Question

Find the vector area of a triangle OAB where OA = a, OB = b, and they are inclined at an angle $\theta$. Also, find the vector area of a triangle whose vertices are the points A, B, and C.
(this question has multiple correct options)
A) $\dfrac{1}{2}(a\times b)$
B) $\dfrac{1}{2}(b\times a)$
C) \[\dfrac{1}{2}\left( a\times b+b\times c+c\times a \right)\]
D) None of these

Answer 1

Hint: To solve this question, what we will do is first we will take O as the origin. Now, let the position vectors of A, B, and C be a, b, and c respectively and we will find the position vector of BC= c−b and BA= a−b and then using the cross product of vectors, we will find out the area of triangle.

Complete step-by-step solution:
We know that, $a\times b=ab \sin \theta$
Again, we know that the area of a triangle with sides a and b inclined at angle θ, is given by \[\dfrac{\text{1}}{\text{2}}\text{ab }\sin \theta\]
Therefore area of $\Delta OAB$
\[\begin{align}
  & \text{=}\dfrac{\text{1}}{\text{2}}\text{OA}\text{.OB }\sin \theta \\
 & \text{=}\dfrac{\text{1}}{\text{2}}\text{ab }\sin \theta \\
 & \text{=}\dfrac{1}{2}\text{a }\!\!\times\!\!\text{ b} \\
\end{align}\]
Hence vector area of triangle OAB is \[\dfrac{1}{2}\text{a }\!\!\times\!\!\text{ b}\]
Therefore option (A) is correct.

Now take O as the origin.
Also, let the position vectors of A, B, and C be a, b and c respectively.
Therefore,
BC= c−b and BA= a−b.
∴ Vector area of $\Delta ABC$ is ​
\[\begin{align}
  & =\dfrac{1}{2}BC\times BA~ \\
 & =\dfrac{1}{2}(c-b)\times (a-b) \\
 & =\dfrac{1}{2}\left[ (c\times a)-(c\times b)-(b\times a)+(b\times b) \right] \\
& =\dfrac{1}{2}(a\times b+b\times c+c\times a)~\ldots \ldots [\because b\times b=0~and~-(c\times b)=b\times c]~ \\
\end{align}\]
Therefore vector area of a triangle whose vertices are the points A, B, and C is given by \[\dfrac{1}{2}(a\times b+b\times c+c\times a)\]
Hence, the correct option is (C).

Note: We can also use an alternative method to solve this question. Now, note that the area of a triangle with sides a and b inclined at angle θ, is given by \[\dfrac{\text{1}}{\text{2}}\text{absin }\!\!\theta\!\!\text{ }\] and also that, the vector area of a triangle with sides a and b is equal to half of the area of a parallelogram whose adjacent sides are a and b ,i.e. \[\dfrac{1}{2}\text{a }\!\!\times\!\!\text{ b}\].