Friday, 8 August 2014

CHAPTER 22 -Worked Out Examples

     Example: 23   

Prove that the area of a quadrilateral ABCD can be given by \dfrac{1}{2}\left| {\overrightarrow {AC}  \times \overrightarrow {BD} } \right|
Solution: 23

The vector area of the quadrilateral can be written as
\vec A = \overrightarrow {{\rm{area}}\;\,(\Delta ABC)}  + \overrightarrow {{\rm{area}}\;\,(\Delta ACD)}
 = \dfrac{1}{2}(\overrightarrow {AB}  \times \overrightarrow {AC} ) + \dfrac{1}{2}(\overrightarrow {AC}  \times \overrightarrow {AD} ) To add the areas their directions must be the same
 = \dfrac{1}{2}(\overrightarrow {AB}  \times \overrightarrow {AC}  + \overrightarrow {AC}  \times \overrightarrow {AD} )
 = \dfrac{1}{2}(\overrightarrow {AB}  \times \overrightarrow {AC}  - \overrightarrow {AD}  \times \overrightarrow {AC} )
 = \dfrac{1}{2}(\overrightarrow {AB}  - \overrightarrow {AD} ) \times \overrightarrow {AC}
 = \dfrac{1}{2}\;\overrightarrow {DB}  \times \overrightarrow {AC}
A = \dfrac{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: 24

area (\Delta ABC) = \dfrac{1}{2}\left| {\overrightarrow {AB}  \times \overrightarrow {AC} } \right|
 = \dfrac{1}{2}\left| {(\vec b - \vec a) \times (\vec c - \vec a)} \right|
 = \dfrac{1}{2}\left| {(\vec b \times \vec c - \vec b \times \vec a - \vec a \times \vec c)} \right|
 = \dfrac{1}{2}\left| {(\vec a \times \vec b + \vec b \times \vec c + \vec c \times \vec a)} \right|
But area (\Delta ABC) also equals \dfrac{1}{2} \times AB \times d
 \Rightarrow  \,\,\,\, \dfrac{1}{2} \times AB \times d = \dfrac{1}{2}\left| {\vec a \times \vec b + \vec b \times \vec c + \vec c \times \vec a} \right|
Since AB equals , we have
d = \dfrac{{\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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