OBJECTIVES

When you finish this section, you should be able to:

1. Represent a vector algebraically (p. 703)
2. Add, subtract, and find scalar multiples of vectors (p. 705)
3. Find the magnitude of a vector (p. 707)
4. Find a unit vector (p. 708)
5. Find a vector in the plane from its direction and magnitude (p. 709)

IN WORDS

Every vector can be represented by a unique vector whose initial point is at the origin.

Finding a Position Vector

Find the unique position vector \(\mathbf{v}\) of a vector whose initial point is \(P_{1}\) and whose terminal point is \(P_{2}.\)

1. \(P_{1}=(-1,2),\quad P_{2}=(4,6)\)
2. \(P_{1}=(3,4,2),\quad P_{2}=(7,-8,5)\)

Figure 27

Solution To find \(\mathbf{v}\), subtract corresponding components.

1. Then \(\mathbf{v}=\langle 4-(-1), 6-2\rangle =\langle 5,4\rangle \), as shown in Figure 27.
2. Here, \(\mathbf{v}=\langle 7-3,-8-4,5-2\rangle =\langle 4,-12,3\rangle\)

NOW WORK

Problems 9(a) and 13(a).

spanDEFINITIONspan Operations with Vectors

If \(\mathbf{v}=\langle v_{1},v_{2}\rangle \) and \(\mathbf{w}=\langle w_{1},w_{2}\rangle \) are two vectors in the plane and if \(a\) is a scalar, then

• \(\mathbf{v}+\mathbf{w}=\langle v_{1}+w_{1}, v_{2}+w_{2}\rangle\)
• \(\mathbf{v}-\mathbf{w}=\langle v_{1}-w_{1}, v_{2}-w_{2}\rangle\)
• \(a\,\mathbf{v}=\langle av_{1}, av_{2}\rangle\)

If \(\mathbf{v}=\langle v_{1}, v_{2}, v_{3}\rangle \) and \(\mathbf{w}=\langle w_{1}, w_{2}, w_{3}\rangle \) are two vectors in space and if \(a\) is a scalar, then

• \(\mathbf{v}+\mathbf{w}=\langle v_{1}+w_{1},v_{2}+w_{2},v_{3}+w_{3}\rangle\)
• \(\mathbf{v}-\mathbf{w}=\langle v_{1}-w_{1},v_{2}-w_{2},v_{3}-w_{3}\rangle\)
• \(a\,\mathbf{v}=\langle av_{1},av_{2},av_{3}\rangle\)

Adding and Subtracting Vectors Algebraically

If \(\mathbf{v}=\langle 2,3,-1\rangle \) and \(\mathbf{w}=\langle -1,-2, 4\rangle \), find:

1. \(\mathbf{v}+\mathbf{w}\)
2. \(\mathbf{w-v}\)
3. \(\dfrac{1}{2}\mathbf{w}\)
4. \(2\mathbf{v}+3\mathbf{w}\)

Solution

(a) \(\mathbf{v} + \mathbf{w} = \langle 2,3,-1\rangle +\langle -1,-2,4\rangle =\left\langle 2+(-1), 3+(-2), -1+4\right\rangle =\langle 1,1,3\rangle\)

(b) \(\mathbf{w-v=} \langle -1,-2,4\rangle - \langle 2,3,-1\rangle =\left\langle -1-2, -2-3, 4-( -1) \right\rangle =\left\langle -3,-5,5\right\rangle \)

(c) \(\dfrac{1}{2}\mathbf{w}=\dfrac{1}{2}\,\langle -1,-2,4\rangle = \left\langle \dfrac{1}{2}( -1), \dfrac{1}{2} ( -2), \dfrac{1}{2}(4) \right\rangle = \left\langle -\dfrac{1}{2},-1,2\right\rangle \)

(d) \(2\mathbf{v} + 3\mathbf{w} = 2\,\langle 2,3,-1\rangle+3\,\langle -1,-2,4\rangle =\langle 4,6,-2\rangle +\langle -3,-6,12\rangle =\langle 1,0,10\rangle\)

NOW WORK

Problem 27.

THEOREM Properties of Vector Addition and Scalar Multiplication

If \(\mathbf{u}\), \(\mathbf{v}\), and \(\mathbf{w}\) are vectors and if \(a\) and \(b\) are scalars, then the following properties hold:

• Commutative property of vector addition: \[\bbox[5px, border:1px solid black, #F9F7ED]{ \mathbf{u}+\mathbf{v}=\mathbf{v}+\mathbf{u}} \]
• Associative property of vector addition: \[\bbox[5px, border:1px solid black, #F9F7ED]{ \mathbf{u}+(\mathbf{v}+\mathbf{w})=(\mathbf{u}+\mathbf{v})+\mathbf{w} } \]
• Additive identity \(\mathbf{0}\): \[\bbox[5px, border:1px solid black, #F9F7ED]{ \mathbf{v}+\mathbf{0}=\mathbf{0}+\mathbf{v}=\mathbf{v} } \]
• Additive inverse property: \[\bbox[5px, border:1px solid black, #F9F7ED]{ \mathbf{v}+(-\mathbf{v})=\mathbf{0} } \]
• Distributive property of scalar multiplication over vector addition: \[\bbox[5px, border:1px solid black, #F9F7ED]{ a( \mathbf{u}+\mathbf{v}) =a\,\mathbf{u}+a\,\mathbf{v} } \]
• Distributive property of scalar multiplication over scalar addition: \[\bbox[5px, border:1px solid black, #F9F7ED]{ ( a+b) \mathbf{v}=a\,\mathbf{v}+b\,\mathbf{v} } \]
• Linearity property of scalar multiplication: \[\bbox[5px, border:1px solid black, #F9F7ED]{ a( b\,\mathbf{v}) =( ab) \mathbf{v} } \]
• Scalar multiplication identity: \[\bbox[5px, border:1px solid black, #F9F7ED]{ 1\mathbf{v}=\mathbf{v}} \]

Proof

Commutative property of vector addition (in the plane) : If \(\mathbf{u}=\langle u_{1},u_{2}\rangle \) and if \(\mathbf{v} =\langle v_{1},v_{2}\rangle \), then \[ \begin{eqnarray*} \mathbf{u}+\mathbf{v}&=&\langle u_{1},u_{2}\rangle +\langle v_{1},v_{2}\rangle =\langle u_{1}+v_{1},u_{2}+v_{2}\rangle =\langle v_{1}+u_{1},v_{2}+u_{2}\rangle\\[4pt] &=&\langle v_{1},v_{2}\rangle +\langle u_{1},u_{2}\rangle =\mathbf{v}+\mathbf{u} \end{eqnarray*} \]

Scalar multiplication identity (in space): If \(\mathbf{v}=\langle v_{1}, v_{2}, v_{3}\rangle \), then \[ \begin{equation*} 1\mathbf{v}=\langle 1v_{1}, 1v_{2}, 1v_{3}\rangle =\langle v_{1}, v_{2}, v_{3}\rangle =\mathbf{v} \end{equation*} \]

spanDEFINITIONspan Parallel Vectors

Let \(\mathbf{v}\) and \(\mathbf{w}\) be two distinct nonzero vectors. If there is a nonzero scalar \(a\) for which \(\mathbf{v}=a\,\mathbf{w}\), then \( \mathbf{v}\) and \(\mathbf{w}\) are called parallel vectors.

Showing Two Vectors Are Parallel

1. The vectors \[ \begin{equation*} \mathbf{v}=\langle 2,3,-1\rangle\qquad\hbox{and}\qquad\mathbf{w}=\langle 4,6,-2\rangle \end{equation*} \] are parallel, since \(\mathbf{v}=\dfrac{1}{2}\mathbf{w}\). In this case, \(\mathbf{v}\) and \(\mathbf{w}\) have the same direction.
2. The vectors \[ \begin{equation*} \mathbf{v}=\left\langle 1,2,3\right\rangle\qquad \hbox{and}\qquad\mathbf{w}=\langle -3,-6,-9\rangle \end{equation*} \] are parallel, since \(\mathbf{w}=-3\,\mathbf{v}\). In this case, \(\mathbf{v}\) and \(\mathbf{w}\) have opposite directions.

NOW WORK

Problem 75.

spanDEFINITIONspan

The magnitude or length of a vector \(\mathbf{v}\), denoted by the symbol \(\Vert \mathbf{v}\Vert \), is the distance from the initial point to the terminal point of \(\mathbf{v}\). \( \Vert \mathbf{v}\Vert \) is also called the norm of the vector \( \mathbf{v}\).

THEOREM Magnitude of a Vector

If \(\mathbf{v}=\langle v_{1},v_{2}\rangle \) is a vector in the plane, the magnitude of \(\mathbf{v}\) is \[\bbox[5px, border:1px solid black, #F9F7ED]{ \Vert \mathbf{v}\Vert \,=\sqrt{v_{1}^{2}+v_{2}^{2}}} \]

If \(\mathbf{v}=\langle v_{1},v_{2},v_{3}\rangle \) is a vector in space, the magnitude of \(\mathbf{v}\) is \[\bbox[5px, border:1px solid black, #F9F7ED]{ \Vert \mathbf{v}\Vert \,=\sqrt{v_{1}^{2}+v_{2}^{2}+v_{3}^{2}} } \]

Finding the Magnitude of a Vector

Find the magnitude of \(\mathbf{v}\) if:

1. \(\mathbf{v} = \langle 3,-4\rangle\)
2. \(\mathbf{v}=\langle 2,3,-1\rangle\)

Solution (a) \(\Vert \mathbf{v}\Vert = \sqrt{v_{1}^{2}+v_{2}^{2}} = \sqrt{3^{2}+(-4)^{2}} = \sqrt{9+16} = 5\)

(b) \(\Vert \mathbf{v}\Vert = \sqrt{v_{1}^{2} + v_{2}^{2} + v_{3}^{2}} = \sqrt{2^{2} + 3^{2} + (-1)^{2}} = \sqrt{4 + 9 + 1} = \sqrt{14}\)

NOW WORK

Problem 21.

THEOREM Magnitude of the Scalar Multiple of a Vector

If \(a\) is scalar and \(\mathbf{v}\) is a vector, then \[\bbox[5px, border:1px solid black, #F9F7ED]{ \Vert a \mathbf{v}\Vert = |a| \Vert \mathbf{v}\Vert} \]

Proof

Suppose \(\mathbf{v} = \langle v_{1}, v_{2}\rangle\). Then the scalar multiple of \(a\) and \(\mathbf{v}\) is \(a\mathbf{v} = \langle av_{1}, av_{2}\rangle\) and \[ \begin{eqnarray*} \Vert a\mathbf{v}\Vert &=& \sqrt{(av_{1})^{2} + (av_{2})^{2}} = \sqrt{a^{2}v_{1}^{2} + a^{2}v_{2}^{2}} = \sqrt{a^{2}\big(v_{1}^{2} + v_{2}^{2}\big)}\\[4pt] &=&\sqrt{a^{2}}\sqrt{v_{1}^{2} + v_{2}^{2}} = |a| \Vert v\Vert \end{eqnarray*} \]

spanDEFINITIONspan

A vector with magnitude \(1\) is called a unit vector.

THEOREM Unit Vector in the Direction of \(\mathbf{v}\)

For any nonzero vector \(\mathbf{v}\), the vector \[\bbox[5px, border:1px solid black, #F9F7ED]{ \mathbf{u}=\dfrac{\mathbf{v}}{\Vert \mathbf{v\Vert }}} \]

is the unit vector that has the same direction as \(\mathbf{v}\).

Proof

Since \(\dfrac{1}{\Vert \mathbf{v}\Vert}\) is a positive scalar, the vector \(\mathbf{u} = \dfrac{1}{\Vert \mathbf{v}\Vert} \mathbf{v}\) has the same direction as \(\mathbf{v}\). Next we show that the magnitude of \(\mathbf{u}\) is \(1\).

\[ \begin{eqnarray*} \Vert \mathbf{u}\Vert =\left\Vert \dfrac{\mathbf{v}}{\Vert \mathbf{v\Vert }} \right\Vert =\left\Vert \dfrac{1}{\Vert \mathbf{v\Vert }}\mathbf{v} \right\Vert =\dfrac{1}{\Vert \mathbf{v\Vert }}\left\Vert \mathbf{v} \right\Vert =1 \end{eqnarray*} \]

Since \(\Vert \mathbf{u}\Vert =1, \mathbf{u}\) is a unit vector in the direction of \(\mathbf{v}\).

Normalizing a Vector

Normalize each vector. That is, find a unit vector \(\mathbf{u}\) that has the same direction as:

1. \(\mathbf{v} = \langle 3,-4\rangle\)
2. \(\mathbf{v} = \langle -1,2,-2\rangle\)

Solution (a) Since \(\mathbf{v} = \langle 3,-4\rangle\), then \(\Vert \mathbf{v}\Vert = \sqrt{9 + 16} = 5\). The unit vector \(\mathbf{u}\) in the same direction as \(\mathbf{v}\) is \[ \begin{eqnarray*} \mathbf{u} = \dfrac{\mathbf{v}}{\Vert \mathbf{v\Vert }} = \dfrac{\mathbf{v}}{5} = \left\langle \dfrac{3}{5}, -\dfrac{4}{5}\right\rangle \end{eqnarray*} \]

(b) Since \(\mathbf{v} = \langle -1,2,-2\rangle\), then \(\Vert \mathbf{v}\Vert = \sqrt{1 + 4 + 4} = 3\). The unit vector \(\mathbf{u}\) in the same direction as \(\mathbf{v}\) is \[ \begin{eqnarray*} \mathbf{u} = \dfrac{\mathbf{v}}{\Vert \mathbf{v\Vert }} = \dfrac{\mathbf{v}}{3} = \left\langle -\dfrac{1}{3}, \dfrac{2}{3}, -\dfrac{2}{3}\right\rangle \end{eqnarray*} \]

NOW WORK

Problem 43.

THEOREM Writing a Vector in Terms of Standard Basis Vectors

Every vector \(\mathbf{v}\) = (\(v_{1},v_{2}\ \)) in the plane can be written in terms of the standard basis vectors \(\mathbf{i}\) and \(\mathbf{j}\) as \[\bbox[5px, border:1px solid black, #F9F7ED]{ \mathbf{v} = v_{1}\mathbf{i} + v_{2} \mathbf{j}} \]

Every vector \(\mathbf{v} = \langle v_{1}, v_{2}, v_{3}\rangle \) in space can be written in terms of the standard basis vectors \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\) as \[\bbox[5px, border:1px solid black, #F9F7ED]{ \mathbf{v} = v_{1} \mathbf{i} + v_{2} \mathbf{j} + v_{3} \mathbf{k}} \]

Proof

\[\mathbf{v} = \left\langle v_{1}, v_{2}\right\rangle = \left\langle v_{1}, 0\right\rangle + \left\langle 0, v_{2}\right\rangle = v_{1} \left\langle 1, 0\right\rangle + v_{2} \left\langle 0, 1\right\rangle = v_{1} \mathbf{i} + v_{2} \mathbf{j}\]

Writing a Vector Using Standard Basis Vectors

1. \(\langle 4, 3\rangle = 4 \mathbf{i} + 3 \mathbf{j}\)
2. \(\langle 2, -3\rangle = \dot{2}\mathbf{i} + (-3\dot{)}\mathbf{j} = 2 \mathbf{i} - 3 \mathbf{j}\)
3. \(\langle 1, -3, 2\rangle = \mathbf{i}-3 \mathbf{j} + 2 \mathbf{k}\)
4. \(\langle 0, 3, 4\rangle = 0 \mathbf{i} + 3 \mathbf{j} + 4 \mathbf{k} = 3 \mathbf{j} + 4 \mathbf{k}\)

NOW WORK

Problems 9(b) and 13(b).

Working with Standard Basis Vectors

1. If \(\mathbf{v} = 2 \mathbf{i} - 3 \mathbf{j}\) and \(\mathbf{w} = -\mathbf{i} + 2 \mathbf{j}\), find \(4\mathbf{v}-\mathbf{w}.\)
2. If \(\mathbf{v} = 2\mathbf{i} + 3 \mathbf{j} -\mathbf{k}\) and \(\mathbf{w} = -\mathbf{i}-2\mathbf{j}+4\mathbf{k}\), find \(2 \mathbf{v}-3 \mathbf{w}.\)

Solution

(a) \[\begin{align*} 4\mathbf{v} - \mathbf{w} = 4 (2 \mathbf{i} - 3 \mathbf{j}) - (-\mathbf{i} + 2 \mathbf{j}) = (8 \mathbf{i} - 12 \mathbf{j}) + (\mathbf{i} - 2 \mathbf{j}) = 9 \mathbf{i} - 14 \mathbf{j} \end{align*}\]

(b) \[\begin{align*} 2 \mathbf{v} - 3 \mathbf{w} &= 2(2 \mathbf{i} + 3 \mathbf{j} - \mathbf{k}) - 3(-\mathbf{i} - 2 \mathbf{j} + 4 \mathbf{k}) \\ & =(4 \mathbf{i} + 6 \mathbf{j} - 2 \mathbf{k}) + (3 \mathbf{i} + 6 \mathbf{j} - 12 \mathbf{k}) = 7 \mathbf{i} + 12 \mathbf{j} - 14 \mathbf{k} \end{align*}\]

NOW WORK

Problems 55 and 65.

THEOREM

If \(\mathbf{v}\) is a nonzero vector, and if \(\theta \) is the angle between \(\mathbf{v}\) and the positive \(x\)-axis, then \[\bbox[5px, border:1px solid black, #F9F7ED]{ \mathbf{v}=\left\Vert \mathbf{v}\right\Vert \left( \cos \theta \,\mathbf{i}+\sin \theta \,\mathbf{j}\right)} \]

NOTE

Wind direction is measured from where it blows.

Using Standard Basis Vectors to Model a Problem

An airplane has an air speed of \(400\) km\(/\)h and is headed east. There is a northwesterly wind of \(80\) km\(/\)h. (Northwesterly winds blow toward the southeast.)

1. Find a vector representing the velocity of the airplane in the air.
2. Find a vector representing the wind velocity.
3. Find a vector representing the velocity of the airplane relative to the ground.
4. Find the true speed of the airplane.

Figure 31

Solution We use a two-dimensional coordinate system with the direction north along the positive \(y\)-axis. Then the direction east is along the positive \(x\)-axis.

Using a scale of \(1~\hbox{unit} = 1\) km\(/\)h, define \[ \begin{eqnarray*} \mathbf{v}_{\mathrm{a}} &=& \hbox{Velocity of the airplane in the air}\\[2pt] \mathbf{v}_{\mathrm{w}} &=& \hbox{Velocity of the wind}\\[2pt] \mathbf{v}_{\mathrm{g}} &=& \hbox{Velocity of the airplane relative to the ground} \end{eqnarray*} \]

Figure 32

Figure 31 shows the vectors \(\mathbf{v}_{a},\) \(\mathbf{v}_{w},\) \(\mathbf{v}_{g}.\)

1. The vector \(\mathbf{v}_{\rm{a}}\) has magnitude \(400\) and direction \(\mathbf{i}\), so \(\mathbf{v}_{\mathrm{a}}\) = 400i.
2. The vector \(\mathbf{v}_{\mathrm{w}}\) has magnitude \(80\) and makes an angle of \(\dfrac{7\pi}{4}\) with the positive \(x\)-axis, as shown in Figure 32. Since the magnitude of \(\mathbf{v}_{\rm{w}}\) is 80, \[ \mathbf{v}_{\rm{w}} = 80\left(\cos \dfrac{7\pi}{4}\mathbf{i} + \sin \dfrac{7\pi}{4}\mathbf{j}\right) = 80 \left(\dfrac{\sqrt{2}}{2}\mathbf{i} - \dfrac{\sqrt{2}}{2}\mathbf{j}\right) = 40\sqrt{2}\mathbf{i} - 40\sqrt{2}\mathbf{j} \]
3. The velocity \(\mathbf{v}_{\mathrm{g}}\) of the airplane relative to the ground is the resultant of \(\mathbf{v}_{\rm{a }}\) and \( \mathbf{v}_{\rm{w}}.\) \[ \mathbf{v}_{\mathrm{g}} = \mathbf{v}_{\mathrm{a}} + \mathbf{v}_{\mathrm{w}} = 400\ \mathbf{i} + \left[ 40\sqrt{2} \mathbf{i} - 40\sqrt{2} \mathbf{j}\right] = (400 + 40\sqrt{2}) \mathbf{i} - 40\sqrt{2} \mathbf{j} \]
4. The true speed of the airplane is the magnitude of the vector \(\mathbf{v}_{\mathrm{g}}.\) \[ \left\Vert \mathbf{v}_{\mathrm{g}}\right\Vert =\sqrt{(400 + 40\sqrt{2})^{2} + (40\sqrt{2})^{2}}\approx 460.060 \]

The true speed of the airplane is approximately \(460.060 \,\rm{km/h}\).

NOW WORK

Problem 93.

Finding the Tension in Cable

A commercial air conditioner weighing 8600 lb is suspended from two cables, as shown in Figure 33. What is the tension in each cable?

Solution We represent the cables and the air conditioner with a force diagram, as shown in Figure 34.

The tension in each cable is the magnitude of \(\Vert\mathbf{F}_{1}\Vert \) and \(\Vert\mathbf{F}_{2}\Vert,\) respectively. The weight of the air conditioner is \(\Vert\mathbf{F}_{3}\Vert = 8600~\text{lb}\). \[ \begin{eqnarray*} \mathbf{F}_{1} &=& \Vert\mathbf{F}_{1}\Vert \left( \cos 45^{\circ}\mathbf{i} + \sin 45^{\circ}\mathbf{j}\right) = \Vert\mathbf{F}_{1}\Vert \left(\dfrac{\sqrt{2}}{2} \mathbf{i} + \dfrac{\sqrt{2}}{2} \mathbf{j}\right) \\[4pt] \mathbf{F}_{2} &=& \Vert \mathbf{F}_{2}\Vert \left(\cos 140^{\circ} \mathbf{i} + \sin 140^{\circ} \mathbf{j}\right) \\[4pt] \mathbf{F}_{3} &=& -8600 \mathbf{j} \end{eqnarray*} \]

Figure 34

For the air conditioner to be in static equilibrium, the sum of the forces acting on it must equal \(\mathbf{0}\). That is, \[ \begin{eqnarray*} \mathbf{F}_{1} + \mathbf{F}_{2} + \mathbf{F}_{3} &=& \left(\dfrac{\sqrt{2}}{2} \Vert\mathbf{F}_{1}\Vert \mathbf{i} + \dfrac{\sqrt{2}}{2} \Vert\mathbf{F}_{1}\Vert \mathbf{j}\right)\\[6pt] &&+ \left(\cos 140^{\circ} \Vert\mathbf{F}_{2}\Vert \mathbf{i} + \sin 140^{\circ} \Vert\mathbf{F}_{2}\Vert \mathbf{j}\right) -8600\mathbf{j} = \mathbf{0} \end{eqnarray*} \]

Since both the \(\mathbf{i}\) and \(\mathbf{j}\) components equal \(0,\) we have a system of two equations: \[ \begin{equation*} \left\{ \begin{array}{rcl@{\hspace*{5pc}}r@{}} \dfrac{\sqrt{2}}{2}\Vert\mathbf{F}_{1}\Vert + \cos 140^{\circ} \Vert\mathbf{F}_{2}\Vert &=& 0 & \hbox{(1)} \\ \dfrac{\sqrt{2}}{2}\Vert \mathbf{F}_{1}\Vert + \sin 140^{\circ} \Vert\mathbf{F}_{2}\Vert -8600 &=& 0 & \hbox{(2)} \end{array} \right. \end{equation*} \]

We solve equation (1) for \(\left\Vert \mathbf{F}_{1}\right\Vert\): \[ \begin{equation*} \Vert\mathbf{F}_{1}\Vert = -\dfrac{2\cos 140^{\circ} \Vert\mathbf{F}_{2}\Vert}{\sqrt{2}} \tag{3} \end{equation*} \]

and substitute the result into equation (2). \[ \begin{eqnarray*} \dfrac{\sqrt{2}}{2}\left(-\dfrac{2\cos 140^{\circ} \Vert\mathbf{F}_{2}\Vert}{\sqrt{2}}\right) + \sin 140^{\circ} \Vert\mathbf{F}_{2}\Vert -8600 &=& 0 \\[4pt] \left[-\cos 140^{\circ} + \sin 140^{\circ} \right] \Vert\mathbf{F}_{2}\Vert &=& 8600 \\[4pt] \Vert \mathbf{F}_{2}\Vert &=& \dfrac{8600}{\sin 140^{\circ} - \cos 140^{\circ}}\approx 6104 \end{eqnarray*} \]

Then from (3) \[ \begin{equation*} \Vert\mathbf{F}_{1}\Vert = -\dfrac{2\cos 140^{\circ} \Vt\mathbf{F}_{2}\Vert}{\sqrt{2}}\approx -\dfrac{2\cos 140^{\circ}}{\sqrt{2}}(6104) \approx 6613 \end{equation*} \]

NOW WORK

Problem 97.