Let n = (-2, 1). Decompose the vector g = (0, -9.8) into the sum of two orthogonal vectors, one parallel to n and the other orthogonal to n.
I'm not sure how to go about doing this.
From $\displaystyle \vec n = (-2,1)$ you'll get the orthogonal vector as $\displaystyle \vec o = (1,2)$ since $\displaystyle \vec n \cdot \vec o = 0$.
Now you are asked to find values of $\displaystyle s, t \in \mathbb{R}$ such that
$\displaystyle s \cdot (-2,1) + t \cdot (1,2) = (0, -9.8)$
Can you handle it from here?
Hello, RedKMan!
Are you familiar with "projections"?Let $\displaystyle \vec n \,=\,\langle-2, 1\rangle$
Decompose the vector $\displaystyle \vec g \,=\, \langle0, -9.8\rangle$ into the sum of two orthogonal vectors,
one parallel to $\displaystyle \vec n$, the other orthogonal to $\displaystyle \vec n$
Given vectors $\displaystyle \vec u, \vec v$
The projection of u onto v is given by: .$\displaystyle \overrightarrow{w_1} \;=\;\frac{\vec u \cdot\vec v}{|\vec v|^2}\,\vec v$
. . This is the component of $\displaystyle \vec u$ that is parallel to $\displaystyle \vec v.$
And: .$\displaystyle \overrightarrow{w_2} \;=\;\vec u -\overrightarrow w_1$
. . This is the component of $\displaystyle \vec u$ that is orthogonal to $\displaystyle \vec v.$
We have: .$\displaystyle \vec u \:=\:\langle 0,-9.8\rangle,\;\vec v \:=\:\langle-2,1\rangle$
. . $\displaystyle \overrightarrow{w_1} \;=\;\frac{\langle0,-9.8\rangle\cdot\langle-2,1\rangle}{(\sqrt{(-2)^2 + 1^2})^2}\,\langle-2,1\rangle \;=\;\frac{-9.8}{5}\langle-2,1\rangle \;=\;\langle3.92,-1.96\rangle $
. . $\displaystyle \overrightarrow{w_2} \;=\;\langle 0.-9.8\rangle - \langle3.92,-1.96\rangle \;=\;\langle-3.92,-7.84\rangle $
Therefore: .$\displaystyle \boxed{\;\vec g \;\;=\;\;\langle 3.92,-1.96\rangle + \langle -3.92,-7.84\rangle\;} $