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

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

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Vectors
grade 11 | Questions Set 1
Vectors
grade 11 | Answers Set 1
Vectors
grade 11 | Questions Set 2
Vectors
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Three Dimensional Geometry
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