# Find the Area of the Parallelogram Determined by the Vector 3 ^ I + ^ J − 2 ^ K and ^ I − 3 ^ J + 4 ^ K - Mathematics

Sum

Find the area of the parallelogram determined by the vector $3 \hat{ i } + \hat{ j } - 2 \hat{ k } \text{ and } \hat{ i } - 3 \hat{ j } + 4 \hat{ k }$ .

#### Solution

$\text{ Let } :$
$\vec{a} = 3 \hat{ i } + \hat{ j } - 2 \hat{ k }$
$\vec{b} = 1 \hat{ i } - 3 \hat{ j } + 4 \hat{ k }$
$\vec{a} \times \vec{b} = \begin{vmatrix}\hat{ i } & \hat{ j } & \hat{ k } \\ 3 & 1 & - 2 \\ 1 & - 3 & 4\end{vmatrix}$
$= \hat{ i } \left( 4 - 6 \right) - \hat{ j } \left( 12 + 2 \right) + \hat{ k } \left( - 9 - 1 \right)$
$= - 2 \hat{ i } - 14 \hat{ j } - 10 \hat{ k }$
$\text{ Area of the parallelogram } =\left| \vec{a} \times \vec{b} \right|$
$= \sqrt{\left( - 2 \right)^2 + \left( - 14 \right)^2 + \left( - 10 \right)^2}$
$= \sqrt{300}$
$= 10\sqrt{3} \text{ sq. units }$

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#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 25 Vector or Cross Product
Exercise 25.1 | Q 8.3 | Page 29