Question

Prove that the vector area of triangle whose vertices are a b c is 1/2(b*c+c*a+a*b)

Answer 1

Answer:

Area of triangle ABC

$=\frac{1}{2}[\vec{a}*\vec{b}+\vec{b}*\vec{c}+\vec{c}*\vec{a}]$

Step-by-step explanation:

Let O be the origin.

Let the position vectors of the vertices A, B and C be $\vec{a}, \vec{b}\:and\:\vec{c}$ respectively.

That is

$\vec{OA}=\vec{a}$

$\vec{OB}=\vec{b}$

$\vec{OC}=\vec{c}$

Now

$\vec{AB}=\vec{OB}-\vec{OA}=\vec{b}-\vec{a}$

$\vec{AC}=\vec{OC}-\vec{OA}=\vec{c}-\vec{a}$

Vector Area of tirangle ABC

$=\frac{1}{2}[\vec{AB}*\vec{AC}]$

$=\frac{1}{2}[(\vec{b}-\vec{a})*(\vec{c}-\vec{a})]$

$=\frac{1}{2}[\vec{b}*\vec{c}-\vec{b}*\vec{a}-\vec{a}*\vec{c}+\vec{a}*\vec{a}]$

$=\frac{1}{2}[\vec{b}*\vec{c}+\vec{a}*\vec{b}+\vec{c}*\vec{a}]$

$=\frac{1}{2}[\vec{a}*\vec{b}+\vec{b}*\vec{c}+\vec{c}*\vec{a}]$

Answer 2

Let O be the point of reference.

Let the position vectors of the vertices A, B and C of the triangle ABC be a, b and c respectively with respect to O.

Vector area of triangle ABC is 1/2(AB x BC) = 1/2[(OB-OA)x(OC-OB)]

= 1/2[(b-a)x(c-b)] = 1/2(b x c -b x b - a x c + a x b)

= 1/2( a x b + b x c + c x a)