Question

Find the position vector of the a point lying on the line joining the points whose position vectors are $\hat{i}+\hat{j}-\hat{k}$ and $\hat{i}-\hat{j}+\hat{k}$. $$$$

Answer 1

Hint: We recall the definition of position vectors. We use the fact that the position vector of any point P that lies on the line joining two points A,B whose positions vector are known as $\overrightarrow{a},\overrightarrow{b}$ is given by $\overrightarrow{a}+\lambda \overrightarrow{d}$ where $\lambda$ is any real scalar and $\overrightarrow{d}=\overrightarrow{b}-\overrightarrow{a}$ is the distance vector.$$$$

Complete step by step answer:
We can represent any point in the space $P(x,y,z)$ as vector with original O as the initial point and P as the final point in terms of orthogonal unit vectors $\hat{i}$,$\hat{j}$ and $\hat{k}$ as $\overrightarrow{OP}=\overrightarrow{p}=x\hat{i}+y\hat{j}+\hat{k}$. This vector is called position vector, location vector or radius vector. We have the rough figure of the position vector $\overrightarrow{p}$ below. $$$$
 

If there are $\overrightarrow{a}=(a_1\hat{i}+a_2\hat{j}+a_3\hat{k}),\overrightarrow{b}=(b_1\hat{i}+b_2\hat{j}+b_3\hat{k})$ be two position vectors of two points then the position vector of any point on the line joining the two points is given with a real scalar $\lambda$as;
$$\begin{array}{rl}& \overrightarrow{p}=\overrightarrow{a}+\lambda \overrightarrow{d}\\ & \Rightarrow \overrightarrow{p}=\overrightarrow{a}+\lambda (\overrightarrow{b}-\overrightarrow{a})\\ & \Rightarrow \overrightarrow{p}=(a_1+\lambda (b_1-a_1))\hat{i}+(a_2+\lambda b_2-a_2)\hat{j}+(a_3+\lambda (b_3-a_3))\hat{k}\end{array}$$
We are given in the question the position vectors of two points as $\hat{i}+\hat{j}-\hat{k}$ and $\hat{i}-\hat{j}+\hat{k}$. Let us denote the two points as A and B and the position vectors as $\overrightarrow{a}=\hat{i}+\hat{j}-\hat{k}$ and $\overrightarrow{b}=\hat{i}-\hat{j}+\hat{k}$.So we have
$$\begin{array}{rl}& \overrightarrow{a}=(a_1\hat{i}+a_2\hat{j}+a_3\hat{k})=\hat{i}+\hat{j}-\hat{k}\\ & \overrightarrow{b}=(b_1\hat{i}+b_2\hat{j}+b_3\hat{k})=\hat{i}-\hat{j}+\hat{k}\end{array}$$
We compare the respective components and we get;
$$\begin{array}{rl}& \Rightarrow a_1=1,a_2=1,a_3=-1\\ & \Rightarrow b_1=1,b_2=-1,b_3=1\end{array}$$
Let P be any point on the line segment joining A and B.The position vector of P is;
$$\begin{array}{rl}& \overrightarrow{p}=\overrightarrow{a}+\lambda \overrightarrow{b}=(a_1+\lambda (b_1-a_1))\hat{i}+(a_2+\lambda (b_2-a_2))\hat{j}+(a_3+\lambda (b_3-a_3))\hat{k}\\ & \Rightarrow \overrightarrow{p}=(1+\lambda \cdot 0)\hat{i}+(1+\lambda (-2))\hat{j}+(-1+\lambda \cdot 2)\hat{k}\\ & \Rightarrow \overrightarrow{p}=\hat{i}+(1-2\lambda )\hat{j}+(-1+2\lambda )\hat{k}\end{array}$$

Note: We also know that $\hat{i}$,$\hat{j}$ and $\hat{k}$ are orthogonal unit vectors (vectors with magnitude 1) along $x,y$ and $z$ axes respectively. If we want to find the position vector of midpoint , it is is given by ${\displaystyle \frac{\overrightarrow{a}+\overrightarrow{b}}{2}}$ and the position vector of any point that divides the line segment AB at a ratio $m:n$ is given by ${\displaystyle \frac{m\overrightarrow{a}+n\overrightarrow{b}}{m+n}}$. We note that the position vector of the point system reference is unique.