Find the Area of the Parallelogram Whose Diagonals Are 3 ^ I + 4 ^ J and ^ I + ^ J + ^ K - Mathematics

Sum

Find the area of the parallelogram whose diagonals are  $3 \hat{ i } + 4 \hat{ j } \text{ and } \hat{ i } + \hat{ j } + \hat{ k }$

Solution

$\text{ Let: }$
$\vec{a} = 3 \hat{ i } + 4 \hat{ j } + 0 \hat{ k }$
$\vec{b} = \hat{ i } + \hat{ j } + \hat{ k }$
$\therefore \vec{a} \times \vec{b} = \begin{vmatrix}\hat{ i } & \hat{ j } & \hat{ k } \\ 3 & 4 & 0 \\ 1 & 1 & 1\end{vmatrix}$
$= \left( 4 - 0 \right) \hat{ i } - \left( 3 - 0 \right) \hat{ j } + \left( 3 - 4 \right) \hat{ k}$
$= 4 \hat{ i } - 3 \hat{ j } - \hat{ k }$
$\Rightarrow \left| \vec{a} \times \vec{b} \right| = \sqrt{4^2 + \left( - 3 \right)^2 + \left( - 1 \right)^2}$
$= \sqrt{26}$
$\text{ Area of the parallelogram } =\frac{1}{2}\left| \vec{a} \times \vec{b} \right|$
$= \frac{\sqrt{26}}{2} \text{ sq. units }$

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APPEARS IN

RD Sharma Class 12 Maths
Chapter 25 Vector or Cross Product
Exercise 25.1 | Q 9.3 | Page 30