Answer:
[tex]b = (q,r , 5q-2\sqrt{26} )[/tex]
Step-by-step explanation:
From the question we are told that
The vector given is
[tex]a = (5,0,-1)[/tex]
Also [tex]comp_a b = 2[/tex]
Generally
[tex]comp_a b = \frac{a \cdot b}{|a|}[/tex]
Here [tex]|a|[/tex] is the magnitude of a which is mathematically represented as
[tex]|a| = \sqrt{5^2 + 0^2 +(-1)^2 }[/tex]
=> [tex]|a| = \sqrt{26}[/tex]
b is vector which we will assume to have the following parameters
[tex]b = (q , r , x)[/tex]
So
[tex]comp_a b = \frac{(5,0,-1) \cdot (q,r,x)}{\sqrt{26} } = 2[/tex]
=> [tex]\frac{5q + 0 -x}{\sqrt{26} } = 2[/tex]
=> [tex]x = 5q-2\sqrt{26}[/tex]
Hence the vector be can be mathematically represented as
[tex]b = (q,r , 5q-2\sqrt{26} )[/tex]
This means that the vector b is more than one value since it is made up of variable
This means that if
[tex]q = 1\\r=1\\x =1[/tex]
Then
[tex]b = (1,1,5-2\sqrt{26} )[/tex]
A quantity has magnitude but not position, that's why in this the vector [tex]b =(0,0, -2\sqrt{26}).[/tex]
If[tex]a =(5,0,-1)[/tex] Our goal is to determine a vector b that will allow comp to work properly [tex]_{a}b = 2.[/tex]
Allow the vector to flow naturally [tex]b =(x, y,z)[/tex]
Recalling
comp[tex]_{a}b =\frac{a.b}{|a|}[/tex]
Therefore,
comp [tex]_{a}b= 2 \\\\[/tex]
[tex]\to \frac{a.b}{|a|}= 2\\\\[/tex]
[tex]\to \frac{(5,0,-1)\cdot (x,y,z)}{ |(5,0,-1)|} =2\\\\\to \frac{5x+0-z}{5^2+0+(-1)^2}=2\\\\ \to \frac{5x-z}{\sqrt{26}}= 2 \\\\\to 5x-z = 2\sqrt{26}\\\\ \to z=5x-2\sqrt{26}\\\\[/tex]
As a result, the vector b can be expressed as,
[tex]\to b =(x,y,z) =(x, y, 5x-2\sqrt{26}) \\\\[/tex]
As a result, this aforementioned form can be found in an endless number of vectors.
One of them is [tex]b =(0,0, -2\sqrt{26}).[/tex]
Find out more about the vector here:
brainly.com/question/13322477
