Question

Show that the area of the triangle contained between the vectors a and b is one half of the magnitude of a × b.

Answer 1

Let ${\displaystyle \overrightarrow{\text{OA}}=\overrightarrow{\text{a}};\overrightarrow{\text{OB}}=\overrightarrow{\text{b}},\angle \text{AOB}=\theta }$

As shown in the figure, ΔOAB is formed between the vectors ${\displaystyle \overrightarrow{\text{a}}}$ and ${\displaystyle \overrightarrow{\text{b}}}$ while OACB is a parallelogram.

From the definition,

${\displaystyle \overrightarrow{\text{a}}\times \overrightarrow{\text{b}}=\left|\overrightarrow{\text{a}}\right|\left|\overrightarrow{\text{b}}\right|\sin \theta \hat{\text{n}}}$

where ${\displaystyle \hat{\text{n}}}$ is the unit vector perpendicular to both vectors ${\displaystyle \overrightarrow{\text{a}}}$ and ${\displaystyle \overrightarrow{\text{b}}}$

∴ ${\displaystyle |\overrightarrow{\text{a}}\times \overrightarrow{\text{b}}|=\left|\overrightarrow{\text{a}}\right|\left|\overrightarrow{\text{b}}\right|\sin \theta }$

= OA . OB sin θ    ...(1)

In the figure, BD is perpendicular to OA.

∴ ${\displaystyle \text{sin} \theta =\frac{\text{BD}}{\text{OB}}}$  या  ${\displaystyle \text{BD}=\text{OB} \text{sin} \theta }$

From equation (1),

${\displaystyle |\overrightarrow{\text{a}}\times \overrightarrow{\text{b}}|=\text{OA}.\text{BD}}$

= Base of parallelogram × Perpendicular distance between parallel sides

= Area of parallelogram OACB

But area of ΔOAB = ${\displaystyle \frac{1}{2}}$ area of parallelogram OACB

= ${\displaystyle \frac{1}{2}|\overrightarrow{\text{a}}\times \overrightarrow{\text{b}}|}$

Thus, the area of the triangle contained between the vectors a and b is one half of the magnitude of a × b.