Question

Any vector can be written as a unit vector multiplied by the magnitude of the vector (a positive scalar). Write each of the following vectors as the magnitude of the vector times the appropriate unit vector:
a. ( 0.00330, 0, -0.00330)
b. (0, -676, 0)
c. (0.00316, 0, -0.00316)

Answer 1

The correct magnitude of the vector times the appropriate unit vector of given vectors are given as follows:

a. <0.708, 0, -0.708> $*4.66*10^{-3}$

b. <0, -1, 0>*676

c. $\overrightarrow{C} = <0.707, 0, -0.707>*4.47*10^2$

The general vector can be represented as:

$\underset{A}{\rightarrow}$ = (a, b, c) ......1

the magnitude $| \right\underset{A}{\rightarrow} |$ can be obtained as :

$| \right\underset{A}{\rightarrow} | = \sqrt{a^{2}+ b^{2} +c^{2} }$ .....2

and unit vector would be:

$\hat{u}_A=\dfrac{\overrightarrow{A}}{|\overrightarrow{A}|}$.......3

Solution:

a) ( 0.00330, 0, -0.00330)

here,

$\underset{A}{\rightarrow} = ( 0.00330, 0, -0.00330)$

applying this magnitude to equation two:

$| \right\underset{A}{\rightarrow} | = \sqrt{0.00330^{2}+ 0^{2} +(-0.00330)^{2} }$ = $4.66*10^{-3}$

the unit vector would be:

$\hat{u}_A=\dfrac{\overrightarrow{A}}{|\overrightarrow{A}|}$

$\hat{u}_A= < \frac{0.00330}{4.66*10^{-3}}, \frac{0}{4.66*10^{-3}}, \frac{-0.00330}{4.66*10^{-3}}>$

$\hat{u}_A= <708, 0, -708>$

Thus, the vector ${\overrightarrow{A}$ would be - <0.708, 0, -0.708> $*4.66*10^{-3}$

B) (0, -676, 0)

Solving simailarly like A),

Vector $| \right\underset{B}{\rightarrow} | = \sqrt{0^{2}+ -676^{2} +0^{2} }$ = 676

the unit vector would be:

$\hat{u}_B=\dfrac{\overrightarrow{B}}{|\overrightarrow{B}|}\\\hat{u}_B= < \frac{0}{676}, \frac{-676}{676}, \frac{0}{676}>$

Thus,  the vector $\overrightarrow{B}$ = <0, -1, 0>*676

C) (0.00316, 0, -0.00316)

Attempting the third vector similarly we will get magnitude:

$| \right\underset{C}{\rightarrow} | = \sqrt{0.00316^{2}+ 0^{2} +(-0.00316)^{2} }$ = $*4.47*10^{-3}$

Then,

$\hat{u}_C=\dfrac{\overrightarrow{C}}{|\overrightarrow{C}|}\\\hat{u}_C= < \frac{0.00316}{4.47*10^2}, \frac{0}{4.47*10^2}, \frac{0.00316}{4.47*10^2}>$

The vector $\overrightarrow{C} = <0.707, 0, -0.707>*4.47*10^2$

Learn more about vector unit:

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