Find the Area of the Parallelogram Whose Diagonals Are 2 ^ I + 3 ^ J + 6 ^ K and 3 ^ I − 6 ^ J + 2 ^ K - Mathematics

Sum

Find the area of the parallelogram whose diagonals are $2 \hat{ i } + 3 \hat{ j } + 6 \hat{ k } \text{ and } 3 \hat{ i } - 6 \hat{ j } + 2 \hat{ k }$

Solution

$\text{ Let } :$
$\vec{a} = 2 \hat{ i } + 3 \hat{ j } + 6 \hat{ k }$
$\vec{b} = 3 \hat{ i } - 6 \hat{ j } + 2 \hat{ k }$
$\therefore \vec{a} \times \vec{b} = \begin{vmatrix}\hat{ i } & \hat{ j } & \hat{ k } \\ 2 & 3 & 6 \\ 3 & - 6 & 2\end{vmatrix}$
$= \left( 6 + 36 \right) \hat{ i } - \left( 4 - 18 \right) \hat[{ j } + \left( - 12 - 9 \right) \hat{ k }$
$= 42 \hat{ i } + 14 \hat{ j } - 21 \hat{ k }$
$\Rightarrow \left| \vec{a} \times \vec{b} \right| = \sqrt{{42}^2 + {14}^2 + \left( - 21 \right)^2}$
$= \sqrt{2401}$
$= 49$
$\text{ Area of the parallelogram } =\frac{1}{2}\left| \vec{a} \times \vec{b} \right|$
$= \frac{49}{2} \text{ sq. units }$

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APPEARS IN

RD Sharma Class 12 Maths
Chapter 25 Vector or Cross Product
Exercise 25.1 | Q 9.4 | Page 30