JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Vector Algebra

Let P, Q, R and S be the points on the plane with position vectors

- 2 \hat{i} - \hat{j}, 4 \hat{i}, 3 \hat{i} + 3 \hat{j}

and

- 3 \hat{i} + 2 \hat{j}

respectively. The quadrilateral PQRS must be a

Answer: (A)

Solution

Position vectors:

P = (- 2, - 1)

Q = (4, 0)

R = (3, 3)

S = (- 3, 2)

.

Vectors:

P Q = (6, 1), S R = (6, 1)

P S = (- 1, 3), QR = (- 1, 3)

Since

P Q = S R

and

P S = QR

, PQRS is a parallelogram.

Check if it's a rectangle:

P Q \cdot P S = 6 (- 1) + 1 (3) = - 6 + 3 = - 3 \neq = 0

So it's not a rectangle.

Check if it's a rhombus:

∣ P Q ∣ = 36 + 1 = 37

∣ P S ∣ = 1 + 9 = 10

Since

∣ P Q ∣ \neq = ∣ P S ∣

, it's not a rhombus.

Therefore, PQRS is a parallelogram which is neither a rhombus nor a rectangle.

JEE Advanced · 2010 · 11 Apr 2010 · Paper 1 · Single correct option