# Examples On Vector Cross Product Set-1

Go back to  'Vectors and 3-D Geometry'

Example – 23

Prove that the area of a quadrilateral ABCD can be given by $$\frac{1}{2}\left| {\overrightarrow {AC} \times \overrightarrow {BD} } \right|$$

Solution:

The vector area of the quadrilateral can be written as

\begin{align}& \vec A\; = \overrightarrow {{\text{area}}\;\,(\Delta ABC)} + \overrightarrow {{\text{area}}\;\,(\Delta ACD)} \hfill \\\\&\quad= \frac{1}{2}(\overrightarrow {AB} \times \overrightarrow {AC} ) + \frac{1}{2}(\overrightarrow {AC} \times \overrightarrow {AD} )\left\{ {To{\text{ }}add{\text{ }}the{\text{ }}areas{\text{ }}their{\text{ }}directions{\text{ }}must{\text{ }}be{\text{ }}the{\text{ }}same} \right\} \hfill \\\\&\quad = \frac{1}{2}(\overrightarrow {AB} \times \overrightarrow {AC} + \overrightarrow {AC} \times \overrightarrow {AD} ) \hfill \\\\&\quad = \frac{1}{2}(\overrightarrow {AB} \times \overrightarrow {AC} - \overrightarrow {AD} \times \overrightarrow {AC} ) \hfill \\\\&\quad = \frac{1}{2}(\overrightarrow {AB} - \overrightarrow {AD} ) \times \overrightarrow {AC} \hfill \\\\&\quad= \frac{1}{2}\;\overrightarrow {DB} \times \overrightarrow {AC} \hfill \\ & \end{align}

Thus,

$A = \frac{1}{2}\left| {\overrightarrow {AC} \times \overrightarrow {BD} } \right|$

Example – 24

Find the perpendicular distance of  $$C(\vec c)$$ from the segment joining $$A(\vec a)\,\,and\,\,B(\vec b)$$, in terms of  $$\vec a,\vec b,\vec c,$$.

Solution:

Now,

\begin{align}&area(\Delta ABC) = \frac{1}{2}\left| {\overrightarrow {AB} \times \overrightarrow {AC} } \right| \hfill \\\\&\qquad\qquad\qquad = \frac{1}{2}\left| {(\vec b - \vec a) \times (\vec c - \vec a)} \right| \hfill \\\\&\qquad\qquad\qquad = \frac{1}{2}\left| {(\vec b \times \vec c - \vec b \times \vec a - \vec a \times \vec c)} \right| \hfill \\\\&\qquad\qquad\qquad= \frac{1}{2}\left| {(\vec a \times \vec b + \vec b \times \vec c + \vec c \times \vec a)} \right| \hfill \\\\ &\qquad\qquad\qquad\end{align}

But area $$(\Delta ABC)$$ also equals $$\frac{1}{2} \times AB \times d$$

$\Rightarrow \quad \frac{1}{2} \times AB \times d = \frac{1}{2}\left| {\vec a \times \vec b + \vec b \times \vec c + \vec c \times \vec a} \right|$

Since AB equals $$\left| {(\vec a - \vec b)} \right|$$ , we have

$d = \frac{{\left| {(\vec a \times \vec b + \vec b \times \vec c + \vec c \times \vec a)} \right|}}{{\left| {\vec a - \vec b} \right|}}$

Vectors
grade 11 | Questions Set 1
Vectors