Question

If the position vectors of A, B, C, D are \( \vec{A} = \hat{i} + 2\hat{j} + 2\hat{k}, \vec{B} = 2\hat{i} - \hat{j}, \vec{C} = \hat{i} + \hat{j} + 3\hat{k}, \vec{D} = 4\hat{j} + 5\hat{k} \), then the quadrilateral ABCD is a:

• A. square
• B. rectangle
• C. rhombus
• D. parallelogram

Hint:

For quadrilaterals in vector form:
- Equal side lengths and non-right angles suggest a rhombus.
- Check dot product to confirm angles.

Answer 1

The Correct Option is C

Step 1: Find the side vectors
Let: \[ \vec{AB} = \vec{B} - \vec{A} = (2\hat{i} - \hat{j}) - (\hat{i} + 2\hat{j} + 2\hat{k}) = \hat{i} - 3\hat{j} - 2\hat{k} \] \[ \vec{BC} = \vec{C} - \vec{B} = (\hat{i} + \hat{j} + 3\hat{k}) - (2\hat{i} - \hat{j}) = -\hat{i} + 2\hat{j} + 3\hat{k} \] \[ \vec{CD} = \vec{D} - \vec{C} = (4\hat{j} + 5\hat{k}) - (\hat{i} + \hat{j} + 3\hat{k}) = -\hat{i} + 3\hat{j} + 2\hat{k} \] \[ \vec{DA} = \vec{A} - \vec{D} = (\hat{i} + 2\hat{j} + 2\hat{k}) - (4\hat{j} + 5\hat{k}) = \hat{i} - 2\hat{j} - 3\hat{k} \] Step 2: Check lengths of all sides
Find magnitudes: \[ |\vec{AB}| = \sqrt{1^2 + (-3)^2 + (-2)^2} = \sqrt{1 + 9 + 4} = \sqrt{14} \] \[ |\vec{BC}| = \sqrt{(-1)^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14} \] \[ |\vec{CD}| = \sqrt{(-1)^2 + 3^2 + 2^2} = \sqrt{1 + 9 + 4} = \sqrt{14} \] \[ |\vec{DA}| = \sqrt{1^2 + (-2)^2 + (-3)^2} = \sqrt{1 + 4 + 9} = \sqrt{14} \] All sides are equal \( \Rightarrow \) quadrilateral is equilateral. Step 3: Check angles
Compute dot product of adjacent sides: \[ \vec{AB} . \vec{BC} = (1)(-1) + (-3)(2) + (-2)(3) = -1 -6 -6 = -13 \neq 0 \] So no angle is \(90^\circ\), hence not a square or rectangle. Step 4: Conclusion
Since all sides are equal and adjacent sides are not perpendicular, it is a **rhombus**.