Question

Explain the resolution of vectors in three dimensions. Draw the diagram to explain it properly.

Answer 1

Resolution of a vector in Three Dimensions
Consider that the vector \(\overrightarrow{O P}=\vec{A}\) represents a vector in space. In order to express it in the form of three mutually perpendicular components, construct a rectangular parallel OTBCDEP with three edges along the three co-ordinate axes OX, OY and OZ. Let \(\vec i ,\vec j\) and \(\vec k\) be the unit vectors along OX, OY and OZ respectively as shown in the fig.

Then according to the polygon law of addition of vectors, we have
\(\overrightarrow{O P}=\overrightarrow{O T}+\overrightarrow{T B}+\overrightarrow{B P}\) ……….. (i)
If the three sides of the rectangular figure respectively are OT = A_x, OC = A_y and OE = A_z
Then, \(\overrightarrow{O T}=A_{x} \hat{i}\), \(\overrightarrow{O C}=A_{y} \hat{j}\) and \(\overrightarrow{O E}=A_{z} \hat{k}\)
Since, \(\overrightarrow{T B}=\overrightarrow{O C}=A_{y} \hat{j}\) and
\(\overrightarrow{B P}=\overrightarrow{O E}=\overrightarrow{C D}=A_{z} \hat{k}\),
Then the equation (i) becomes
\(\vec{A}=A_{x} \hat{i}+A_{y} \hat{j}+A_{z} \hat{k}\) ……………… (ii)
The equation (ii) expresses the vector  oriented in space (three dimensions) in terms of its three rectangular components \(A_{x} \hat{i}\), \(A_{y} \hat{j}\) and \(A_{z} \hat{k}\). 

Magnitude of \(\vec A\) :
In triangle OBP
OP² = OB² + BP²
In triangle OCB
OB² = OC² + CB² = OC² + OT²
∴ OP² = OC² + OT² + BP²
A² = \(A_{y}^{2}+A_{x}^{2}+A_{z}^{2}\)
or A = \(\sqrt{A_{x}^{2}+A_{y}^{2}+A_{z}^{2}}\) …………… (iii)
Thus the magnitude of a vector is equal to the square root of the sum of the squares of the magnitude of its rectangular components.