For the analytical method of vector addition and subtraction, we use some simple geometry and trigonometry, instead of using a ruler and protractor as we did for graphical methods. The analytical method is more accurate than the graphical method, which is limited by the precision of the drawing. For a refresher on the definitions of the sine, cosine, and tangent of an angle, see Figure 4.6(a)

Figure-4.6(a)

Figure 4.6(a)shows, For a right triangle, the sine, cosine, and tangent of $\theta$ are defined in terms of the adjacent side, the opposite side, or the hypotenuse. In this figure, \(x\) is the adjacent side, \(y\) is the opposite side, and \(h\) is the hypotenuse.

Since, by definition, \(\cos \theta=x / h\), we can find the length \(x\) if we know hand \(\theta\) by using \(X=\) hcos \(\theta\). Similarly, we can find the length of \(y\) by using \(y=h \sin \theta\). These trigonometric relationships are useful for adding vectors.

When a vector acts in more than one dimension, it is useful to break it down into its \(\mathrm{x}\) and \(\mathrm{y}\) components. For a two-dimensional vector, a component is a piece of a vector that points in either the \(x\) – or \(y\)-direction. Every 2 -d vector can be expressed as a sum of its \(x\) and \(y\) components.

For example, given a vector like \(A\) in Figure \(4.6(b)\), we may want to find what two perpendicular vectors, \(A_{x}\) and \(A_{y}\), add to produce it. In this example, \(A_{x}\) and \(A_{y}\) form a right triangle, meaning that the angle between them is 90 degrees.

\(A_{x}\) and \(A_{y}\) are defined to be the components of \(A_{\text {along }}\) the \(x\) – and \(y\)-axes. The three vectors, \(A_{1} A_{x}\), and \(A_{y}\), form a right triangle.
\(
\mathbf{A}_{\mathbf{x}}+\mathbf{A}_{\mathbf{y}}=\mathbf{A}
\) components, we use the following relationships for a right triangle:
\(
A_{x}=A \cos \theta
\)
and
\(
A_{y}=A \sin \theta
\)
where \(A_{X}\) is the magnitude of \(\mathbf{A}\) in the \(x\)-direction, \(A_{y}\) is the magnitude of \(\mathbf{A}\) in the \(y\)-direction, and \(\theta\) is the angle of the resultant with respect to the \(x\)-axis, as shown in Figure 4.6(b).

Figure-4.6(b)

Figure-4.6(b): The magnitudes of the vector components \(\mathbf{A}_{x}\) and \(\mathbf{A}_{y}\) can be related to the resultant vector \(\mathbf{A}\) and the angle \(\theta\) with trigonometric identities. Here we see that \(A_{x}=A \cos \theta\) and \(A_{y}=A \sin \theta\).

Analytical Method of Vector Addition and Subtraction

Calculating a resultant vector (or vector addition) is the reverse of breaking the resultant down into its components. If the perpendicular components \(A_{x}\) and \(A_{y}\) of a vector \(A\) are known, then we can find \(A\) analytically. How do we do this? Since, by definition,
\(
\tan \theta=y / x\left(\text { or in this case } \tan \theta=A_{y} / A_{x}\right)
\)
we solve for \(\theta\) to find the direction of the resultant.
\(
\theta=\tan ^{-1}\left(A_{y} / A_{x}\right)
\)
Since this is a right triangle, the Pythagorean theorem \(\left(x^{2}+y^{2}=h^{2}\right)\) for finding the hypotenuse applies. In this case, it becomes
\(
A^{2}=A_{x}^{2}+A_{y}^{2}
\)
Solving for A gives
\(
A=\sqrt{A_{x}^{2}+A_{y}{ }^{2}}
\)
In summary, to find the magnitude \(A_{\text {and direction }} \theta\) of a vector from its perpendicular components \(A_{x}\) and \(A_{y}\), as illustrated in Figure \(4.6(c)\), we use the following relationships:
\(
\begin{aligned}
A=& \sqrt{A_{x}{ }^{2}+A_{y}{ }^{2}} \\
& \theta=\tan ^{-1}\left(A_{y} / A_{x}\right)
\end{aligned}
\)

Figure-4.6(c)

Figure-4.6(c) shows, the magnitude and direction of the resultant vector \(\mathbf{A}\) can be determined once the horizontal components \(\mathbf{A}_{x}\) and \(\mathbf{A}_{y}\) have been determined.

Sometimes, the vectors added are not perfectly perpendicular to one another. An example of this is the case below, where the vectors \(A\) and \(B\) are added to produce the resultant \(R\), as illustrated in figure below.

In the above Figure 4.6(d), Vectors \(\mathbf{A}\) and \(\mathbf{B}\) are two legs of a walk, and \(\mathbf{R}\) is the resultant or total displacement. You can use analytical methods to determine the magnitude and direction of \(\mathbf{R}\).

If \(A\) and \(B\) represent two legs of a walk (two displacements), then \(R\) is the total displacement. The person taking the walk ends up at the tip of \(R\). There are many ways to arrive at the same point. The person could have walked straight ahead first in the \(x\)-direction and then in the \(y\)-direction. Those paths are the \(x\)-and \(y\)-components of the resultant, \(R_{x}\) and \(R_{y}\). If we know \(R_{x}\) and \(R_{y}\), we can find \(R^{2}\) and \(\theta\) using the equations \(R=\sqrt{R_{x}^{2}+R_{y}^{2}}\) and \(\theta=\tan ^{-1}\left(R_{y} / R_{x}\right)\)

Step-1: Draw in the \(x\) and \(y\) components of each vector (including the resultant) with a dashed line. Use the equations \(A_{x}=A \cos \theta\) and \(A_{y}=A \sin \theta\) to find the components. In the Figure shown below, these components are \(A_{x}, A_{y}, B_{x}\), and \(B_{y}\). Vector \(A\) makes an angle of \(\theta_{A}\) with the \(x\)-axis, and vector \(B\) makes and angle of \(\theta_{B}\) with its own \(x\)-axis (which is slightly above the \(x\)-axis used by vector \(A\) ).

In Figure \(4.6(e)\) to add vectors \(A\) and \(B\), first determine the horizontal and vertical components of each vector. These are the dotted vectors \(A_{x}, A_{y},  B_{x}, B_{y}\) shown in the image.

Step-2: Find the \(x\) component of the resultant by adding the \(x\) component of the vectors
\(
\mathrm{R}_{\mathrm{x}}=\mathrm{A}_{\mathrm{x}}+\mathrm{B}_{\mathrm{x}}
\)
and find the \(y\) component of the resultant (as illustrated in the figure shown below by adding the \(y\) component of the vectors.
\(
R_{y}=A_{y}+B_{y}
\)

In the above Figure, the vectors \(A_{x}\) and \(B_{x}\) add to give the magnitude of the resultant vector in the horizontal direction, \(R_{x}\). Similarly, the vectors \(A_{y}\) and \(B_{y}\) add to give the magnitude of the resultant vector in the vertical direction, \(R_{y}\).
Now that we know the components of \(\mathrm{R}\), we can find its magnitude and direction.
Step-3: To get the magnitude of the resultant R, use the Pythagorean theorem.
\(
R=\sqrt{R_{x}^{2}+R_{y}^{2}}
\)
Step-4: To get the direction of the resultant
\(
\theta=\tan ^{-1}\left(R_{y} / R_{x}\right)
\)

Vectors in the 3-dimensional plane

So far we have considered a vector lying in an \(x\) – \(y\) plane. The same procedure can be used to resolve a general vector \(\mathbf{A}\) into three components along \(x_{-}, y\)-, and \(z\)-axes in three dimensions. If \(\alpha, \beta\), and \(\gamma\) are the angles between \(A\) and the \(x\)-, \(y\)-, and \(z\)-axes, respectively [Fig. \(4.9(\mathrm{~d})\) ], we have

\(
\mathrm{A}_{\mathrm{x}}=\mathrm{A} \cos \alpha, \mathrm{A}_{\mathrm{y}}=\mathrm{A} \cos \beta, \mathrm{A}_{\mathrm{z}}=\mathrm{A} \cos \gamma
\)
In general, we have
\(
\mathbf{A}=A_x \hat{\mathbf{i}}+A_y \hat{\mathbf{j}}+A_z \hat{\mathbf{k}}
\)
The magnitude of vector \(A\) is
\(
A=\sqrt{A_x^2+A_y^2+A_z^2}
\)
A position vector \(\mathbf{r}\) can be expressed as
\(
\mathbf{r}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}
\)
where \(x, y\), and \(z\) are the components of \(r\) along \(x-, y^{-}\), z-axes, respectively.

Example 1: Find the magnitude and direction of the resultant of two vectors A and B in terms of their magnitudes and angle \(\theta\) between them.

Answer: Let \(\mathbf{O P}\) and \(\mathbf{O Q}\) represent the two vectors A and B making an angle \(\theta\) (Fig. 4.10). Then, using the parallelogram method of vector addition, OS represents the resultant vector \(\mathbf{R}\) :
\(
\mathbf{R}=\mathbf{A}+\mathbf{B}
\)
\(S N\) is normal to \(O P\) and \(P M\) is normal to \(O S\). From the geometry of the figure,
\(
\begin{aligned}
& O S^2=O N^2+S N^2 \\
& \text { but } O N=O P+P N=A+B \cos \theta \\
& \mathrm{S} N=B \sin \theta \\
& O S^2=(A+B \cos \theta)^2+(B \sin \theta)^2 \\
& \text { or, } R^2=A^2+B^2+2 A B \cos \theta \\
& R=\sqrt{A^2+B^2+2 A B \cos \theta} \dots(4.24a)\\
&
\end{aligned}
\)
In \(\Delta \mathrm{OSN}, \quad \mathrm{SN}=\mathrm{OS} \sin \alpha=R \sin \alpha\), and in \(\Delta \mathrm{PSN}, \quad \mathrm{SN}=\mathrm{PS} \sin \theta=B \sin \theta\)
Therefore, \(R \sin \alpha=B \sin \theta\)
or, \(\frac{R}{\sin \theta}=\frac{B}{\sin \alpha} \dots(4.24b) \)
Similarly,
\(
\mathrm{PM}=A \sin \alpha=B \sin \beta
\)
or, \(\frac{A}{\sin \beta}=\frac{B}{\sin \alpha} \dots(4.24c) \)
Combining Eqs. (4.24b) and (4.24c), we get
\(
\frac{R}{\sin \theta}=\frac{A}{\sin \beta}=\frac{\mathrm{B}}{\sin \alpha} \dots(4.24d)
\)
Using Eq. (4.24d), we get:
\(
\sin \alpha=\frac{B}{R} \sin \theta \dots(4.24e)
\)
where \(R\) is given by Eq. (4.24a).
or, \(\tan \alpha=\frac{S N}{O P+P N}=\frac{B \sin \theta}{A+B \cos \theta} \dots(4.24f) \)
Equation (4.24a) gives the magnitude of the resultant and Eqs. \((4.24 \mathrm{e})\) and (4.24f) its direction. Equation (4.24a) is known as the Law of cosines and Eq. (4.24d) as the Law of sines.

Example 2: A motorboat is racing towards north at \(25 \mathrm{~km} / \mathrm{h}\) and the water current in that region is \(10 \mathrm{~km} / \mathrm{h}\) in the direction of \(60^{\circ}\) east of south. Find the resultant velocity of the boat.

Answer: The vector \(\mathbf{v}_{\mathrm{b}}\) representing the velocity of the motorboat and the vector \(\mathbf{v}_{\mathrm{c}}\) representing the water current are shown in Fig. \(4.11 \mathrm{in}\) directions specified by the problem. Using the parallelogram method of addition, the resultant \(\mathbf{R}\) is obtained in the direction shown in the figure.

We can obtain the magnitude of \(\mathbf{R}\) using the Law of cosine :
\(
\begin{aligned}
& R=\sqrt{v_{\mathrm{b}}^2+v_{\mathrm{c}}^2+2 v_{\mathrm{b}} v_{\mathrm{c}} \cos 120^{\circ}} \\
= & \sqrt{25^2+10^2+2 \times 25 \times 10(-1 / 2)} \cong 22 \mathrm{~km} / \mathrm{h}
\end{aligned}
\)
To obtain the direction, we apply the Law of sines
\(
\begin{aligned}
& \frac{R}{\sin \theta}=\frac{v_c}{\sin \phi} \text { or, } \sin \phi=\frac{v_c}{R} \sin \theta \\
& =\frac{10 \times \sin 120^{\circ}}{21.8}=\frac{10 \sqrt{3}}{2 \times 21.8} \cong 0.397 \\
& \phi \cong 23.4^{\circ}
\end{aligned}
\)