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title: Untitled

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![](https://i.imgur.com/JGj9JRF.png)

$$
Area(\vec v_2,b\cdot\vec v_1+c\cdot\vec w) = b\cdot Area(\vec v_2,\vec v_1)+c\cdot Area(\vec v_2,\vec w)
$$

$h_1$ is the length of the hight of the red area

$h_2$ is the length of the height of the yellow area

$h_3$ is the length of the height of the green area.

$\vec h_1$ is the length of the hight of the red area

$\vec h_2$ is the length of the height of the yellow area

$\vec h_3$ is the length of the height of the green area.

**We want to show $h_3=b\cdot h_1+c\cdot h_2$ correct?**

Because
$$
Area(\vec v_2,b\cdot\vec v_1+c\cdot\vec w) = ||\vec v_2||\cdot h_3
$$


$$
\vec h_1=\vec v_1-\frac{\vec v_1\cdot \vec v_2}{\vec v_2\cdot \vec v_2}\vec v_2
$$

$$
\vec h_2=\vec w-\frac{\vec w\cdot \vec v_2}{\vec v_2\cdot \vec v_2}\vec v_2
$$

$$
\vec h_3=(b\cdot v_1+c\cdot \vec w) -\frac{(b\cdot v_1+c\cdot \vec w)\cdot \vec v_2}{\vec v_2\cdot \vec v_2}\vec v_2
$$

Because $\vec h_1\cdot \vec v_2=\vec h_2\cdot \vec v_2=\vec h_3\cdot \vec v_2=0$, then all those vectors $\vec h_1$, $\vec h_2,\vec h_3$ are colinear, meaning that $\cos \theta(\vec h_1,\vec h_2)=1$.

Take dot product with $\vec h_1$, we have
$$
b\cdot \vec h_1\cdot \vec h_1+c\cdot \vec h_2\cdot \vec h_1 = \vec h_3\cdot \vec h_1.
$$

$$
\vec h_2\cdot \vec h_1=||\vec h_2||||\vec h_1||\cos(\theta(\vec h_1,\vec h_2))
$$

Therefore



Therefore, 
$$
h_3=b\cdot h_1+c\cdot h_2.
$$

From expressions, we see that
$$
\vec h_3=c\cdot \vec h_2+b\cdot \vec h_1
$$



$$
h_1=||\vec h_1||
$$