Question

Find the position vector of point \( C \) which divides the line segment joining points \( A \) and \( B \) having position vectors \( \hat{i} + 2\hat{j} - \hat{k} \) and \( -\hat{i} + \hat{j} + \hat{k} \), respectively, in the ratio 4:1 externally. Further, find \( | \overrightarrow{AB} | : | \overrightarrow{BC} | \).

Hint:

For external division, the ratio formula includes subtraction in the denominator: \( \frac{m\vec{B} - n\vec{A}}{m-n} \).

Answer 1

1. Position vector of \( C \): The formula for the position vector of a point dividing a line segment externally in the ratio \( m:n \) is: \[ \vec{r}_C = \frac{m\vec{r}_B - n\vec{r}_A}{m-n}. \] Here, \( \vec{r}_A = \hat{i} + 2\hat{j} - \hat{k} \), \( \vec{r}_B = -\hat{i} + \hat{j} + \hat{k} \), \( m = 4 \), and \( n = 1 \). Substitute: \[ \vec{r}_C = \frac{4(-\hat{i} + \hat{j} + \hat{k}) - 1(\hat{i} + 2\hat{j} - \hat{k})}{4-1}. \] Simplify: \[ \vec{r}_C = \frac{-4\hat{i} + 4\hat{j} + 4\hat{k} - \hat{i} - 2\hat{j} + \hat{k}}{3} = \frac{-5\hat{i} + 2\hat{j} + 5\hat{k}}{3}. \] Thus: \[ \vec{r}_C = -\frac{5}{3} \hat{i} + \frac{2}{3} \hat{j} + \frac{5}{3} \hat{k}. \] 2. Find \( |\overrightarrow{AB}| : |\overrightarrow{BC}| \): First, calculate \( \overrightarrow{AB} = \vec{r}_B - \vec{r}_A \): \[ \overrightarrow{AB} = (-\hat{i} + \hat{j} + \hat{k}) - (\hat{i} + 2\hat{j} - \hat{k}) = -2\hat{i} - \hat{j} + 2\hat{k}. \] Magnitude: \[ | \overrightarrow{AB} | = \sqrt{(-2)^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3. \] Now calculate \( \overrightarrow{BC} = \vec{r}_C - \vec{r}_B \): \[ \overrightarrow{BC} = \left(-\frac{5}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{5}{3}\hat{k}\right) - (-\hat{i} + \hat{j} + \hat{k}). \] Simplify: \[ \overrightarrow{BC} = \left(-\frac{5}{3} + 1\right)\hat{i} + \left(\frac{2}{3} - 1\right)\hat{j} + \left(\frac{5}{3} - 1\right)\hat{k} = \frac{-2}{3}\hat{i} - \frac{1}{3}\hat{j} + \frac{2}{3}\hat{k}. \] Magnitude: \[ | \overrightarrow{BC} | = \sqrt{\left(-\frac{2}{3}\right)^2 + \left(-\frac{1}{3}\right)^2 + \left(\frac{2}{3}\right)^2} = \sqrt{\frac{4}{9} + \frac{1}{9} + \frac{4}{9}} = \sqrt{\frac{9}{9}} = 1. \] Thus: \[ |\overrightarrow{AB}| : |\overrightarrow{BC}| = 3:1. \] Final Answer: \[ \vec{r}_C = -\frac{5}{3} \hat{i} + \frac{2}{3} \hat{j} + \frac{5}{3} \hat{k}, \quad |\overrightarrow{AB}| : |\overrightarrow{BC}| = 3:1. \]