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UPSC NDA I 2016 Mathematics Solved Paper

Question Numbers: 58-60

Consider a parallelogram whose vertices are A (1, 2), B (4, y), C (x, 6) and D (3, 5) taken in order

Question : 60 of 120

Marks: +1, -0

What is the area of the parallelogram?

Solution:

Concept:

Area of a parallelogram formed by two adjacent vectors is equal to the magnitude of their cross product.

Explanation:

Let the vertices be

A(1,2)

B(4,3)

, and

D(3,5)

Take

\vec{AB}

and

\vec{AD}

as the adjacent sides.

Compute

\vec{AB} = (4-1)\hat{i} + (3-2)\hat{j} = 3\hat{i} + \hat{j}

Compute

\vec{AD} = (3-1)\hat{i} + (5-2)\hat{j} = 2\hat{i} + 3\hat{j}

Find the cross product:

\vec{AB} \times \vec{AD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 0 \\ 2 & 3 & 0 \end{vmatrix} = (0-0)\hat{i} - (0-0)\hat{j} + (9-2)\hat{k} = 7\hat{k}.

The magnitude is

|7\hat{k}| = 7

Hence the area of the parallelogram is

7

square units.

Answer:

7

square units (Option D).

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