4.5 Resolution of vectors

RESOLUTION OF VECTORS

Figure 4f shows a vector \(\vec{a}=\overrightarrow{O A}\) in the \(X\)-Y plane drawn from the origin \(O\). The vector makes an angle \(\alpha\) with the \(X\)-axis and \(\beta\) with the \(Y\)-axis. Draw perpendicular \(A B\) and \(A C\) from \(A\) to the \(X\) and \(Y\) axes respectively. The length \(O B\) is called the projection of
\(\overrightarrow{O A}\) on \(X\)-axis. Similarly \(O C\) is the projection of \(\overrightarrow{O A}\) on \(Y\)-axis. According to the rules of vector addition
\(
\vec{a}=\overrightarrow{O A}=\overrightarrow{O B}+\overrightarrow{O C}
\)
Thus, we have resolved the vector \(\vec{a}\) into two parts, one along \(O X\) and the other along \(O Y\). The magnitude of the part along \(O X\) is \(O B=a \cos \alpha\) and the magnitude of the part along \(O Y\) is \(O C=a \cos \beta\). If \(\vec{i}\) and \(\vec{j}\) denote vectors of unit magnitude along \(O X\) and \(O Y\) respectively, we get
\(
\begin{aligned}
\overrightarrow{O B} &=a \cos \alpha \vec{i} \text { and } \overrightarrow{O C}=a \cos \beta \vec{j} \\
\text { so that } \quad \vec{a} &=a \cos \alpha \vec{i}+a \cos \beta \vec{j} .
\end{aligned}
\)

Figure 4f

If the vector \(\vec{a}\) is not in the \(X-Y\) plane, it may have nonzero projections along \(X, Y, Z\) axes and we can resolve it into three parts i.e., along the \(X, Y\) and \(Z\) axes. If \(\alpha, \beta, \gamma\) be the angles made by the vector \(\vec{a}\) with the three axes respectively, we get
\(
\vec{a}=a \cos \alpha \vec{i}+a \cos \beta \vec{j}+a \cos \gamma \vec{k}
\)
where \(\vec{i}, \vec{j}\) and \(\vec{k}\) are the unit vectors along \(X, Y\) and \(Z\) axes respectively. The magnitude \((a \cos \alpha)\) is called the component of \(\vec{a}\) along \(X\)-axis, \((a \cos \beta)\) is called the component along \(Y\)-axis and \((a \cos \gamma)\) is called the component along \(Z\)-axis. In general, the component of a vector \(\vec{a}\) along a direction making an angle \(\theta\) with it is \(a \cos \theta\) (figure 4g ) which is the projection of \(\vec{a}\) along the given direction.

Figure 4g

We can easily add two or more vectors if we know their components along the rectangular coordinate axes. Let us have
\(\begin{array}{ll}\vec{a}=a_{x} \vec{i}+a_{y} \vec{j}+a_{z} \vec{k} \\ \vec{b} =b_{x} \vec{i}+b_{y} \vec{j}+b_{z} \vec{k} \\ \text { and } \vec{c}=c_{x} \vec{i}+c_{y} \vec{j}+c_{z} \vec{k} \\ \text { then } \vec{a}+\vec{b}+\vec{c}=\left(a_{x}+b_{x}+c_{x}\right) \vec{i}+\left(a_{y}+b_{y}+c_{y}\right) \vec{j}+\left(a_{z}+b_{z}+c_{z}\right) \vec{k} .\end{array}\)
If all the vectors are in the \(X-Y\) plane then all the \(z\) components are zero and the resultant is simply
\(
\vec{a}+\vec{b}+\vec{c}=\left(a_{x}+b_{x}+c_{x}\right) \vec{i}+\left(a_{y}+b_{y}+c_{y}\right) \vec{j} .
\)
This is the sum of two mutually perpendicular vectors of magnitude \(\left(a_{x}+b_{x}+c_{x}\right)\) and \(\left(a_{y}+b_{y}+c_{y}\right)\). The resultant can easily be found to have a magnitude
\(
\sqrt{\left(a_{x}+b_{x}+c_{x}\right)^{2}+\left(a_{y}+b_{y}+c_{y}\right)^{2}}
\)
making an angle \(\alpha\) with the \(X\)-axis where
\(
\tan \alpha=\frac{a_{y}+b_{y}+c_{y}}{a_{x}+b_{x}+c_{x}} .
\)

Example 1:  A force of \(10.5 \mathrm{~N}\) acts on a particle along a direction making an angle of \(37^{\circ}\) with the vertical. Find the component of the force in the vertical direction.

Answer: The component of the force in the vertical direction will be
\(
\begin{aligned}
F_{\perp} & =F \cos \theta=(10.5 \mathrm{~N})\left(\cos 37^{\circ}\right) \\
& =(10.5 \mathrm{~N}) \frac{4}{5}=8.40 \mathrm{~N} .
\end{aligned}
\)

DOT PRODUCT OR SCALAR PRODUCT OF TWO VECTORS

The dot product (also called scalar product) of two vectors \(\vec{a}\) and \(\vec{b}\) is defined as
\(
\vec{a} \cdot \vec{b}=a b \cos \theta
\)
where \(a\) and \(b\) are the magnitudes of \(\vec{a}\) and \(\vec{b}\) respectively and \(\theta\) is the angle between them. The dot product between two mutually perpendicular vectors is zero as \(\cos 90^{\circ}=0\).
The dot product is commutative and distributive.
\(
\begin{gathered}
\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{a} \\
\vec{a} \cdot(\vec{b}+\vec{c})=\vec{a} \cdot \vec{b}+\vec{a} \cdot \vec{c} .
\end{gathered}
\)

Figure 4h

Dot Product of Two Vectors in terms of the Components along the Coordinate Axes

Consider two vectors \(\vec{a}\) and \(\vec{b}\) represented in terms of the unit vectors \(\vec{i}, \vec{j}, \vec{k}\) along the coordinate axes as and \(\quad \vec{b}=b_{x} \vec{i}+b_{y} \vec{j}+b_{z} \vec{k}\).
Then
\(
\begin{aligned}
\vec{a} \cdot \vec{b}=&\left(a_{x} \vec{i}+a_{y} \vec{j}+a_{z} \vec{k}\right) \cdot\left(b_{x} \vec{i}+b_{y} \vec{j}+b_{z} \vec{k}\right) \\
=& a_{x} b_{x} \vec{i} \cdot \vec{i}+a_{x} b_{y} \vec{i} \cdot \vec{j}+a_{x} b_{z} \vec{i} \cdot \vec{k} \\
&+a_{y} b_{x} \vec{j} \cdot \vec{i}+a_{y} b_{y} \vec{j} \cdot \vec{j}+a_{y} b_{z} \vec{j} \cdot \vec{k} \\
&+a_{z} b_{x} \vec{k} \cdot \vec{i}+a_{z} b_{y} \vec{k} \cdot \vec{j}+a_{z} b_{z} \vec{k} \cdot \vec{k} \ldots \quad \text { (i) }\end{aligned}
\)
Since, \(\vec{i}, \vec{j}\) and \(\vec{k}\) are mutually orthogonal,

we have \(\vec{i} \cdot \vec{j}=\vec{i} \cdot \vec{k}=\vec{j} \cdot \vec{i}=\vec{j} \cdot \vec{k}=\vec{k} \cdot \vec{i}=\vec{k} \cdot \vec{j}=0\).
Also, \(\quad \vec{i} \cdot \vec{i}=1 \times 1 \cos 0=1\).
Similarly, \(\vec{j} \cdot \vec{j}=\vec{k} \cdot \vec{k}=1\).
Using these relations in equation (i) we get
\(
\vec{a} \cdot \vec{b}=a_{x} b_{x}+a_{y} b_{y}+a_{z} b_{z}
\)

Example 2: The work done by a force \(\vec{F}\) during a displacement \(\vec{r}\) is given by \(\vec{F} \cdot \vec{r}\). Suppose a force of \(12 \mathrm{~N}\) acts on a particle in the vertically upward direction and the particle is displaced through \(2.0 \mathrm{~m}\) in the vertically downward direction. Find the work done by the force during this displacement.

Answer: The angle between the force \(\vec{F}\) and the displacement \(\vec{r}\) is \(180^{\circ}\). Thus, the work done is
\(
\begin{aligned}
W & =\vec{F} \cdot \vec{r} \\
& =F r \cos \theta \\
& =(12 \mathrm{~N})\left(2^{\cdot} 0 \mathrm{~m}\right)\left(\cos 180^{\circ}\right) \\
& =-24 \mathrm{~N}-\mathrm{m}=-24 \mathrm{~J} .
\end{aligned}
\)

CROSS PRODUCT OR VECTOR PRODUCT OF TWO VECTORS

When two vectors are multiplied with each other and the product of the vectors is also a vector quantity, then the resultant vector is called the cross product of two vectors or the vector product. The resultant vector is perpendicular to the plane containing the two given vectors. 

The cross product or the vector product of two vectors \(\vec{a}\) and \(\vec{b}\), denoted by \(\vec{a} \times \vec{b}\) is itself a vector. The magnitude of this vector is
\(
|\vec{a} \times \vec{b}|=a b \sin \theta    ——– (4.0)
\)

Figure 4i
where \(a\) and \(b\) are the magnitudes of \(\vec{a}\) and \(\vec{b}\) respectively and \(\theta\) is the smaller angle between the two. When two vectors are drawn with both the tails coinciding, two angles are formed between them (figure 4i). One of the angles is smaller than \(180^{\circ}\) and the other is greater than \(180^{\circ}\) unless both are equal to \(180^{\circ}\). The angle \(\theta\) used in equation (4.0) is the smaller one. If both the angles are equal to \(180^{\circ}\), \(\sin \theta=\sin 180^{\circ}=0\) and hence \(|\vec{a} \times \vec{b}|=0\). Similarly if \(\theta=0, \sin \theta=0\) and \(|\vec{a} \times \vec{b}|=0\). The cross product of two parallel vectors is zero.

RIGHT-HAND RULE CROSS PRODUCT OF TWO VECTORS

Figure 4j

The direction of \(\vec{a} \times \vec{b}\) is perpendicular to both \(\vec{a}\) and \(\vec{b}\). Thus, it is perpendicular to the plane formed by \(\vec{a}\) and \(\vec{b}\). Draw the two vectors \(\vec{a}\) and \(\vec{b}\) with both the tails coinciding (figure 4j). Now place your stretched right palm perpendicular to the plane of \(\vec{a}\) and \(\vec{b}\) in such a way that the fingers are along the vector \(\vec{a}\) and when the fingers are closed they go towards \(\vec{b}\). The direction of the thumb gives direction of arrow to be put on the vector \(\vec{a} \times \vec{b}\). This is known as the right hand thumb rule.

The right-hand thumb rule makes the cross-product noncommutative. That is
\(
\vec{a} \times \vec{b}=-\vec{b} \times \vec{a} .
\)
The cross product follows the distributive law
\(
\vec{a} \times(\vec{b}+\vec{c})=\vec{a} \times \vec{b}+\vec{a} \times \vec{c} .
\)
It does not follow the associative law
\(
\vec{a} \times(\vec{b} \times \vec{c}) \neq(\vec{a} \times \vec{b}) \times \vec{c} .
\)
When we choose a coordinate system any two perpendicular lines may be chosen as \(X\) and \(Y\) axes. However, once \(X\) and \(Y\) axes are chosen, there are two possible choices of \(Z\)-axis. The \(Z\)-axis must be perpendicular to the \(X-Y\) plane. But the positive direction of \(Z\)-axis may be defined in two ways. We choose the positive direction of \(Z\)-axis in such a way that
\(
\vec{i} \times \vec{j}=\vec{k} .
\)
Such a coordinate system is called a right-handed system. In such a system
\(
\vec{j} \times \vec{k}=\vec{i} \quad \text { and } \vec{k} \times \vec{i}=\vec{j} .
\)
Of course \(\vec{i} \times \vec{i}=\vec{j} \times \vec{j}=\vec{k} \times \vec{k}=0\).

Example: 3

The vector \(\vec{A}\) has a magnitude of 5 units, \(\vec{B}\) has a magnitude of 6 units, and the cross product of \(\vec{A}\) and \(\vec{B}\) has a magnitude of 15 unit. Find the angle between \(\vec{A}\) and \(\vec{B}\).
Solution: If the angle between \(\vec{A}\) and \(\vec{B}\) is \(\theta\), the cross product will have a magnitude
\(
|\vec{A} \times \vec{B}|=A B \sin \theta
\)
\(
15=5 \times 6 \sin \theta
\)
\(
\sin \theta=\frac{1}{2} \text {. }
\)
Thus, \(\theta=30^{\circ} \text { or, } 150^{\circ}.\)

Cross Product of Two Vectors in terms of the Components along the Coordinate Axes

Let \(\quad \vec{a}=a_{x} \vec{i}+a_{y} \vec{j}+a_{z} \vec{k}\) and
\(
\vec{b}=b_{x} \vec{i}+b_{y} \vec{j}+b_{z} \vec{k}
\)

Then \(\vec{a} \times \vec{b}=\left(a_{x} \vec{i}+a_{y} \vec{j}+a_{z} \vec{k}\right) \times\left(b_{x} \vec{i}+b_{y} \vec{j}+b_{z} \vec{k}\right)\)
\(
\begin{aligned}
=& a_{x} b_{x} \vec{i} \times \vec{i}+a_{x} b_{y} \vec{i} \times \vec{j}+a_{x} b_{z} \vec{i} \times \vec{k} \\
&+a_{y} b_{x} \vec{j} \times \vec{i}+a_{y} b_{y} \vec{j} \times \vec{j}+a_{y} b_{z} \vec{j} \times \vec{k} \\
&+a_{z} b_{x} \vec{k} \times \vec{i}+a_{z} b_{y} \vec{k} \times \vec{j}+a_{z} b_{z} \vec{k} \times \vec{k} \\
=& a_{x} b_{y} \vec{k}+a_{x} b_{z}(-\vec{j})+a_{y} b_{x}(-\vec{k})+a_{y} b_{z}(\vec{i}) \\
&+a_{z} b_{x}(\vec{j})+a_{z} b_{y}(-\vec{i}) \\
=&\left(a_{y} b_{z}-a_{z} b_{y}\right) \vec{i}+\left(a_{z} b_{x}-a_{x} b_{z}\right) \vec{j} \\
&+\left(a_{x} b_{y}-a_{y} b_{x}\right) \vec{k} .
\end{aligned}
\)

Zero Vector

A null or zero vector is a vector with zero magnitude. Since the magnitude is zero, we don’t have to specify its direction. The direction of a zero vector is indeterminate. We can write this vector as \(\overrightarrow{0}\). The concept of a zero vector is also helpful when we consider the vector product of parallel vectors. If \(\vec{A} \| \vec{B}\), the vector \(\vec{A} \times \vec{B}\) is zero vector. For any vector \(\vec{A}\)
\(
\begin{aligned}
&\vec{A}+\overrightarrow{0}=\vec{A} \\
&\vec{A} \times \overrightarrow{0}=\overrightarrow{0}
\end{aligned}
\)
and for any number \(\lambda\),
\(
\lambda \overrightarrow{0}=\overrightarrow{0} .
\)