Answer:
The value is [tex]u = < \frac{2\sqrt{2}}{10} , \frac{3\sqrt{2}}{10} , - \frac{\sqrt{2}}{2}>[/tex]
Step-by-step explanation:
From the question we are told that
The vector is a=<-4,-3,5>
Generally the unit vector is [tex]u = ay[/tex]
Here y represent the y-coordinate
So
[tex]u =y <-4,-3,5>[/tex]
=> [tex]u = <-4y,-3y,5y>[/tex]
Generally the resultant of a unit vector is 1
So
[tex]|u| = \sqrt{ (-4y)^2 + (-3y)^2 + (5y)^2} = 1 [/tex]
Hence
[tex]|u| = \sqrt{ 16y^2 + 9y^2 + 25y^2} = 1 [/tex]
Taking the square of both sides
[tex] 16y^2 + 9y^2 + 25y^2 = 1 [/tex]
=> [tex] 50y^2 = 1 [/tex]
=> [tex] y = \pm \frac{1}{\sqrt{50}} [/tex]
=> [tex] y = \pm \frac{1}{5 \sqrt{2}} [/tex]
Rationalizing
=> [tex] y = \pm \frac{\sqrt{2}}{10} [/tex]
Given that the first coordinate is positive
[tex] y = - \frac{\sqrt{2}}{10} [/tex]
Hence the unit vector is
[tex]u = <-4(- \frac{\sqrt{2}}{10}),-3(- \frac{\sqrt{2}}{10}),5(- \frac{\sqrt{2}}{10})>[/tex]
=> [tex]u = < \frac{2\sqrt{2}}{10} , \frac{3\sqrt{2}}{10} , - \frac{\sqrt{2}}{2}>[/tex]
