Resolving a vector into Components

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The following figure shows a vector \(\overrightarrow {AB} \) of length \(5\sqrt 2 \) units, inclined at an angle of 450 to the positive x-direction:

Vector with magnitude and 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:

Two components of vector

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 1200 to the positive x-direction:

Two components of vector example 1

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

Observe the following figure:

Two components of vector example 2

\(\overrightarrow {PR} \) 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:

Two components of vector example 3

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:

Two components of vector example 4

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 \(\overrightarrow a \):

Two components of vector example 5

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 \(\left| {\overrightarrow r } \right| = \sqrt {{x^2} + {y^2}} \).

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