Question

If a point \( C \) divides the line segment joining the points with position vectors \( 2\mathbf{i} - 3\mathbf{j} + 2\mathbf{k} \) and \( 3\mathbf{i} - \mathbf{j} - 2\mathbf{k} \) in the ratio \( 2:3 \), then the distance of \( C \) from the point with position vector \( 2\mathbf{i} - \mathbf{j} + \mathbf{k} \) is

• A. \( \frac{7}{5} \)
• B. \( \frac{4}{5} \)
• C. \( \frac{1}{5} \)
• D. \( \frac{3}{5} \)

Hint:

Use the section formula to find the position vector of point \( C \). Then, find the vector representing the line segment joining the point with the given position vector and point \( C \). The magnitude of this vector gives the required distance.

Answer 1

The Correct Option is A

Let the position vectors of the two points be \( \mathbf{a} = 2\mathbf{i} - 3\mathbf{j} + 2\mathbf{k} \) and \( \mathbf{b} = 3\mathbf{i} - \mathbf{j} - 2\mathbf{k} \).
The point \( C \) divides the line segment joining these points in the ratio \( 2:3 \).
Using the section formula for internal division, the position vector of \( C \), denoted by \( \mathbf{c} \), is given by: $$ \mathbf{c} = \frac{m\mathbf{b} + n\mathbf{a}}{m + n} $$ Here, \( m = 2 \) and \( n = 3 \).
$$ \mathbf{c} = \frac{2(3\mathbf{i} - \mathbf{j} - 2\mathbf{k}) + 3(2\mathbf{i} - 3\mathbf{j} + 2\mathbf{k})}{2 + 3} $$ $$ \mathbf{c} = \frac{(6\mathbf{i} - 2\mathbf{j} - 4\mathbf{k}) + (6\mathbf{i} - 9\mathbf{j} + 6\mathbf{k})}{5} $$ $$ \mathbf{c} = \frac{(6 + 6)\mathbf{i} + (-2 - 9)\mathbf{j} + (-4 + 6)\mathbf{k}}{5} $$ $$ \mathbf{c} = \frac{12\mathbf{i} - 11\mathbf{j} + 2\mathbf{k}}{5} = \frac{12}{5}\mathbf{i} - \frac{11}{5}\mathbf{j} + \frac{2}{5}\mathbf{k} $$ Let the position vector of the point from which the distance of \( C \) is to be found be \( \mathbf{d} = 2\mathbf{i} - \mathbf{j} + \mathbf{k} \).
The vector \( \vec{DC} \) is given by \( \mathbf{c} - \mathbf{d} \): $$ \vec{DC} = \left( \frac{12}{5}\mathbf{i} - \frac{11}{5}\mathbf{j} + \frac{2}{5}\mathbf{k} \right) - (2\mathbf{i} - \mathbf{j} + \mathbf{k}) $$ $$ \vec{DC} = \left( \frac{12}{5} - 2 \right)\mathbf{i} + \left( -\frac{11}{5} - (-1) \right)\mathbf{j} + \left( \frac{2}{5} - 1 \right)\mathbf{k} $$ $$ \vec{DC} = \left( \frac{12 - 10}{5} \right)\mathbf{i} + \left( \frac{-11 + 5}{5} \right)\mathbf{j} + \left( \frac{2 - 5}{5} \right)\mathbf{k} $$ $$ \vec{DC} = \frac{2}{5}\mathbf{i} - \frac{6}{5}\mathbf{j} - \frac{3}{5}\mathbf{k} $$ The distance of \( C \) from the point with position vector \( \mathbf{d} \) is the magnitude of the vector \( \vec{DC} \): $$ |\vec{DC}| = \sqrt{\left( \frac{2}{5} \right)^2 + \left( -\frac{6}{5} \right)^2 + \left( -\frac{3}{5} \right)^2} $$ $$ |\vec{DC}| = \sqrt{\frac{4}{25} + \frac{36}{25} + \frac{9}{25}} = \sqrt{\frac{4 + 36 + 9}{25}} = \sqrt{\frac{49}{25}} = \frac{7}{5} $$