Question

If a and b are two vectors along two sides of a triangle whose area is 20 squares units.Find the area of another triangle with vectors along its two sides are 3a+b and 2a+3b

Answer 1

$\textbf{Given:}$

$\text{ \vec{a} and \vec{b} are two vectors along two}$

$\text{sides of a triangle whose area is 20 square units}$

$\textbf{To find:}$

$\text{The area of another triangle with vectors along}$

$\text{its two sides are 3\vec{a}+\vec{b} and 2\vec{a}+3\vec{b} }$

$\textbf{Soluition:}$

$\text{Area of triangle having sides \vec{a} and \vec{b} is 20 square units}$

$\implies\,\dfrac{1}{2}|\vec{a}{\times}\vec{b}|=20$

$\implies|\vec{a}{\times}\vec{b}|=40$

$\text{Now,}$

$\text{Area of triangle having sides 3\vec{a}+\vec{b} and 2\vec{a}+3\vec{b} }$

$=\dfrac{1}{2}|(3\vec{a}+\vec{b}){\times}(\vec{a}+3\vec{b})|$

$=\dfrac{1}{2}|3(\vec{a}{\times}\vec{a})+9(\vec{a}{\times}\vec{b})+(\vec{b}{\times}\vec{a})+3(\vec{b}{\times}\vec{b})|$

$=\dfrac{1}{2}|3(\vec{0})+9(\vec{a}{\times}\vec{b})-(\vec{a}{\times}\vec{b})+3(\vec{0})|$

$=\dfrac{1}{2}|9(\vec{a}{\times}\vec{b})-(\vec{a}{\times}\vec{b})|$

$=\dfrac{1}{2}|8(\vec{a}{\times}\vec{b})|$

$=\dfrac{1}{2}[8|\vec{a}{\times}\vec{b}|]$

$=4|\vec{a}{\times}\vec{b}|$

$=4(40)$

$=160\;\text{square units}$

$\textbf{Answer:}$

$\textbf{The area of another triangle with vectors along}$

$\textbf{its two sides are 3\vec{a}+\vec{b} and 2\vec{a}+3\vec{b} is 160 square units}$