4.0 Components of a Vector

  Vectors

Let us consider the points $A(1, 0, 0)$, $B(0, 1, 0)$ and $C(0, 0, 1)$ on the $x$-axis, $y$-axis and $z$-axis respectively. Then, it is clear from the figure that

$$\left| {OA} \right| = \left| {OB} \right| = \left| {OC} \right| = 1$$

The vectors $OA$, $OB$ and $OC$, each having magnitude $1$, are called unit vectors along the axes $OX$, $OY$ and $OZ$, respectively, and denoted by $\widehat {i}$ , $\widehat j$ and $\widehat k$, respectively.

Now, let us consider the position vector $\overrightarrow {OP} $ of a point $P(x, y, z)$ as shown in figure. Let ${P_1}$ be the foot of the perpendicular from $P$ on the plane $XOY$. We thus, see that ${P_1}P$ is parallel to $z$-axis. As $\widehat {i}$, $\widehat j$ and $\widehat k$ are the unit vectors along the $x$, $y$ and $z$-axes, respectively, and by the definition of the coordinates of $P$, we have

$$\begin{equation} \begin{aligned} \overrightarrow {{P_1}P} = \overrightarrow {OR} = z\widehat k \\ \overrightarrow {Q{P_1}} = \overrightarrow {OS} = y\widehat j \\ \overrightarrow {OQ} = x\widehat i \\\end{aligned} \end{equation} $$

Then from figure, we can write as $$O{P_1} = OQ + Q{P_1} = x\widehat i + y\widehat j$$$$OP = O{P_1} + {P_1}P = x\widehat i + y\widehat j + z\widehat k$$

Hence, the position vector of $P$ with respect to $O$ is given by $\overrightarrow {OP} $ i.e., $$\overrightarrow r = x\hat i + y\hat j + z\hat k$$

This type of form of any vector is called its component form, where, $x$, $y$ and $z$ are called as the scalar components of $\overrightarrow {OP} $, and $x\widehat i + y\widehat j + z\widehat k$ are called the vector components of $\overrightarrow {OP} $ along the respective axes. $x$, $y$ and $z$ are also termed as rectangular components.

The length of any vector $\overrightarrow {OP} $ = $x\widehat i + y\widehat j + z\widehat k$ can be determined by applying the Pythagoras theorem. Consider the right angle triangle $OQ{P_1}$ as shown in figure. $$\left| {\overrightarrow {O{P_1}} } \right| = \sqrt {{{\left| {\overrightarrow {OQ} } \right|}^2} + {{\left| {\overrightarrow {Q{P_1}} } \right|}^2}} = \sqrt {{x^2} + {y^2}} $$ and in the right angle triangle $O{P_1}P$, we have $$\overrightarrow {OP} = \sqrt {{{\left| {\overrightarrow {O{P_1}} } \right|}^2} + {{\left| {\overrightarrow {{P_1}P} } \right|}^2}} = \sqrt {({x^2} + {y^2}) + {z^2}} $$ Hence, the length of any vector $\overrightarrow r $ = $x\widehat i + y\widehat j + z\widehat k$ is given by $$\left| {x\widehat i + y\widehat j + z\widehat k} \right|$$$$\sqrt {{x^2} + {y^2} + {z^2}} $$