example 1

• (a) If \({\bf a}=3{\bf i}+{\bf j}-2{\bf k}\) and \({\bf b}={\bf i}-{\bf j}+{\bf k}\), calculate \({\bf a}\,{ \cdot}\,{\bf b}\).
• (b) Calculate \((2{\bf i}+{\bf j}-{\bf k})\,{ \cdot}\,(3{\bf k}-2{\bf j})\).

solution

• (a) \({\bf a}\,{ \cdot}\,{\bf b}=3\,{ \cdot}\,1+1\,{ \cdot}\,(-1)+(-2)\,{ \cdot}\, 1=3-1-2=0\).
• (b) \((2{\bf i}+{\bf j}-{\bf k})\,{ \cdot}\,(3{\bf k}-2{\bf j})=(2{\bf i}+{\bf j}-{\bf k})\,{ \cdot}\, (0{\bf i}-2{\bf j}+3{\bf k})=2\,{ \cdot}\, 0-1\,{ \cdot}\, 2-1\,{ \cdot}\, 3=-5\).

example 2

• (a) Normalize \({\bf v}=2{\bf i}+3{\bf j}-\frac{1}{2}{\bf k}\).
• (b) Find unit vectors \({\bf a},{\bf b}\), and \({\bf c}\) in the plane such that \({\bf b}+{\bf c}={\bf a}\).

solution

• (a) We have \(\|{\bf v}\|={\textstyle\sqrt{2^2+3^2+(1/2)^2}}=(1/2)\sqrt{53}\), so the normalization of \({\bf v}\) is \[ {\bf u}=\frac{1}{\|{\bf v}\|}\, {\bf v}=\frac{4}{\sqrt{53}}\, {\bf i}+ \frac{6}{\sqrt{53}}\, {\bf j}-\frac{1}{\sqrt{53}}\, {\bf k}. \]
• (b) Because all three vectors are to have length 1, a triangle with sides \({\bf a},{\bf b}\), and \({\bf c}\) must be equilateral, as in Figure 11.32. If we orient the triangle as in the figure and we take \({\bf a}={\bf i}\), then necessarily \[ {\bf b}=\frac{1}{2}{\bf i}+ \frac{\sqrt{3}}{2}{\bf j},\qquad \hbox{and}\qquad {\bf c}=\frac{1}{2}{\bf i}-\frac{\sqrt{3}}{2} {\bf j}. \] Note that indeed \(\|{\bf a}\|=\|{\bf b}\|=\|{\bf c}\|=1\) and that \({\bf b}+{\bf c}={\bf a}\).

Figure 11.32: The vectors a, b, and c are represented by the sides of an equilateral triangle.

Inner Product, Length, and Distance

Letting \({\bf a}=a_1{\bf i}+a_2{\bf j}+a_3{\bf k}\) and \({\bf b}=b_1{\bf i}+b_2{\bf j}+b_3{\bf k}\), their inner product is \[ {\bf a}\,{ \cdot}\,{\bf b} = a_1b_1+a_2b_2+a_3b_3, \] while the length of a is \[ \|{\bf a}\| = \sqrt{{\bf a}\, { \cdot}\,{\bf a}} = \sqrt{a_1^2+a_2^2+a_3^2}. \]

To normalize a vector \({\bf a}\), form the vector \[ \frac{{\bf a}}{\|{\bf a}\|}. \]

The distance between the endpoints of \({\bf a}\) and \({\bf b}\) is \(\|{\bf a}-{\bf b}\|\), and the distance between P and Q is \(\|\overrightarrow{\rm PQ\vphantom{^\prime}}\|\).

example 3

Find the distance from the endpoint of the vector \({\bf i}\), that is, the point \((1, 0, 0)\), to the endpoint of the vector \({\bf j}\), that is, the point \((0,1,0).\)

solution

\(\|{\bf j}-{\bf i}\|=\sqrt{(0-1)^2+(1-0)^2+(0-0)^2}=\sqrt{2}.\)

Theorem 1

Let \({\bf a}\) and \({\bf b}\) be two vectors in \({\mathbb R}^3\) and let \(\theta\), where \(0\le \theta\le \pi ,\) be the angle between them (Figure 11.35). Then \[ {\bf a}\,{ \cdot}\,{\bf b}=\|{\bf a}\| \|{\bf b}\| \cos \theta. \]

proof

If we apply the law of cosines from trigonometry to the triangle with one vertex at the origin and adjacent sides determined by the vectors \({\bf a}\) and \({\bf b}\) (as in Figure 11.35 below ), it follows that \[ \|{\bf b-a}\|^2=\|{\bf a}\|^2+\|{\bf b}\|^2-2\|{\bf a}\|\|{\bf b}\|\cos \theta. \]

Figure 11.35: The vectors a, b, and the angle \(\theta \) between them; the geometry for Theorem 1 and its proof.

Because \(\|{\bf b}-{\bf a}\|^2=({\bf b}-{\bf a})\,{ \cdot}\, ({\bf b}-{\bf a}),\|{\bf a}\|^2= {\bf a}\,{ \cdot}\,{\bf a}\), and \(\|{\bf b}\|^2={\bf b}\,{ \cdot}\,{\bf b}\), we can rewrite the preceding equation as \[ ({\bf b}-{\bf a})\,{ \cdot}\, ({\bf b}-{\bf a})={\bf a}\,{ \cdot}\,{\bf a}+ {\bf b}\,{ \cdot}\,{\bf b}-2\|{\bf a}\| \|{\bf b}\| \cos \theta. \]

We can also expand \(({\bf b}-{\bf a})\, { \cdot}\, ({\bf b}-{\bf a})\) as follows: \begin{eqnarray*} ({\bf b}-{\bf a}) \,{ \cdot}\,( {\bf b}-{\bf a}) &=&{\bf b}\,{ \cdot}\, ({\bf b}-{\bf a})-{\bf a}\,{ \cdot}\, ({\bf b}-{\bf a})\\ &=&{\bf b}\,{ \cdot}\,{\bf b}-{\bf b}\,{ \cdot}\,{\bf a}- {\bf a}\,{ \cdot}\,{\bf b}+{\bf a}\,{ \cdot}\,{\bf a}\\ &=&{\bf a}\,{ \cdot}\,{\bf a}+{\bf b}\,{ \cdot}\,{\bf b}-2 {\bf a}\,{ \cdot}\,{\bf b}. \end{eqnarray*} Thus, \[ {\bf a}\,{ \cdot}\,{\bf a}+{\bf b}\,{ \cdot}\,{\bf b}-2{\bf a}\,{ \cdot}\,{\bf b} ={\bf a}\,{ \cdot}\,{\bf a}+{\bf b}\,{ \cdot}\,{\bf b} -2\|{\bf a}\| \|{\bf b}\|\cos \theta. \] That is, \[ {\bf a} \,{ \cdot}\,{\bf b}=\|{\bf a}\| \|{\bf b}\|\cos \theta. \]

example 4

Find the angle between the vectors \({\bf i}+{\bf j}+{\bf k}\) and \({\bf i}+{\bf j}-{\bf k}\) (see Figure 11.36).

Figure 11.36: Finding the angle between \({\bf a} = {\bf i} + {\bf j} + {\bf k} \hbox { and } {\bf b} = {\bf i} + {\bf j} - {\bf k}\).

solution Using Theorem \(1,\) we have \[ ({\bf i}+{\bf j}+{\bf k}) \,{ \cdot}\, ( {\bf i}+{\bf j}-{\bf k})= \|{\bf i}+{\bf j}+{\bf k}\| \|{\bf i}+{\bf j}-{\bf k}\|\cos \theta, \] and so \[ 1+1-1=(\sqrt{3})(\sqrt{3})\cos \theta. \] Hence, \[ \cos \theta={\textstyle \frac{1}{3}}. \] That is, \[ \theta=\cos^{-1} ({\textstyle \frac{1}{3}})\approx 1.23 \hbox{ radians } (71^\circ). \]

Corollary Cauchy–Schwarz Inequality

For any two vectors a and b\({,}\) we have \[ |{\bf a}\,{ \cdot}\,{\bf b}|\le \|{\bf a}\| \|{\bf b}\| \] with equality if and only if \({\bf a}\) is a scalar multiple of \({\bf b},\) or one of them is \({\bf 0}\).

proof

If a is not a scalar multiple of \({\bf b}\), then \(\theta\), the angle between them, is not zero or \(\pi\), and so \(|{\cos \theta} | < 1\), and thus the inequality holds; in fact, if a and b are both nonzero, strict inequality holds in this case. When \({\bf a}\) is a scalar multiple of \({\bf b}\), then \(\theta=0\) or \(\pi\) and \(|{\cos \theta}|=1\), so equality holds in this case.

example 5

Verify the Cauchy–Schwarz inequality for \({\bf a}=-{\bf i}+{\bf j}+{\bf k}\) and \({\bf b}=\) \(3{\bf i}+{\bf k}\).

solution The dot product is \({\bf a}\,{ \cdot}\,{\bf b}=-3+0+1=-2\), so \(|{\bf a}\,{ \cdot}\,{\bf b}|=2\). Also, \(\|{\bf a}\|=\sqrt{1+1+1}=\sqrt{3}\) and \(\|{\bf b}\|=\sqrt{9+1}=\sqrt{10}\), and it is true that \(2\leq \sqrt{3}\,{ \cdot}\,\sqrt{10}\) because \(\sqrt{3}\,{ \cdot}\,\sqrt{10}> \sqrt{3}\,{ \cdot}\,\sqrt{3}=3\geq 2\).

example 6

The vectors \({\bf i}_\theta=(\cos \theta){\bf i}+(\sin \theta){\bf j}\) and \({\bf j}_\theta=-(\sin \theta){\bf i}+(\cos \theta){\bf j}\) are orthogonal, because \[ {\bf i}_{\theta}\,{ \cdot}\, {\bf j}_{\theta}=-\kern-1pt\cos \theta \sin \theta+\sin\theta\cos\theta=0. \]

Here, \({\bf i}_{\theta}\) is the rotation of \({\bf i}\), \({\theta}^{\circ}\) counterclockwise. Also, \({\bf j}_{\theta}\) is the rotation of \({\bf j}\), \({\theta}^{\circ}\) counterclockwise (see Figure 11.37).

Figure 11.37: The vectors \({\bf i}_{\theta} \) and \({\bf j}_{\theta }\) are orthogonal and of unit length, that is, they are orthonormal.

example 7

Let \({\bf a}\) and \({\bf b}\) be two nonzero orthogonal vectors. If \({\bf c}\) is a vector in the plane spanned by \({\bf a}\) and \({\bf b}\), then there are scalars \(\alpha\) and \(\beta\) such that \({\bf c}=\alpha {\bf a}+\beta {\bf b}\). Use the inner product to determine \(\alpha\) and \(\beta\) (see Figure 11.38).

Figure 11.38: The geometry for finding \(\alpha \) and \(\beta \), where c \(=\) \(\alpha\) a \(+\) \(\beta\) b.

solution Taking the inner product of \({\bf a}\) and \({\bf c}\), we have \[ {\bf a}\,{ \cdot}\,{\bf c}= {\bf a} \,{ \cdot}\,(\alpha {\bf a}+\beta {\bf b}) =\alpha {\bf a}\,{ \cdot}\,{\bf a}+\beta {\bf a}\,{ \cdot}\,{\bf b}. \] Because \({\bf a}\) and \({\bf b}\) are orthogonal, \({\bf a}\,{ \cdot}\,{\bf b}=0\), and so \[ \alpha=\frac{{\bf a}\,{ \cdot}\,{\bf c}}{{\bf a}\,{ \cdot}\,{\bf a}}= \frac{{\bf a}\,{ \cdot}\,{\bf c}} {\|{\bf a}\|^2}. \] Similarly, \[ \beta=\frac{{\bf b}\,{ \cdot}\,{\bf c}}{{\bf b}\,{ \cdot}\,{\bf b}}= \frac{{\bf b}\,{ \cdot}\,{\bf c}} {\|{\bf b}\|^2}. \]

Orthogonal Projection

The orthogonal projection of \({\bf v}\) on \({\bf a}\) is the vector \[ {\bf p}=\frac{{\bf a}\,{ \cdot}\,{\bf v}}{\|{\bf a}\|^2}\,{\bf a}. \]

example 8

Find the orthogonal projection of \({\bf i}+{\bf j}\) on \({\bf i}-2{\bf j}\).

solution With \({\bf a}={\bf i}-2{\bf j}\) and \({\bf v}={\bf i}+{\bf j}\), the orthogonal projection of \({\bf v}\) on \({\bf a}\) is \[ \frac{{\bf a}\,{ \cdot}\,{\bf v}}{{\bf a}\,{ \cdot}\,{\bf a}}\,{\bf a}=\frac{1-2}{1+4}\,({\bf i}-2{\bf j})=-\frac{1}{5}\,({\bf i}-2{\bf j}) \] (see Figure 11.40).

Figure 11.40: The orthogonal projection of v on a equals \(-\frac{1}{5}\)a.

Theorem 2 Triangle Inequality

For vectors \({\bf a}\) and \({\bf b}\) in space, \[ \|{\bf a}+{\bf b}\|\leq \|{\bf a}\|+\|{\bf b}\|. \]

proof

While this result may be clear geometrically, it is useful to give a proof using the Cauchy–Schwarz inequality, as it will generalize to n-dimensional vectors. We consider the square of the left-hand side: \[ \|{\bf a} +{\bf b}\|^2=({\bf a}+{\bf b})\,{ \cdot}\,({\bf a}+{\bf b})= \|{\bf a}\|^2+2{\bf a}\,{ \cdot}\,{\bf b}+\|{\bf b}\|^2. \]

Figure 11.41: This figure illustrates that \(\| OQ\| \le \| OR\| +\| RQ\| \) or, in vector notation, that \(\| {\bf a} + {\bf b}\| \le \| {\bf a}\| + \| {\bf b}\|\), which is the triangle inequality.

By the Cauchy–Schwarz inequality, we have \[ \|{\bf a}\|^2+2{\bf a}\,{ \cdot}\,{\bf b}+\|{\bf b}\|^2 \leq \|{\bf a}\|^2+2\|{\bf a}\|\|{\bf b}\|+\|{\bf b}\|^2 = (\|{\bf a}\|+\|{\bf b}\|)^2. \] Thus, \[ \|{\bf a}+{\bf b}\|^2\leq (\|{\bf a}\|+\|{\bf b}\|)^2; \] taking square roots proves the result.

example 9

• (a) Verify the triangle inequality for \({\bf a}={\bf i}+{\bf j}\) and \({\bf b}=2{\bf i}+{\bf j}+{\bf k}\).
• (b) Prove that \(\|{\bf u}-{\bf v}\|\leq \|{\bf u}-{\bf w}\|+\|{\bf w}-{\bf v}\|\) for any vectors \({\bf u},{\bf v}\), and \({\bf w}\). Illustrate with a figure in which \({\bf u},{\bf v}\), and \({\bf w}\) have the same base point.

solution

• (a) We have \({\bf a}+{\bf b}=3{\bf i}+2{\bf j}+{\bf k}\), so \(\|{\bf a}+{\bf b}\|= \sqrt{9+4+1}=\sqrt{14}\). On the other hand, \(\|{\bf a}\|=\sqrt{2}\) and \(\|{\bf b}\|=\sqrt{6}\), so the triangle inequality asserts that \(\sqrt{14}\leq\sqrt{2}+\sqrt{6}\). The numbers bear this out: \(\sqrt{14}\approx 3.74\), while \(\sqrt{2}+\sqrt{6}\approx 1.41+ 2.45=3.86\).
• (b) We find that \({\bf u}-{\bf v}=({\bf u}-{\bf w})+({\bf w}-{\bf v})\), so the result follows from the triangle inequality with \({\bf a}\) replaced by \({\bf u}-{\bf w}\) and \({\bf b}\) replaced by \({\bf w}-{\bf v}\). Geometrically, we are considering the shaded triangle in Figure 11.42.

Figure 11.42: Illustrating the inequality \(\|\) u \(-\) v \(\| \le \|\) u \(-\) w \(\| +\|\) w \(-\) v \(\|\).

example 10

Suppose that two navigators who cannot see one another but can communicate by radio wish to determine the relative position of their ships. Explain how they can do this if they can each determine their displacement vector to the same lighthouse.

solution Let P\(_1\) and P\(_2\) be the positions of the ships, and let Q be the position of the lighthouse. The displacement of the lighthouse from the \(i\)th ship is the vector \({\bf d}_i\) joining P\(_i\) to Q. The displacement of the second ship from the first is the vector \({\bf d}\) joining P\(_1\) to P\(_2\). We have \({\bf d}+{\bf d}_2={\bf d}_1\) (Figure 11.43), and so \({\bf d}={\bf d}_1-{\bf d}_2\). That is, the displacement from one ship to the other is the difference between the displacements from the ships to the lighthouse.

Figure 11.43: Vector methods can be used to locate objects.

Displacement and Velocity

If an object has a (constant) velocity vector \({\bf v}\), then in \(t\) units of time the resulting displacement vector of the object is \({\bf d}=t{\bf v}\); thus, after time \(t = 1\), the displacement vector equals the velocity vector. See Figure 11.46.

example 11

A bird is flying in a straight line with velocity vector \(10{\bf i}+6{\bf j}+{\bf k}\) (in kilometers per hour). Suppose that \((x,y)\) are its coordinates on the ground and \(z\) is its height above the ground.

• (a) If the bird is at position \((1, 2, 3)\) at a certain moment, what is its location 1 hour later? 1 minute later?
• (b) How many seconds does it take the bird to climb \(10\) meters\(?\)

solution

• (a) The displacement vector from (1, 2, 3) after 1 hour is given by \(10{\bf i}+6{\bf j}+{\bf k}\), so the new position is \((1, 2, 3) + (10, 6, 1)=(11, 8, 4)\). After 1 minute, the displacement vector from \((1,2,3)\) is \(\frac{1}{60}(10{\bf i}+ 6{\bf j} + {\bf k})=\frac{1}{6}{\bf i}+ \frac{1}{10}{\bf j}+\frac{1}{60}{\bf k}\), and so the new position is \((1,2,3)+(\frac{1}{6},\frac{1}{10},\frac{1}{60}) =(\frac{7}{6},\frac{21}{10},\frac{181}{60})\).
• (b) After \(t\) seconds \(({=t}/3600\) hours), the displacement vector from (1, 2, 3) is \((t/3600)\) \((10{\bf i}+6{\bf j}+{\bf k})=(t/360){\bf i}+(t/600){\bf j} +(t/3600) {\bf k}\). The increase in altitude is the \(z\) component, namely, \(t/3600\). This will equal 10 m \(({=\frac{1}{100}}\hbox{ km})\) when \(t/3600=\frac{1}{100}\), that is, when \(t=36\hbox{ s}\).

example 12

Physical forces have magnitude and direction and may thus be represented by vectors. If several forces act at once on an object, the resultant force is represented by the sum of the individual force vectors. Suppose that forces \({\bf i}+{\bf k}\) and \({\bf j}+{\bf k}\) are acting on a body. What third force F must we impose to counteract the two–that is, to make the total force equal to zero?

solution The force \({\bf F}\) should be chosen so that \(({\bf i}+{\bf k})+ (\,{\bf j}+{\bf k})+{\bf F}={\bf 0}\); that is, \({\bf F}=-({\bf i}+{\bf k})-(\kern1pt{\bf j}+{\bf k})=-{\bf i}-{\bf j}-2{\bf k}\). (Recall that 0 is the zero vector, the vector whose components are all zero.)

EXAMPLE 13

What is the minimum force you must apply to pull a 20-kg wagon up a frictionless ramp inclined at an angle \(\theta = 15^\circ\)?

Figure 11.47: The angle between \(F_{\mathrm{g}}\) and \(F_{||}\) is \(90^\circ-\theta\).

Solution Let \(F_{\textrm{g}}\) be the force on the wagon due to gravity. It has magnitude \(20g\) newtons with \(g = 9.8\). Referring to Figure 11.47, we decompose \(F_{\textrm{g}}\) as a sum \[ F_{\textrm{g}} = F_{||} + F_{\perp} \]

where \(F_{||}\) is the projection along the ramp and \(F_\perp\) is the “normal force” orthogonal to the ramp. The normal force \(F_\perp\) is canceled by the ramp pushing back against the wagon in the normal direction, and thus (because there is no friction), you need only pull against \(F_{||}\).

Notice that the angle between \(F_{\mathrm{g}}\) and the ramp is the complementary angle \(90^\circ-\theta\). Since \(F_{||}\) is parallel to the ramp, the angle between \(F_{\textrm{g}}\) and \(F_{||}\) is also \(90^\circ-\theta\), or \(75^\circ\), and \[ \parallel{F_{||}}\parallel = \parallel{F_{\textrm{g}}}\parallel\cos(75^\circ) \approx 20(9.8) (0.26) \approx 51 \textrm{N} \]

Since gravity pulls the wagon down the ramp with a \(51\)-newton force, it takes a minimum force of 51 newtons to pull the wagon up the ramp.