Geometrical Interpretation of Product of Vectors - Practice Questions & MCQ

Area of Parallelogram

If and , are two non-zero, non-parallel vectors represented by AD and AB respectively and let Ө be the angle between them.

$$
\begin{aligned}
& \text { In } \triangle \mathrm{ADE}, \quad \sin \theta=\frac{D E}{A D} \\
& \Rightarrow \quad D E=A D \sin \theta=|\overrightarrow{\mathbf{a}}| \sin \theta \\
& \text { Area of parallelogram } \mathrm{ABCD}=\mathrm{AB} \cdot \mathrm{DE} \\
& \text { Thus, } \\
& \text { Area of parallelogram } \mathrm{ABCD}=|\vec{b}||\vec{a}| \sin \theta=|\vec{a} \times \vec{b}|
\end{aligned}
$$

Area of Triangle
If $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathbf{b}}$, are two non-zero, non-parallel vectors represented as the adjacent sides of a triangle then its area is given as $\frac{1}{2}|\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}|$

The area of a triangle is ½ (Base) x (Height)

From the figure,

Area of triangle $\mathrm{ABC}=\frac{1}{2} \mathrm{AB} \cdot \mathrm{CD}$
But $\mathrm{AB}=|\overrightarrow{\mathbf{b}}|$ (as given), and $\mathrm{CD}=|\overrightarrow{\mathbf{a}}| \sin \theta$
Thus, Area of triangle $\mathrm{ABC}=\frac{1}{2}|\overrightarrow{\mathbf{b}}||\overrightarrow{\mathbf{a}}| \sin \theta=\frac{1}{2}|\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}|$

NOTE:
1. The area of a parallelogram with diagonals $\overrightarrow{\mathbf{d}}_1$ and $\overrightarrow{\mathbf{d}}_2$ is $\frac{1}{2}\left|\overrightarrow{\mathbf{d}}_1 \times \overrightarrow{\mathbf{d}}_2\right|$.
2. The area of a plane quadrilateral $A B C D$ with AC and BD as diagonal is $\frac{1}{2}|\overrightarrow{\mathbf{A C}} \times \overrightarrow{\mathbf{B D}}|$.
3. If $\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{c}}$ are position vectors of a $\triangle A B C$, then its area is $\frac{1}{2}|(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}})+(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}})+(\overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{a}})|$.