# Resolving a vector into Components

The following figure shows a vector \(\overrightarrow {AB} \) of length \(5\sqrt 2 \) units, inclined at an angle of 45^{0} to the positive *x*-direction:

Can we express this vector using the \(\hat i\)- \(\hat j\) system?

Observe the following figure, where we have *resolved* the vector \(\overrightarrow {AB} \) into two *components* \(\overrightarrow {AC} \) and \(\overrightarrow {CB} \), where \(\overrightarrow {AC} \) is parallel to the *x*-axis, and \(\overrightarrow {CB} \) is parallel to the *y*-axis:

We note that:

\[\begin{align}&\left| {\overrightarrow {AC} } \right| = 5\sqrt 2 \cos {45^0} = 5\sqrt 2 \times \frac{1}{{\sqrt 2 }} = 5\\\,\,\, \qquad\qquad \qquad\qquad &\qquad\qquad\qquad\qquad\;\;\Rightarrow \,\,\,\overrightarrow {AC} = 5\hat i\\&\left| {\overrightarrow {CB} } \right| = 5\sqrt 2 \sin {45^0} = 5\sqrt 2 \times \frac{1}{{\sqrt 2 }} = 5\\\,\,\, \qquad\qquad \qquad \qquad &\qquad\qquad\qquad\qquad\;\Rightarrow \,\,\,\overrightarrow {CB} = 5\hat j\end{align}\]Thus,

\[\overrightarrow {AB} = \overrightarrow {AC} + \overrightarrow {CB} = 5\hat i + 5\hat j\]

Let us see another example of resolving a vector into its components (which simply means finding the \(\hat i\)-component and the \(\hat j\)-component). The following figure shows a vector \(\overrightarrow {PQ} \) of length 6 units inclined at an angle of 120^{0} to the positive *x*-direction:

What will be the components of \(\overrightarrow {PQ} \)?

Observe the following figure:

\(\overrightarrow {PR} \) is the \(\hat i\)-component of \(\overrightarrow {PQ} \), while \(\overrightarrow {RQ} \) is the \(\hat j\)-component of \(\overrightarrow {PQ} \). We have:

\[\begin{array}{l}\left| {\overrightarrow {PR} } \right| = 6\cos {60^0} = 6\left( {\frac{1}{2}} \right) = 3\\ \qquad \qquad \qquad \quad \Rightarrow \,\,\,\overrightarrow {PR} = - 3\hat i\\\left| {\overrightarrow {RQ} } \right| = 6\sin {60^0} = 6\left( {\frac{{\sqrt 3 }}{2}} \right) = 3\sqrt 3 \\ \qquad \qquad \qquad \quad \Rightarrow \,\,\,\overrightarrow {RQ} = 3\sqrt 3 \hat j\end{array}\]

We can now specify \(\overrightarrow {PQ} \) using the \(\hat i\)- \(\hat j\)system:

\[\overrightarrow {PQ} = \overrightarrow {PR} + \overrightarrow {RQ} = - 3\hat i + 3\sqrt 3 \hat j\]

The technique of resolving a vector into its components is extensively used in the study of Physics, and must be practiced as much as possible.

**Example 1:** A vector is inclined at an angle of \( - {30^0}\) to the positive *x*-direction, and has a length of 10 units. Specify the vector using the \(\hat i\)- \(\hat j\)system.

**Solution:** Consider the following figure:

We have:

\[\begin{array}{l}\left| {\overrightarrow {AC} } \right| = 10\cos {30^0} = 10\left( {\frac{{\sqrt 3 }}{2}} \right) = 5\sqrt 3 \\\left| {\overrightarrow {CB} } \right| = 10\sin {30^0} = 10\left( {\frac{1}{2}} \right) = 5\end{array}\]

Thus,

\[\overrightarrow {AB} = \overrightarrow {AC} + \overrightarrow {CB} = 5\sqrt 3 \widehat i - 5\widehat j\]

**Example 2:** A force of magnitude *F* is acting on a box at an angle of \(\theta \)to the horizontal. Take the horizontal right direction as the \(\widehat i\) direction, and the vertical up direction as the \(\widehat j\) direction. Specify the force \(\overrightarrow F \) using this \(\widehat i\)- \(\widehat j\) system.

**Solution:** The vector components of the force \(\overrightarrow F \) are shown below:

We have:

\[\overrightarrow {{F_1}} = F\cos \theta \widehat i,\,\,\,\overrightarrow {{F_2}} = F\sin \theta \widehat j\]

Thus,

\[\overrightarrow F = \overrightarrow {{F_1}} + \overrightarrow {{F_2}} = F\cos \theta \widehat i + F\sin \theta \widehat j\]

**Example 3:** A vector \(\overrightarrow a \) is specified in the \(\widehat i\)- \(\widehat j\) system as

\[\overrightarrow a = 3\widehat i - 5\widehat j\]

Find the magnitude of \(\overrightarrow a \), and its inclination with the positive *x*-direction.

**Solution:** The following figure shows the vector \(\overrightarrow a \):

Using the Pythagoras Theorem, we have:

\[\left| {\overrightarrow a } \right| = \sqrt {{3^2} + {5^2}} = \sqrt {34} \]

The angle \(\theta \) is given by

\[\tan \theta = \frac{5}{3}\,\,\, \Rightarrow \,\,\,\theta = {\tan ^{ - 1}}\left( {\frac{5}{3}} \right)\]

Thus, the angle of inclination of this vector with the horizontal is \( - {\tan ^{ - 1}}\left( {\frac{5}{3}} \right)\).

We note that in general, a vector \(\overrightarrow r = x\widehat i + y\widehat j\) will have a magnitude of \(\left| {\overrightarrow r } \right| = \sqrt {{x^2} + {y^2}} \).