## Resolving a Vector into Perpendicular Components

Analytical techniques and right triangles go hand-in-hand in physics because (among other things) motions along perpendicular directions are independent. We very often need to separate a vector into perpendicular components. For example, given a vector like \(\mathbf{A}\) in the figure below, we may wish to find which two perpendicular vectors, \({\mathbf{A}}_{x}\) and \({\mathbf{A}}_{y}\), add to produce it.

\({\mathbf{A}}_{x}\) and \({\mathbf{A}}_{y}\) are defined to be the components of \(\mathbf{A}\) along the *x*– and *y*-axes. The three vectors \(\mathbf{A}\), \({\mathbf{A}}_{x}\), and \({\mathbf{A}}_{y}\) form a right triangle:

\({\mathbf{A}}_{x}{\mathbf{ + A}}_{y}\mathbf{ = A}\mathbf{.}\)

Note that this relationship between vector components and the resultant vector holds only for vector quantities (which include both magnitude and direction). The relationship does not apply for the magnitudes alone. For example, if \({\mathbf{\text{A}}}_{x}=3\text{ m}\) east, \({\mathbf{\text{A}}}_{y}=4\text{ m}\) north, and \(\mathbf{\text{A}}=5\text{ m}\) north-east, then it is true that the vectors \({\mathbf{A}}_{x}{\mathbf{ + A}}_{y}\mathbf{ = A}\). However, it is *not* true that the sum of the magnitudes of the vectors is also equal. That is,

\(3\text{ m} + 4\text{ m} \ne 5\)

Thus,

\({A}_{x}+{A}_{y}\ne A\)

If the vector \(\mathbf{A}\) is known, then its magnitude \(A\) (its length) and its angle \(\theta \) (its direction) are known. To find \({A}_{x}\) and \({A}_{y}\), its *x*– and *y*-components, we use the following relationships for a right triangle.

\({A}_{x}=A\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta \)

and

\({A}_{y}=A\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta \text{.}\)

Suppose, for example, that \(\mathbf{A}\) is the vector representing the total displacement of the person walking in a city considered in Two-Dimensional Motion: Walking in a City and Vector Addition: Head-to-Tail Method.

Then \(A=10.3\) blocks and \(\theta =29.1º\), so that

\({A}_{x}=A\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta =\left(\text{10.3 blocks}\right)\left(\text{cos}\phantom{\rule{0.25em}{0ex}}29.1º\right)=\text{9.0 blocks}\)

\({A}_{y}=A\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta =\left(\text{10.3 blocks}\right)\left(\text{sin}\phantom{\rule{0.25em}{0ex}}29.1º\right)=\text{5.0 blocks}\text{.}\)