{PQ}\) and \(\vec{PR}\) of two vectors a and b with the same initial point \(P\). If \(S\) is the foot of the perpendicular from \(R\) to the line containing \(\vec{PQ}\), then the vector with representation \(\vec{PS}\) is called the vector projection of b onto a and is denoted by proj\(_\mathbf{a}\)b.

The scalar projection of b onto a (also called the component of b along a) is defined to be the signed magnitude of the vector projection, which is the number \(|\mathbf{b}|\cos\theta\), where \(\theta\) is the angle between a and b. (See Figure 5.) This is denoted by comp\(_\mathbf{a}\)b. Observe that it is negative if \(\pi/2 < \theta \le \pi\). The equation \[ \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta = |\mathbf{a}|(|\mathbf{b}|\cos\theta) \] shows that the dot product of a and b can be interpreted as the length of a times the scalar projection of b onto a. Since \[ |\mathbf{b}|\cos\theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}|} = \frac{\mathbf{a}}{|\mathbf{a}|} \cdot \mathbf{b} \] the component of b along a can be computed by taking the dot product of b with the unit vector in the direction of a. We summarize these ideas as follows.

Scalar projection of b onto a: \(\text{comp}_\mathbf{a}\mathbf{b} = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}|}\) Vector projection of b onto a: \(\text{proj}_\mathbf{a}\mathbf{b} = \left(\frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}|}\right)\frac{\mathbf{a}}{|\mathbf{a}|} = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}|^2}\mathbf{a}\)

Notice that the vector projection is the scalar projection times the unit vector in the direction of a.

EXAMPLE 6 Find the scalar projection and vector projection of \(\mathbf{b} = \langle 1, 1, 2 \rangle\) onto \(\mathbf{a} = \langle -2, 3, 1 \rangle\).

One use of projections occurs in physics in calculating work. We know that the work done by a constant force \(F\) in moving an object through a distance \(d\) as \(W = Fd\), but this applies only when the force is directed along the line of motion of the object. Suppose, however, that the constant force is a vector \(\mathbf{F} = \vec{PR}\) pointing in some other direction, as in Figure 6. If the force moves the object from \(P\) to \(Q\), then the displacement vector is \(\mathbf{D} = \vec{PQ}\). The work done by this force is defined to be the product of the component of the force along D and the distance moved: \[ W = (|\mathbf{F}|\cos\theta)|\mathbf{D}| \] But then, from Theorem 3, we have \[ W = |\mathbf{F}||\mathbf{D}|\cos\theta = \mathbf{F} \cdot \mathbf{D} \tag{12} \] Thus the work done by a constant force F is the dot product \(\mathbf{F} \cdot \mathbf{D}\), where D is the displacement vector.

EXAMPLE 7 A wagon is pulled a distance of 100 m along a horizontal path by a constant force of 70 N. The handle of the wagon is held at an angle of \(35^\circ\) above the horizontal. Find the work done by the force.

Which of the following expressions are meaningful? Which are meaningless? Explain. (a) \((\mathbf{a} \cdot \mathbf{b}) \cdot \mathbf{c}\) (b) \((\mathbf{a} \cdot \mathbf{b})\mathbf{c}\) (c) \(|\mathbf{a}|(\mathbf{b} \cdot \mathbf{c})\) (d) \(\mathbf{a} \cdot (\mathbf{b} + \mathbf{c})\) (e) \(\mathbf{a} \cdot \mathbf{b} + \mathbf{c}\) (f) \(|\mathbf{a}| \cdot (\mathbf{b} + \mathbf{c})\)

Find \(\mathbf{a} \cdot \mathbf{b}\). \(\mathbf{a} = \langle 5, -2 \rangle, \mathbf{b} = \langle 3, 4 \rangle\)

Find \(\mathbf{a} \cdot \mathbf{b}\). \(\mathbf{a} = \langle 1.5, 0.4 \rangle, \mathbf{b} = \langle -4, 6 \rangle\)

Find \(\mathbf{a} \cdot \mathbf{b}\). \(\mathbf{a} = \langle 6, -2, 3 \rangle, \mathbf{b} = \langle 2, 5, -1 \rangle\)

Find \(\mathbf{a} \cdot \mathbf{b}\). \(\mathbf{a} = \langle 4, 1, \frac{1}{4} \rangle, \mathbf{b} = \langle 6, -3, -8 \rangle\)

Find \(\mathbf{a} \cdot \mathbf{b}\). \(\mathbf{a} = \langle p, -p, 2p \rangle, \mathbf{b} = \langle 2q, q, -q \rangle\)

Find \(\mathbf{a} \cdot \mathbf{b}\). \(\mathbf{a} = 2\mathbf{i} + \mathbf{j}, \mathbf{b} = \mathbf{i} - \mathbf{j} + \mathbf{k}\)

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

Find \(\mathbf{a} \cdot \mathbf{b}\). \(|\mathbf{a}| = 7, |\mathbf{b}| = 4\), the angle between a and b is \(30^\circ\)

Find \(\mathbf{a} \cdot \mathbf{b}\). \(|\mathbf{a}| = 80, |\mathbf{b}| = 50\), the angle between a and b is \(3\pi/4\)

If u is a unit vector, find \(\mathbf{u} \cdot \mathbf{v}\) and \(\mathbf{u} \cdot \mathbf{w}\). (A diagram with an equilateral triangle with side length 1 is shown, with vectors u, v, w forming the sides).

If u is a unit vector, find \(\mathbf{u} \cdot \mathbf{v}\) and \(\mathbf{u} \cdot \mathbf{w}\). (A diagram with a square with side length 1 is shown, with vectors u, v, w).

1. Show that \(\mathbf{i} \cdot \mathbf{j} = \mathbf{j} \cdot \mathbf{k} = \mathbf{k} \cdot \mathbf{i} = 0\).
2. Show that \(\mathbf{i} \cdot \mathbf{i} = \mathbf{j} \cdot \mathbf{j} = \mathbf{k} \cdot \mathbf{k} = 1\).

A street vendor sells \(a\) hamburgers, \(b\) hot dogs, and \(c\) soft drinks on a given day. He charges $2 for a hamburger, $1.50 for a hot dog, and $1 for a soft drink. If \(\mathbf{A} = \langle a, b, c \rangle\) and \(\mathbf{P} = \langle 2, 1.5, 1 \rangle\), what is the meaning of the dot product \(\mathbf{A} \cdot \mathbf{P}\)?

Find the angle between the vectors. (First find an exact expression and then approximate to the nearest degree.) \(\mathbf{a} = \langle 4, 3 \rangle, \mathbf{b} = \langle 2, -1 \rangle\)

Find the angle between the vectors. (First find an exact expression and then approximate to the nearest degree.) \(\mathbf{a} = \langle -2, 5 \rangle, \mathbf{b} = \langle 5, 12 \rangle\)

Find the angle between the vectors. (First find an exact expression and then approximate to the nearest degree.) \(\mathbf{a} = \langle 1, -4, 1 \rangle, \mathbf{b} = \langle 0, 2, -2 \rangle\)

Find the angle between the vectors. (First find an exact expression and then approximate to the nearest degree.) \(\mathbf{a} = \langle -1, 3, 4 \rangle, \mathbf{b} = \langle 5, 2, 1 \rangle\)

Find the angle between the vectors. (First find an exact expression and then approximate to the nearest degree.) \(\mathbf{a} = 4\mathbf{i} - 3\mathbf{j} + \mathbf{k}, \mathbf{b} = 2\mathbf{i} - \mathbf{k}\)

Find the angle between the vectors. (First find an exact expression and then approximate to the nearest degree.) \(\mathbf{a} = 8\mathbf{i} - \mathbf{j} + 4\mathbf{k}, \mathbf{b} = 4\mathbf{j} + 2\mathbf{k}\)

Find, correct to the nearest degree, the three angles of the triangle with the given vertices. \(P(2, 0), Q(0, 3), R(3, 4)\)

Find, correct to the nearest degree, the three angles of the triangle with the given vertices. \(A(1, 0, -1), B(3, -2, 0), C(1, 3, 3)\)

Determine whether the given vectors are orthogonal, parallel, or neither. (a) \(\mathbf{a} = \langle 9, 3 \rangle, \mathbf{b} = \langle -2, 6 \rangle\) (b) \(\mathbf{a} = \langle 4, 5, -2 \rangle, \mathbf{b} = \langle 3, -1, 5 \rangle\) (c) \(\mathbf{a} = -8\mathbf{i} + 12\mathbf{j} + 4\mathbf{k}, \mathbf{b} = 6\mathbf{i} - 9\mathbf{j} - 3\mathbf{k}\) (d) \(\mathbf{a} = 3\mathbf{i} - \mathbf{j} + 3\mathbf{k}, \mathbf{b} = 5\mathbf{i} + 9\mathbf{j} - 2\mathbf{k}\)

Determine whether the given vectors are orthogonal, parallel, or neither. (a) \(\mathbf{u} = \langle -5, 4, -2 \rangle, \mathbf{v} = \langle 3, 4, -1 \rangle\) (b) \(\mathbf{u} = 9\mathbf{i} - 6\mathbf{j} + 3\mathbf{k}, \mathbf{v} = -6\mathbf{i} + 4\mathbf{j} - 2\mathbf{k}\) (c) \(\mathbf{u} = \langle c, c, c \rangle, \mathbf{v} = \langle c, 0, -c \rangle\)

Use vectors to decide whether the triangle with vertices \(P(1, -3, -2), Q(2, 0, -4),\) and \(R(6, -2, -5)\) is right-angled.

Find the values of \(x\) such that the angle between the vectors \(\langle 2, 1, -1 \rangle\) and \(\langle 1, x, 0 \rangle\) is \(45^\circ\).

Find a unit vector that is orthogonal to both \(\mathbf{i} + \mathbf{j}\) and \(\mathbf{i} + \mathbf{k}\).

Find two unit vectors that make an angle of \(60^\circ\) with \(\mathbf{v} = \langle 3, 4 \rangle\).

Find the acute angle between the lines. \(2x - y = 3, 3x + y = 7\)

Find the acute angle between the lines. \(x + 2y = 7, 5x - y = 2\)

Find the acute angles between the curves at their points of intersection. (The angle between two curves is the angle between their tangent lines at the point of intersection.) \(y = x^2, y = x^3\)

Find the acute angles between the curves at their points of intersection. (The angle between two curves is the angle between their tangent lines at the point of intersection.) \(y = \sin x, y = \cos x, 0 \le x \le \pi/2\)

Find the direction cosines and direction angles of the vector. (Give the direction angles correct to the nearest degree.) \(\langle 2, 1, 2 \rangle\)

Find the direction cosines and direction angles of the vector. (Give the direction angles correct to the nearest degree.) \(\langle 6, 3, -2 \rangle\)

Find the direction cosines and direction angles of the vector. (Give the direction angles correct to the nearest degree.) \(\mathbf{i} - 2\mathbf{j} - 3\mathbf{k}\)

Find the direction cosines and direction angles of the vector. (Give the direction angles correct to the nearest degree.) \(\langle 1, 1, 1 \rangle\)

Find the direction cosines and direction angles of the vector. (Give the direction angles correct to the nearest degree.) \(\langle c, c, c \rangle\), where \(c > 0\)

If a vector has direction angles \(\alpha = \pi/4\) and \(\beta = \pi/3\), find the third direction angle \(\gamma\).

Find the scalar and vector projections of b onto a. \(\mathbf{a} = \langle -5, 12 \rangle, \mathbf{b} = \langle 4, 6 \rangle\)

Find the scalar and vector projections of b onto a. \(\mathbf{a} = \langle 1, 4 \rangle, \mathbf{b} = \langle 2, 3 \rangle\)

Find the scalar and vector projections of b onto a. \(\mathbf{a} = \langle 4, 7, -4 \rangle, \mathbf{b} = \langle 3, -1, 1 \rangle\)

Find the scalar and vector projections of b onto a. \(\mathbf{a} = \langle -1, 4, 8 \rangle, \mathbf{b} = \langle 12, 1, 2 \rangle\)

Find the scalar and vector projections of b onto a. \(\mathbf{a} = 3\mathbf{i} - 3\mathbf{j} + \mathbf{k}, \mathbf{b} = 2\mathbf{i} + 4\mathbf{j} - \mathbf{k}\)

Find the scalar and vector projections of b onto a. \(\mathbf{a} = \mathbf{i} + 2\mathbf{j} + 3\mathbf{k}, \mathbf{b} = 5\mathbf{i} - \mathbf{k}\)

Show that the vector \(\text{orth}_\mathbf{a}\mathbf{b} = \mathbf{b} - \text{proj}_\mathbf{a}\mathbf{b}\) is orthogonal to a. (It is called an orthogonal projection of b.)

For the vectors in Exercise 40, find \(\text{orth}_\mathbf{a}\mathbf{b}\) and illustrate by drawing the vectors a, b, \(\text{proj}_\mathbf{a}\mathbf{b}\), and \(\text{orth}_\mathbf{a}\mathbf{b}\).

If \(\mathbf{a} = \langle 3, 0, -1 \rangle\), find a vector b such that \(\text{comp}_\mathbf{a}\mathbf{b} = 2\).

Suppose that a and b are nonzero vectors. (a) Under what circumstances is \(\text{comp}_\mathbf{a}\mathbf{b} = \text{comp}_\mathbf{b}\mathbf{a}\)? (b) Under what circumstances is \(\text{proj}_\mathbf{a}\mathbf{b} = \text{proj}_\mathbf{b}\mathbf{a}\)?

Find the work done by a force \(\mathbf{F} = 8\mathbf{i} - 6\mathbf{j} + 9\mathbf{k}\) that moves an object from the point \((0, 10, 8)\) to the point \((6, 12, 20)\) along a straight line. The distance is measured in meters and the force in newtons.

A tow truck drags a stalled car along a road. The chain makes an angle of \(30^\circ\) with the road and the tension in the chain is 1500 N. How much work is done by the truck in pulling the car 1 km?

A sled is pulled along a level path through snow by a rope. A 30-lb force acting at an angle of \(40^\circ\) above the horizontal moves the sled 80 ft. Find the work done by the force.

Use a scalar projection to show that the distance from a point \(P_1(x_1, y_1)\) to the line \(ax + by + c = 0\) is \[ \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}} \] Use this formula to find the distance from the point \((-2, 3)\) to the line \(3x - 4y + 5 = 0\).

If \(\mathbf{r} = \langle x, y, z \rangle, \mathbf{a} = \langle a_1, a_2, a_3 \rangle,\) and \(\mathbf{b} = \langle b_1, b_2, b_3 \rangle\), show that the vector equation \((\mathbf{r} - \mathbf{a}) \cdot (\mathbf{r} - \mathbf{b}) = 0\) represents a sphere, and find its center and radius.

A molecule of methane, CH\(_4\), is structured with the four hydrogen atoms at the vertices of a regular tetrahedron and the carbon atom at the centroid. The bond angle is the angle formed by the H-C-H combination; it is the angle between the lines that join the carbon atom to two of the hydrogen atoms. Show that the bond angle is about \(109.5^\circ\). [Hint: Take the vertices of the tetrahedron to be the points (1, 0, 0), (0, 1, 0), (0, 0, 1), and (1, 1, 1), as shown in the figure. Then the centroid is \((\frac{1}{2}, \frac{1}{2}, \frac{1}{2})\).]

If \(\mathbf{c} = |\mathbf{a}|\mathbf{b} + |\mathbf{b}|\mathbf{a}\), where a, b, and c are all nonzero vectors, show that c bisects the angle between a and b.

Use Theorem 3 to prove the Cauchy-Schwarz Inequality: \(|\mathbf{a} \cdot \mathbf{b}| \le |\mathbf{a}||\mathbf{b}|\)

The Triangle Inequality for vectors is \(|\mathbf{a} + \mathbf{b}| \le |\mathbf{a}| + |\mathbf{b}|\) (a) Give a geometric interpretation of the Triangle Inequality. (b) Use the Cauchy-Schwarz Inequality from Exercise 61 to prove the Triangle Inequality. [Hint: Use the fact that \(|\mathbf{a} + \mathbf{b}|^2 = (\mathbf{a} + \mathbf{b}) \cdot (\mathbf{a} + \mathbf{b})\) and use Property 3 of the dot product.]

The Parallelogram Law states that \(|\mathbf{a} + \mathbf{b}|^2 + |\mathbf{a} - \mathbf{b}|^2 = 2|\mathbf{a}|^2 + 2|\mathbf{b}|^2\) (a) Give a geometric interpretation of the Parallelogram Law. (b) Prove the Parallelogram Law. (See the hint in Exercise 62.)

Show that if \(\mathbf{u} + \mathbf{v}\) and \(\mathbf{u} - \mathbf{v}\) are orthogonal, then the vectors u and v must have the same length.

If \(\theta\) is the angle between vectors a and b, show that \(\text{proj}_\mathbf{a}(\text{proj}_\mathbf{a}\mathbf{b}) = (\mathbf{a} \cdot \mathbf{b})\cos^2\theta\)