1.0 Introduction

Let $P(x,y,z)$ be any point in space and $\widehat i$, $\widehat j$ and $\widehat k$ be the unit vectors along $OX$, $OY$ and $OZ$ respectively.

Therefore, the position vector of point $P$ is written as $\overrightarrow r $ i.e., $$\begin{equation} \begin{aligned} \overrightarrow r = \overrightarrow {OP} = \overrightarrow {OM} + \overrightarrow {MP} \\ \overrightarrow r = \left( {\overrightarrow {OA} + \overrightarrow {AM} } \right) + \overrightarrow {MP} \\ \overrightarrow r = \overrightarrow {OA} + \overrightarrow {OB} + \overrightarrow {OC} \\ \overrightarrow r = x\widehat i + y\widehat j + z\widehat k \\\end{aligned} \end{equation} $$

Question 1. Find the locus of a point which moves such that the sum of its distances from points $A(0,0, - \alpha )$ and $B(0,0,\alpha )$ is constant.

Solution: Let the variable point be $P(h,k,l)$ and it is given that $$PA + PB = {\text{constant}} = 2b({\text{say)}}$$ Therefore, $$\begin{equation} \begin{aligned} \sqrt {{{\left( {h - 0} \right)}^2} + {{\left( {k - 0} \right)}^2} + {{\left( {l - ( - \alpha )} \right)}^2}} + \sqrt {{{\left( {h - 0} \right)}^2} + {{\left( {k - 0} \right)}^2} + {{\left( {l - \alpha } \right)}^2}} = 2b \\ \sqrt {{h^2} + {k^2} + {{\left( {l + \alpha } \right)}^2}} = 2b - \sqrt {{h^2} + {k^2} + {{\left( {l - \alpha } \right)}^2}} \\\end{aligned} \end{equation} $$ Squaring both sides, we get $$\begin{equation} \begin{aligned} {h^2} + {k^2} + {\left( {l + \alpha } \right)^2} = 4{b^2} + {h^2} + {k^2} + {l^2} + {\alpha ^2} - 4b\sqrt {{h^2} + {k^2} + {{\left( {l - \alpha } \right)}^2}} \\ {h^2} + {k^2} + {l^2}\left( {1 - \frac{{{\alpha ^2}}}{{{b^2}}}} \right) = {b^2} - {\alpha ^2} \\ \frac{{{h^2}}}{{{b^2} - {\alpha ^2}}} + \frac{{{k^2}}}{{{b^2} - {\alpha ^2}}} + \frac{{{l^2}}}{{{b^2}}} = 1 \\\end{aligned} \end{equation} $$ Therefore, the required locus is $$\frac{{{x^2}}}{{{b^2} - {\alpha ^2}}} + \frac{{{y^2}}}{{{b^2} - {\alpha ^2}}} + \frac{{{z^2}}}{{{b^2}}} = 1$$

Let us assume a point $P(x,y,z)$ in the plane and the vector joining origin $O$ and a point $P$ i.e., $\overrightarrow {OP} $ makes an angle of $\alpha $, $\beta $ and $\gamma $ (called direction angles) with the positive directions of the co-ordinate axes $OX$, $OY$ and $OZ$ respectively as shown in figure.

Then $\cos \alpha $, $\cos \beta $ and $\cos \gamma $are the direction cosines of $\overrightarrow {OP} $ which are denoted by $l$, $m$ and $n$ respectively i.e., $$\begin{equation} \begin{aligned} l = \cos \alpha \\ m = \cos \beta {\text{ }}\left[ {0 \leqslant \alpha ,\beta ,\gamma \leqslant \pi } \right] \\ n = \cos \gamma \\\end{aligned} \end{equation} $$

Now, the direction cosines of $x$-axis can be find out by putting $\alpha = {0^ \circ }$ and $\beta = \gamma = {90^ \circ }$ i.e., $$\begin{equation} \begin{aligned} (\cos {0^ \circ },\cos {90^ \circ },\cos {90^ \circ }) \\ (1,0,0) \\\end{aligned} \end{equation} $$

Similarly, the direction cosines of $y$-axis can be find out by putting $\beta = {0^ \circ }$ and $\alpha = \gamma = {90^ \circ }$ i.e., $$\begin{equation} \begin{aligned} (\cos {90^ \circ },\cos {0^ \circ },\cos {90^ \circ }) \\ (0,1,0) \\\end{aligned} \end{equation} $$

Similarly, the direction cosines of $z$-axis can be find out by putting $\gamma = {0^ \circ }$ and $\alpha = \beta = {90^ \circ }$ i.e., $$\begin{equation} \begin{aligned} (\cos {90^ \circ },\cos {90^ \circ },\cos {0^ \circ }) \\ (0,0,1) \\\end{aligned} \end{equation} $$

Note: If $l$, $m$ and $n$ are direction cosines of a vector, then ${l^2} + {m^2} + {n^2} = 1$.

Proof: Let us assume the distance of point $P$ from origin be $r$ i.e., $$\begin{equation} \begin{aligned} \left| {\overrightarrow {OP} } \right| = r \\ \Rightarrow \sqrt {{x^2} + {y^2} + {z^2}} = r \\\end{aligned} \end{equation} $$

On squaring both sides, we get $${x^2} + {y^2} + {z^2} = {r^2}$$

and the co-ordinates of point $P$ is also written as $$\begin{equation} \begin{aligned} x = r\cos \alpha = rl \\ y = r\cos \beta = rm \\ z = r\cos \gamma = rn \\\end{aligned} \end{equation} $$

Therefore, $$\begin{equation} \begin{aligned} {l^2}{r^2} + {m^2}{r^2} + {n^2}{r^2} = {r^2} \\ \Rightarrow {l^2} + {m^2} + {n^2} = 1 \\\end{aligned} \end{equation} $$

The direction ratios are the three numbers say $a$, $b$ and $c$ which are proportional to the direction cosines $l$, $m$ and $n$ respectively of a vector joining origin $O$ and a point $P$ i.e., $\overrightarrow {OP} $. $$\begin{equation} \begin{aligned} \frac{l}{a} = \frac{m}{b} = \frac{n}{c} = k \\ \Rightarrow (l,m,n) = (ka,kb,kc) \\\end{aligned} \end{equation} $$

To find the direction cosines if direction ratios is given, use the relation ${l^2} + {m^2} + {n^2} = 1$ and put $l=ka$, $m=kb$ and $n=kc$, we get $$\begin{equation} \begin{aligned} {a^2}{k^2} + {b^2}{k^2} + {c^2}{k^2} = 1 \\ {k^2} = \frac{1}{{{a^2} + {b^2} + {c^2}}} \\ k = \pm \frac{1}{{\sqrt {{a^2} + {b^2} + {c^2}} }} \\\end{aligned} \end{equation} $$ Therefore, the direction cosines are $$l = \pm \frac{a}{{\sqrt {{a^2} + {b^2} + {c^2}} }},{\text{ }}m = \pm \frac{b}{{\sqrt {{a^2} + {b^2} + {c^2}} }},{\text{ }}n = \pm \frac{c}{{\sqrt {{a^2} + {b^2} + {c^2}} }}$$

Note: The direction ratios and direction cosines of two parallel vectors are equal.

Proof: Let us assume two parallel vectors be $\overrightarrow a $ and $\overrightarrow b $ such that $\overrightarrow b = \lambda \overrightarrow a $ for some $\lambda $. If $\overrightarrow a = \overrightarrow {{a_1}} \widehat i + \overrightarrow {{a_2}} \widehat j + \overrightarrow {{a_3}} \widehat k$, then $\overrightarrow b = \overrightarrow {(\lambda {a_1})} \widehat i + \overrightarrow {(\lambda {a_2})} \widehat j + \overrightarrow {(\lambda {a_3})} \widehat k$. Therefore, the direction ratios of $\overrightarrow b $ is ${\lambda {a_1}}$, ${\lambda {a_2}}$, ${\lambda {a_3}}$ i.e., ${{a_1}}$, ${{a_2}}$, ${{a_2}}$. Thus, $\overrightarrow a $ and $\overrightarrow b $ have equal direction ratios and hence equal direction cosines also.

Note: The direction ratios of the line segment joining two points $P({x_1},{y_1},{z_1})$ and $Q({x_2},{y_2},{z_2})$ is ${x_2} - {x_1}$, ${y_2} - {y_1}$, ${z_2} - {z_1}$.

Application of this concept comes when we have to prove that three points $P$, $Q$ and $R$ are collinear. Using this concept, first find the direction ratios of the line segment joining points $P$ and $Q$ and then find the direction ratios of the line segment joining points $Q$ and $R$. If the direction ratios of $PQ$ and $QR$ are proportional then we can say that $PQ$ is parallel to $QR$ but point $Q$ is common to both $PQ$ and $QR$. Therefore, $P$, $Q$ and $R$ are collinear points.