Question

A line passes through the points whose position vectors $\hat{i}+\hat{j}-2\hat{k}$ and $\hat{i}-3\hat{j}+\hat{k}$. Then the position vector of a point on it at a unit distance from the first point is

• A. $\frac{1}{5}(5\hat{i}+\hat{j}-7\hat{k})$
• B. $\frac{1}{5}(5\hat{i}+9\hat{j}-13\hat{k})$
• C. $(\hat{i}-4\hat{j}+3\hat{k})$
• D. $(\hat{i}+4\hat{j}+3\hat{k})$

Answer 1

The correct option is A $\frac{1}{5}(5\hat{i}+\hat{j}-7\hat{k})$

Let points of the position vector $\hat{i}+\hat{j}-2\hat{k}$ and $\hat{i}-3\hat{j}+\hat{k}$ are $A(1,1-2)$ and $(1,-3,1)$

$AB=\sqrt{(1-1{)}^2+(1+3{)}^2+(-2-1{)}^2}=5$

Required points which divide $AB$ in ratio $1:4$

The required position vector of the point

$=\frac{1\times (\overrightarrow{i}-3\overrightarrow{j}+\overrightarrow{k})+4(\overrightarrow{i}+\overrightarrow{j}-2\overrightarrow{k})}{1+4}=\frac{1}{5}(5\overrightarrow{i}+\overrightarrow{j}-7\overrightarrow{k})$