# The Two Adjacent Sides of a Parallelogram Are 2 ^ I − 4 ^ J + 5 ^ K and ^ I − 2 ^ J − 3 ^ K . \ Find the Unit Vector Parallel to One of Its Diagonals. Also, Find Its Area. - Mathematics

Sum

The two adjacent sides of a parallelogram are $2 \hat{ i } - 4 \hat{ j } + 5 \hat{ k } \text{ and } \hat{ i } - 2 \hat{ j } - 3\hat{ k } .$\  Find the unit vector parallel to one of its diagonals. Also, find its area.

#### Solution

$\text{ Suppose } \square ABCD \text{ is the given parallelogram and AC is its diagonal } .$

$\text{ Let } :$

$\vec{AB} = 2 \hat{ i } - 4 \hat{ j } + 5 \hat{ k }$

$\vec{BC} = \hat{ i } - 2 \hat{ j } - 3 \hat{ k }$

$\therefore \text{ Diagonal } \vec{AC} = \vec{AB} + \vec{BC}$

$= 3 \hat{ i } - 6 \hat{ j } + 2 \hat{ k }$

$\Rightarrow \left| \vec{AC} \right| = \sqrt{9 + 36 + 4}$

$= 7$

$\text{ Unit vector parallel to } \vec{AC} =\frac{\vec{AC}}{\left| \vec{AC} \right|}$

$=\frac{3 \hat{ i } - 6 \hat{ j } + 2 \hat{ k } }{7}$

$\text{ Now } ,$

$\vec{AB} \times \vec{BC} = \begin{vmatrix}\hat{ i } & \hat{ j } & \hat{ k } \\ 2 & - 4 & 5 \\ 1 & - 2 & - 3\end{vmatrix}$

$= 22 \hat{ i } + 11 \hat{ j } + 0 \hat{ k }$

$\Rightarrow \left| \vec{AB} \times \vec{AC} \right| = \sqrt{484 + 121}$

$= \sqrt{605}$

$= 11\sqrt{5}$

$Area of triangleABC=\frac{1}{2}\left| \vec{AB} \times \vec{AC} \right|$

$= \frac{11\sqrt{5}}{2}\text{ sq . units }$

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#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 25 Vector or Cross Product
Exercise 25.1 | Q 31 | Page 31