How do you find the area of a triangle given two $2D$ vectors? Question: The absolute value of the determinant of a matrix computes the volume of the shape
defined by its row vectors. In $2D$, this is the area of the parallelogram. What is the area of a triangle with vectors $u = (3, 6)$, $v = (8, 10)$ as sides?
I found a lot of resources of finding the area with $3D$ vectors, but none for $2D$. In this case, I'm not able to follow the steps of finding the cross product first, then magnitude and finally, the area of the triangle. How do I solve this?
 A: Well, the determinant of $2$ vectors $\langle a,b\rangle$ and $\langle c,d\rangle$ is $ad-bc$. That gives you the area of the parallelogram they span, so take half thereof. So I think the area of your triangle is $9$ ($-9$ technically but just take it positive).
A: Here is a geometric interpretation:

$$A=|bc-\frac{(c-a)(b-d)+ab+cd}{2}|=|bc-\frac{bc-cd-ab+ad+ab+cd}{2}|=|\frac{bc-ad}{2}|$$
A: We have a triangle whose sides are the vectors: $u$ and $v$
Let the magnitude of these vectors be $a$ and $b$
Let the angle between the vectors $u$ and $v$ be $\theta$
Then the altitude of this triangle will be $a\sin\theta$
The formula for the area of triangle is $\frac{1}{2}×b×h$ where $b$ is the base and $h$ is altitude.
Substituting the values, we get
$$\frac{1}{2}(u×v)$$ where $×$ is cross product
After solving, you will get a vector $-9\hat{k}$, with $\hat{k}$ being unit normal vector in the direction of $z$-axis
Calculating the magnitude of this vector, we get the area of the triangle, $9$ square units
A: For two 2D vectors: $\vec{a}$ and $\vec{b}$ area of the triangle is $\frac{1}{2} ||\vec{a} \times \vec{b}||$, because $\vec{a} \times \vec{b}$ is a vector perpendicullar to $\vec{a}$ and $\vec{b}$, which length is equal to area of parallelogram created by $\vec{a} + \vec{b}$ and $\vec{b} + \vec{a}$ (if you visualize addition of vectors as putting them tip-to-tail). So you extract the length, and multiply it by $\frac{1}{2}$ to get the area of the triangle.
Then let $\vec{a} = x\hat{i} + y\hat{j} + 0\hat{k}$ and $\vec{b}=u\hat{i} + v\hat{j} + 0\hat{k}$, so $\vec{a} \times \vec{b}$ is
$$det \left(\begin{bmatrix}
\hat{i} & \hat{j} & \hat{k}\\ 
x & y & 0\\ 
u & v & 0 
\end{bmatrix} \right ) = (xv-uy)\hat{k}$$
So $\frac{1}{2} ||\vec{a} \times \vec{b}|| = \frac{1}{2} |(xv-yu)|$
For $\vec{a} = 3\hat{i} + 6\hat{j}$ and $\vec{b} = 8\hat{i} + 10\hat{j}$ the area of the triangle is $\frac{1}{2} |(3\cdot 10 - 6\cdot8)| = \frac{1}{2}|(30-48)|=\frac{1}{2}|-18| = 9$
SO the general formula for the area of the triangle between two 2D vectors $\vec{a}$ and $\vec{b}$ is:
$$Area = \frac{1}{2} |a_{x}b_{y}-a_{y}b_{x}|$$
