Vectors
V ectors and Vector Quantities
A quantity with both magnitude and direction is called a vector quantity. Examples include force, velocity, acceleration, and linear displacement. A vector quantity can be represented by a directed line segment, called a (geometric) vector. The length of the line segment represents the magnitude of the vector; the direction is indicated by the relative positions of the initial point and terminal point of the line segment. (See Fig. 27-1.)
Vectors are indicated by boldface letters. In the figure, P is the initial point of the vector v and Q is the terminal point. Vector v would also be referred to as vector $ $ .
Scalars and Scalar Quantities
A quantity with only magnitude is called a scalar quantity. Examples include mass, length, time, and temperature. The numbers used to measure scalar quantities are called scalars.
Equivalent Vectors
Two vectors are called equivalent if they have the same magnitude and the same direction.
Normally, equivalence is indicated with the equality symbol. In Fig. 27-2, v = w but $ u v $ . Since there are an infinite number of line segments with a given magnitude and direction, there are an infinite number of vectors equivalent to a given vector (sometimes called copies of the vector).
Zero Vector
A zero vector is defined as a vector with zero magnitude and denoted 0. The initial and terminal points of a zero vector coincide; hence a zero vector may be thought of as a single point.
Addition of Vectors
The sum of two vectors is defined in two equivalent ways, the triangle method and the parallelogram method.
TRIANGLE METHOD: Given v and w, $ v + w $ is the vector formed as follows: place a copy of w with initial point coincident with the terminal point of v. Then $ v + w $ has the initial point of v and the terminal point of w.
PARALLELOGRAM METHOD: Given v and w, $ v + w $ is the vector formed as follows: place copies of v and w with the same initial point. Complete the parallelogram (assuming v and w are not parallel line segments). Then $ v + w $ is the diagonal of the parallelogram with this initial point.
Multiplication of a Vector by a Scalar
Given a vector v and a scalar c, the product cv is defined as follows: If c is positive, cv is a vector with the same direction as v and c times the magnitude. If c = 0, then cv = 0v = 0. If c is negative, cv is a vector with the opposite direction from v and lcl times the magnitude.
EXAMPLE 27.1 Given v as shown, draw $ 2v,v $ , and -2v.
The vector 2v has the same direction as v and twice the magnitude. The vector $ v $ has the same direction as v and one-half the magnitude. The vector -2v has the opposite direction from v and twice the magnitude (see Fig. 27-4).
V ector Subtraction
If v is a nonzero vector, -v is the vector with the same magnitude as v and the opposite direction. Then v - w is defined as $ + (-) $ .
EXAMPLE 27.2 Illustrate the relations among v, w, -w, v - w, and $ v + (-w) $ . See Fig. 27-5.
-w has the same magnitude as w and the opposite direction. $ v + (-w) $ is obtained from the triangle method of addition. From the parallelogram method of addition, note that
\[ \mathbf{v}+(-\mathbf{w})+\mathbf{w}=\mathbf{v}\qquad\mathbf{v}-\mathbf{w}+\mathbf{w}=\mathbf{v} \]
Thus, v - w is the vector that must be added to w to obtain v.
Algebraic Vectors
If a vector v is placed in a Cartesian coordinate system such that $ v = $ , where $ P_{1} $ has coordinates $ (x_{1}, y_{1}) $ and $ P_{2} $ has coordinates $ (x_{2}, y_{2}) $ , then the horizontal displacement from $ P_{1} $ to $ P_{2} $ , $ x_{2} - x_{1} $ , is called the horizontal component of v, and the vertical displacement $ y_{2} - y_{1} $ is called the vertical component of v. (See Fig. 27-6.)
Given horizontal and vertical components a and b, then v is completely determined by a and b and is written as the algebraic vector $ = a, b $ . Then $ v = $ , where O is the origin and P has coordinates $ (a, b) $ . There is a one-to-one correspondence between algebraic and geometric vectors; any geometric vector corresponding to $ a, b $ is called a geometric representative of $ a, b $ .
Operations with Algebraic Vectors
Let $ v=v_{1},v_{2}$ and $ w=w_{1},w_{2}$ . Then
\[ \mathbf{v}+\mathbf{w}=\langle\nu_{1}+w_{1},\nu_{2}+w_{2}\rangle \]
\[ \mathbf{-w}=\left\langle-w_{1},-w_{2}\right\rangle \]
\[ \mathbf{v}-\mathbf{w}=\left\langle\boldsymbol{v}_{1}-\boldsymbol{w}_{1},\boldsymbol{v}_{2}-\boldsymbol{w}_{2}\right\rangle \]
\[ c\pmb{\nu}=\left\langle c\nu_{1},c\nu_{2}\right\rangle \]
EXAMPLE 27.3 Given $ a = , -8 $ and $ b = , 2 $ , find $ a + b $ .
\[ \mathbf{a}+\mathbf{b}=\left\langle3,-8\right\rangle+\left\langle5,2\right\rangle=\left\langle3+5,-8+2\right\rangle=\left\langle8,-6\right\rangle \]
Magnitude of an Algebraic Vector
The magnitude of $ v = v_{1}, v_{2} $ is given by
\[ |\mathbf{v}|=\sqrt{v_{1}^{2}+v_{2}^{2}} \]
V ector Algebra
Given vectors u, v, and w, then
\[ \mathbf{v}+\mathbf{w}=\mathbf{w}+\mathbf{v} \]
\[ \mathbf{u}+(\mathbf{v}+\mathbf{w})=(\mathbf{u}+\mathbf{v})+\mathbf{w} \]
\[ \mathbf{v}+\mathbf{0}=\mathbf{v} \]
\[ \mathbf{v}+(-\mathbf{v})=\mathbf{0} \]
\[ c(\mathbf{v}+\mathbf{w})=c\mathbf{v}+c\mathbf{w} \]
\[ (c+d)\mathbf{v}=c\mathbf{v}+d\mathbf{v} \]
\[ (cd)\mathbf{v}=c(d\mathbf{v})=d(c\mathbf{v}) \]
\[ \mathbf{l}\mathbf{v}=\mathbf{v} \]
\[ 0\mathbf{v}=\mathbf{0} \]
V ector Multiplication
Given two vectors \(\mathbf{v} = \langle v_{1}, v_{2} \rangle\) and \(\mathbf{w} = \langle w_{1}, w_{2} \rangle\), the dot product of \(\mathbf{v}\) and \(\mathbf{w}\) is defined as \(\mathbf{v} \cdot \mathbf{w} = v_{1} w_{1} + v_{2} w_{2}\). Note that this is a scalar quantity.
EXAMPLE 27.4 Given $ a = , -8 $ and $ b = , 2 $ , find $ a b $ .
\[ \mathbf{a}\cdot\mathbf{b}=3\cdot5+(-8)2=-1 \]
Angle Between Two Vectors
If two nonzero vectors $ = v_{1}, v_{2} $ and $ = w_{1}, w_{2} $ have geometric representatives $ $ and $ $ , then the angle between v and w is defined as angle VOW (Fig. 27-7).
Theorem on the Dot Product
If $ $ is the angle between two nonzero vectors v and w, then $ v w = |v||w|$ .
Properties of the Dot Product
Given vectors u, v, and w, and a real number, then
\[ \mathbf{u}\cdot\mathbf{v}=\mathbf{v}\cdot\mathbf{u}\mathrm{(c o m m u t a t i v e p r o p e r t y)} \]
\[ (a\mathbf{v})\cdot\mathbf{w}=a(\mathbf{v}\cdot\mathbf{w}) \]
\[ \mathbf{v}\cdot\mathbf{v}\geq0\text{and}\mathbf{v}\cdot\mathbf{v}=0\text{if and only if}\mathbf{v}=\mathbf{0} \]
(associative property)
(distributive property)
(nonzero property)
SOLVED PROBLEMS
27.1. Given vectors v and w as shown, sketch $ v + w $ , 2v, and $ 2v - w $ .
To find $ v + w $ , place a copy of w with initial point coincident with the terminal point of v. Then $ v + w $ has the initial point of v and the terminal point of w.
To find 2v, sketch a vector with the same direction as v and twice the magnitude.
To find $ 2v - w $ , sketch $ -w $ , a vector with the opposite direction to w and half the magnitude, with initial point coincident with the terminal point of 2v. Then $ 2v - w $ has the initial point of 2v and the terminal point of $ -w $ .
27.2. Given $ v = , 3 $ and $ w = , -4 $ , find $ v + w $ , 4v, and $ 2v - w $ .
\[ \mathbf{v}+\mathbf{w}=\left<-5,3\right\rangle+\left<0,-4\right\rangle=\left<-5+0,3+(-4)\right\rangle=\left<-5,-1\right\rangle \]
\[ 4\mathbf{v}=4\langle-5,3\rangle=\langle4(-5),4\cdot3\rangle=\langle-20,12\rangle \]
\[ 2\mathbf{v}-\frac{1}{2}\mathbf{w}=2\langle-5,3\rangle-\frac{1}{2}\langle0,-4\rangle=\langle-10,6\rangle-\langle0,-2\rangle=\langle-10,8\rangle \]
27.3. Given vector $ =v_{1},v_{2}$ , (a) show that the magnitude of v is given by $ ||= $ ; (b) find the angle $ $ formed by vector $ =v_{1},v_{2}$ and the horizontal.
- Draw a copy of v with initial point at the origin; then the terminal point of v is $ P(v_{1}, v_{2}) $ .
From the distance formula,
\[ |\mathbf{v}|=d(O,P)=\sqrt{(v_{1}-0)^{2}+(v_{2}-0)^{2}}=\sqrt{v_{1}^{2}+v_{2}^{2}} \]
- From the definitions of the trigonometric functions as ratios (Chapter 22), since P is a point on the terminal side of angle $ $ ,
\[ \tan\theta=\frac{v_{2}}{v_{1}} \]
Thus, if $ 0 < /2 $ , then $ = ^{-1} $ . Otherwise, $ $ is an angle with this value as reference angle. (If $ v_{1} = 0 $ , then if $ v_{2} > 0 $ , $ $ may be taken as $ /2 + 2n $ ; and if $ v_{2} < 0 $ , $ $ may be taken as $ -/2 + 2n $ .)
27.4 Find $ |v| $ and the angle $ $ formed by vector $ v = v_{1}, v_{2} $ and the horizontal, given
$ v = , 5 ; $
$ v = , -6 . $
$ |v| = = $ . $ = $ . Since if the initial point of v is at the origin, the terminal point (8, 5) is in quadrant I, $ $ may be taken as $ ^{-1} $ .
$ || = = = 6 $ . $ = = 1 $ . Since if the initial point of v is at the origin, the terminal point $ (-6, -6) $ is in quadrant III, $ $ may be taken as any solution of $ = 1 $ in this quadrant, for example, $ 5/4 $ .
27.5. Resolve a vector v into horizontal and vertical components.
See Fig. 27-10. Vector $ =v_{1},v_{2}$ . $ v_{1} $ and $ v_{2} $ are referred to, respectively, as the horizontal and vertical components of v. Since the coordinates of P are $ (v_{1},v_{2}) $ ,
\[ \frac{v_{1}}{|\mathbf{v}|}=\cos\theta\mathrm{a n d}\frac{v_{2}}{|\mathbf{v}|}=\sin\theta, \]
hence $ v_{1} = |v|$ and $ v_{2} = |v|$ are the horizontal and vertical components of v.
27.6. Show that for any two algebraic vectors v and w, v + w = w + v (vector addition is commutative).
Let $ v = v_{1}, v_{2} $ and $ w = w_{1}, w_{2} $ . Then
\[ \mathbf{v}+\mathbf{w}=\langle v_{1},v_{2}\rangle+\langle w_{1},w_{2}\rangle=\langle v_{1}+w_{1},v_{2}+w_{2}\rangle \]
and
\[ \mathbf{w}+\mathbf{v}=\langle w_{1},w_{2}\rangle+\langle v_{1},v_{2}\rangle=\langle w_{1}+\nu_{1},w_{2}+\nu_{2}\rangle. \]
By the commutative law of addition for real numbers, $ v_{1} + w_{1}, v_{2} + w_{2} = w_{1} + v_{1}, w_{2} + v_{2} $ . Hence
\[ \mathbf{v}+\mathbf{w}=\mathbf{w}+\mathbf{v} \]
27.7. Prove that if \(\theta\) is the angle between two nonzero vectors \(\mathbf{v}\) and \(\mathbf{w}\), then \(\mathbf{v} \cdot \mathbf{w} = \left| \mathbf{v} \right| \left| \mathbf{w} \right| \cos \theta\)
First consider the special case when v and w have the same direction. Then $ = 0 $ and w = kv, where k is positive. Hence
\[ \mathbf{v}\cdot\mathbf{w}=\mathbf{v}\cdot k\mathbf{v}=\langle\nu_{1},\nu_{2}\rangle\cdot\langle k\nu_{1},k\nu_{2}\rangle=k\nu_{1}^{2}+k\nu_{2}^{2}, \]
and
\[ \left|\mathbf{v}\right|\mathbf{w}\left|\cos\theta\right.=\sqrt{v_{1}^{2}+v_{2}^{2}}\sqrt{k^{2}v_{1}^{2}+k^{2}v_{2}^{2}}\cos0=k v_{1}^{2}+k v_{2}^{2} \]
Thus \(\mathbf{v}\cdot\mathbf{w}=|\mathbf{v}||\mathbf{w}|\cos\theta\) in this case.
A second special case occurs when v and w have opposite directions. This case is left to the student. Otherwise, take geometric vectors v and w, each with initial point at the origin, and consider the triangle formed by v, w, and v - w. The terminal point of v is $ V(v_{1}, v_{2}) $ and the terminal point of w is $ W(w_{1}, w_{2}) $ . $ - = = v_{1} - w_{1}, v_{2} - w_{2} $ . Then, by the law of cosines applied to triangle VOW,
\[ |\mathbf{v}-\mathbf{w}|^{2}=|\mathbf{v}|^{2}+|\mathbf{w}|^{2}-2|\mathbf{v}||\mathbf{w}|\cos\theta \]
Or, writing in terms of components,
\[ (v_{1}-w_{1})^{2}+(v_{2}-w_{2})^{2}=v_{1}^{2}+v_{2}^{2}+w_{1}^{2}+w_{2}^{2}-2|\mathbf{v}|\mathbf{w}|\cos\theta \]
Simplifying the left side and subtracting and dividing both sides by the same quantity yield, in turn,
\[ \begin{aligned}v_{1}^{2}-2v_{1}w_{1}+w_{1}^{2}+v_{2}^{2}-2v_{2}w_{2}+w_{2}^{2}&=v_{1}^{2}+v_{2}^{2}+w_{1}^{2}+w_{2}^{2}-2|\mathbf{v}|\mathbf{w}|\cos\theta\\-2v_{1}w_{1}-2v_{2}w_{2}&=-2\left|\mathbf{v}\right|\left|\mathbf{w}\right|\cos\theta\\v_{1}w_{1}+v_{2}w_{2}&=\left|\mathbf{v}\right|\left|\mathbf{w}\right|\cos\theta\end{aligned} \]
Since the left side, by definition, is $ v w $ , the proof is complete.
27.8. Find the angle θ between the vectors (5, 6) and (7, −8)
The formula in the previous problem is often written
\[ \cos\theta=\frac{\mathbf{v}\cdot\mathbf{w}}{\left|\mathbf{v}\right|\left|\mathbf{w}\right|} \]
In this case the formula is applied to obtain
\[ \cos\theta=\frac{\left\langle5,6\right\rangle\cdot\left\langle7,-8\right\rangle}{\left|\left\langle5,6\right\rangle\right|\left|\left\langle7,-8\right\rangle\right|}=\frac{5\cdot7+6(-8)}{\sqrt{5^{2}+6^{2}}\sqrt{7^{2}+(-8)^{2}}}=\frac{-13}{\sqrt{61}\sqrt{113}} \]
Thus $ = ^{-1} $ , or, expressed in degrees, $ ^{} $ .
27.9. Prove the commutative property of the dot product.
Let \(\mathbf{u}=\langle u_{1},u_{2}\rangle\) and \(\mathbf{v}=\langle v_{1},v_{2}\rangle\). Then \(\mathbf{u}\cdot\mathbf{v}=u_{1}v_{1}+u_{2}v_{2}\) and \(\mathbf{v}\cdot\mathbf{u}=v_{1}u_{1}+v_{2}u_{2}\). By the commutative law of multiplication for real numbers, \(u_{1}v_{1}=v_{1}u_{1}\) and \(u_{2}v_{2}=v_{2}u_{2}\). Hence
\[ \mathbf{u}\cdot\mathbf{v}=u_{1}v_{1}+u_{2}v_{2}=v_{1}u_{1}+v_{2}u_{2}=\mathbf{v}\cdot\mathbf{u}. \]
27.10. The vector sum of forces is generally called the resultant of the forces. Find the resultant of two forces, a force $ F_{1} $ of 55.0 pounds and a force $ F_{2} $ of 35.0 pounds acting at an angle of $ 120^{} $ to $ F_{1} $ .
Denote the resultant force by R. Sketch a figure (see Fig. 27-12).
Since $ AOB $ is given as $ 120^{} $ , angle $ $ must measure $ 180^{} - 120^{} = 60^{} $ . From the law of cosines applied to triangle OBC,
\[ \begin{aligned}|\mathbf{R}|^{2}&=\left|\mathbf{F_{1}}\right|^{2}+\left|\mathbf{F_{2}}\right|^{2}-2\left|\mathbf{F_{1}}\right|\left|\mathbf{F_{2}}\right|\cos\theta\\&=55^{2}+35^{2}-2\cdot55\cdot35\cos60^{\circ}\\&=2325\end{aligned} \]
Thus $ |R| = = 48.2 $ pounds. This determines the magnitude of the resultant force; since R is a vector, the direction of R must also be determined. From the law of sines applied to triangle OBC,
\[ \frac{\sin AOC}{\left|\mathbf{F}_{2}\right|}=\frac{\sin\theta}{\left|\mathbf{R}\right|} \]
Hence
\[ \angle AOC=\sin^{-1}\frac{\left|\mathbf{F}_{2}\right|\sin\theta}{\left|\mathbf{R}\right|}=\sin^{-1}\frac{35\sin60^{\circ}}{48.2}=38.9^{\circ} \]
SUPPLEMENTARY PROBLEMS
27.11. Let v be a vector with initial point (3,8) and terminal point (1,1). Let w be a vector with initial point (3,-4) and terminal point (0,0). (a) Express v and w in terms of components. (b) Find $ v + w $ , v - w, 3v - 2w, and $ v w $ . (c) Find $ |v| $ , $ |w| $ , and the angle between v and w.
Ans. (a) $ =,-7,=,4; $
$ v + w = , -3 $ , $ v - w = , -11 $ , $ 3v - 2w = , -29 $ , $ v w = -22 $ ;
$ |v| = , |w| = 5 $ , angle = $ ^{-1} $ , or, expressed in degrees $ ^{} $ .
27.12. (a) Show that any vector v can be written as $ |v|,|v|$ .
- Show that a vector parallel to a line with slope m can be written as $ a , m $ for some value of a.
27.13. A unit vector is defined as a vector with magnitude 1. The unit vectors in the positive x and y directions are, respectively, referred to as i and j.
Show that any unit vector can be written as $ ,$ .
Show that any vector $ v = v_{1}, v_{2} $ can be written as $ v_{1}i + v_{2}j $ .
27.14 Two vectors that form an angle of $ /2 $ are called orthogonal.
Show that the dot product of two nonzero vectors is 0 if, and only if, the vectors are orthogonal.
Show that $ ,-6$ and $ ,15$ are orthogonal.
Find a unit vector orthogonal to $ ,-5$ with horizontal component positive.
Ans. (c) $ /,2/$
27.15. (a) Prove the associative property of the dot product;
prove the distributive property of the dot product;
prove the nonzero property of the dot product.
27.16. For v any vector, prove (a) $ v v = |v|^{2} $ ; (b) $ 0 v = 0 $ .
27.17. A force of 46.3 pounds is applied at an angle of $ 34.8^{} $ to the horizontal. Resolve the force into horizontal and vertical components.
Ans. Horizontal: 38.0 pounds; vertical: 26.4 pounds
27.18. A weight of 75 pounds is resting on a surface inclined at an angle of $ 25^{} $ to the ground. Find the components of the weight parallel and perpendicular to the surface.
Ans. 32 pounds parallel to the surface, 68 pounds perpendicular to the surface
27.19. Find the resultant of two forces, one with magnitude 155 pounds and direction N50°W, and a second with magnitude 305 pounds and direction S55°W.
Ans. 376 pounds in the direction S78°W