Question

Let ABCD be a parallelogram and $ 2\bar{i} + \bar{j} $, $ 4\bar{i} + 5\bar{j} + 4\bar{k} $ and $ -\bar{i} - 4\bar{j} - 3\bar{k} $ be the position vectors of the vertices A, B, D respectively. Then the position vector of one of the points of trisection of the diagonal AC is

• A. $ \frac{1}{3} (5\bar{i} + 2\bar{j} - \bar{k}) $
• B. $ \frac{1}{3} (5\bar{i} + 2\bar{j} + \bar{k}) $
• C. $ \frac{1}{3} (5\bar{i} + 4\bar{j} - \bar{k}) $
• D. $ \frac{1}{3} (3\bar{i} + 2\bar{j} + \bar{k}) $

Hint:

Use the property that diagonals of a parallelogram bisect each other or that opposite sides are equal and parallel (in vector form). Apply the section formula for finding points dividing a line segment in a given ratio.

Answer 1

The Correct Option is B

Step 1: Find the position vector of vertex C.
$ \vec{AB} = \bar{b} - \bar{a} = 2\bar{i} + 4\bar{j} + 4\bar{k} $ $ \vec{DC} = \bar{c} - \bar{d} $ $ \bar{c} = \bar{d} + \vec{AB} = (-\bar{i} - 4\bar{j} - 3\bar{k}) + (2\bar{i} + 4\bar{j} + 4\bar{k}) = \bar{i} + \bar{k} $
Step 2: Find the position vectors of the trisection points of AC.
For trisection point P dividing AC in ratio 1:2:
$ \bar{p} = \frac{2\bar{a} + 1\bar{c}}{3} = \frac{2(2\bar{i} + \bar{j}) + (\bar{i} + \mathbf{k})}{3} = \frac{5\bar{i} + 2\bar{j} + \bar{k}}{3} $ For trisection point Q dividing AC in ratio 2:1:
$ \bar{q} = \frac{1\bar{a} + 2\bar{c}}{3} = \frac{(2\bar{i} + \bar{j}) + 2(\bar{i} + \bar{k})}{3} = \frac{4\bar{i} + \bar{j} + 2\mathbf{k}}{3} $
Step 3: Compare with the options.
The position vector $ \bar{1}{3} (5\bar{i} + 2\bar{j} + \bar{k}) $ matches option (2).
Step 4: Conclusion.
The position vector of one of the points of trisection of the diagonal AC is $ \frac{1}{3} (5\bar{i} + 2\bar{j} + \bar{k}) $.