Decompose a Vector

**RedKMan** · August 23rd 2009, 07:01 AM

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.

**earboth** · August 23rd 2009, 07:08 AM

Originally Posted by RedKMan

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 $\vec n = (-2,1)$ you'll get the orthogonal vector as $\vec o = (1,2)$ since $\vec n \cdot \vec o = 0$ .

Now you are asked to find values of $s, t \in \mathbb{R}$ such that

$s \cdot (-2,1) + t \cdot (1,2) = (0, -9.8)$

Can you handle it from here?

**Soroban** · August 23rd 2009, 07:54 AM

Hello, RedKMan!

Let $\vec n \,=\,\langle-2, 1\rangle$
Decompose the vector $\vec g \,=\, \langle0, -9.8\rangle$ into the sum of two orthogonal vectors,
one parallel to $\vec n$ , the other orthogonal to $\vec n$

Are you familiar with "projections"?

Given vectors $\vec u, \vec v$

The projection of u onto v is given by: . $\overrightarrow{w_1} \;=\;\frac{\vec u \cdot\vec v}{|\vec v|^2}\,\vec v$

. . This is the component of $\vec u$ that is parallel to $\vec v.$

And: . $\overrightarrow{w_2} \;=\;\vec u -\overrightarrow w_1$

. . This is the component of $\vec u$ that is orthogonal to $\vec v.$

We have: . $\vec u \:=\:\langle 0,-9.8\rangle,\;\vec v \:=\:\langle-2,1\rangle$

. . $\overrightarrow{w_1} \;=\;\frac{\langle0,-9.8\rangle\cdot\langle-2,1\rangle}{(\sqrt{(-2)^2 + 1^2})^2}\,\langle-2,1\rangle \;=\;\frac{-9.8}{5}\langle-2,1\rangle \;=\;\langle3.92,-1.96\rangle$

. . $\overrightarrow{w_2} \;=\;\langle 0.-9.8\rangle - \langle3.92,-1.96\rangle \;=\;\langle-3.92,-7.84\rangle$

Therefore: . $\boxed{\;\vec g \;\;=\;\;\langle 3.92,-1.96\rangle + \langle -3.92,-7.84\rangle\;}$

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