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UPSC NDA I 2016 Mathematics Solved Paper

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For the next three (3) items that follow :

Consider a parallelogram whose vertices are

A(1,2),B(4,y),C(x,6)

and

D(3,5)

taken in order.

[NDA I 2016]

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Question : 60 of 120

Marks: +1, -0

What is the area of the parallelogram?

$\frac{7}{2}$ square units
$4$ square units
$\frac{11}{2}$ square units
$7$ square units

Solution:

Area of parallelogram = 2 area of Δ ADB

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

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

∴ Area of parallelogram

=2\left[\frac{1}{2}\left|\vec{a}\times\vec{b}\right|\right]=\left|\vec{a}\times\vec{b}\right|

Now;

\vec{a}\times\vec{b}=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\3 & 1 & 0\\2 & 3 & 0\end{vmatrix}=\widehat{7r}

∴ Area

=\left|\vec{a}\times\vec{b}\right|=\left|\widehat{7k}\right|=\sqrt{49}=7

square unit

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