Mathematics: Find the orthogonal decomposition of vector b = 9, 0, 0 with respect to vector a = 4, −5, 0 . (Your instructors prefer angle bracket notation for vectors.)

44 is the answer because the 6 is higher than 5 so you round up...

Find the orthogonal decomposition of vector b =

kingofthesea88

01.10.2020

The orthogonal componentes of $\vec b$ with respect to $\vec a$ are $\vec b_{\parallel} = \left$ and $\vec b_{\perp} = \left$ , respectively.

Step-by-step explanation:

Let be $\vec b =\left$ and $\vec a = \left$ , the component of $\vec b$ parallel to is found by the following expression:

$\vec b_{\parallel}=(\vec b \bullet \hat{a})\cdot \hat{a}$

Where $\bullet$ means for dot product and $\hat{a}$ is the unit vector of $\vec a$ , which is determined by:

$\hat{a} =\frac{\vec a}{\|\vec a\|}$

Where $\|\vec a \|$ is the norm of $\vec a$ , which can be found by Pythagorean Theorem.

The vector of $\vec b$ parallel to vector $\vec a$ is now determined:

$\|\vec a \|=\sqrt{4^{2}+(-5)^{2}+(0)^{2}}$

$\|\vec a\| = \sqrt{41}$

$\hat{a} = \frac{\left}{\sqrt{41}}$

$\hat{a} = \left$

$\vec b \bullet \hat{a} =(9)\cdot \left(\frac{4}{\sqrt{41}} \right)+(0)\cdot \left(-\frac{5}{\sqrt{41}} \right)+(0)\cdot (0)$

$\vec b \bullet \hat{a} = \frac{36}{\sqrt{41}}$

$\vec b_{\parallel} = \frac{36}{\sqrt{41}}\cdot \left$

$\vec b_{\parallel} = \left$

Now, the component of $\vec b$ perpendicular to $\vec a$ is obtained by vector subtraction:

$\vec b_{\perp} = \vec b -\vec b_{\parallel}$

Given that $\vec b =\left$ and $\vec b_{\parallel} = \left$ , the expression for the remaining component is:

$\vec b_{\perp} = \left-\left$

$\vec b_{\perp} = \left$

The orthogonal componentes of $\vec b$ with respect to $\vec a$ are $\vec b_{\parallel} = \left$ and $\vec b_{\perp} = \left$ , respectively.

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