Question

Let a=<-4,-3,5> Find a unit vector in the same direction as having positive first coordinate.

Answer 1

Answer:

The value is $u = < \frac{2\sqrt{2}}{10} , \frac{3\sqrt{2}}{10} , - \frac{\sqrt{2}}{2}>$

Step-by-step explanation:

From the question we are told that

The vector is a=<-4,-3,5>

Generally the unit vector is $u = ay$

Here y represent the y-coordinate

So

$u =y$

=> $u =$

Generally the resultant of a unit vector is 1

So

$|u| = \sqrt{ (-4y)^2 + (-3y)^2 + (5y)^2} = 1$

Hence

$|u| = \sqrt{ 16y^2 + 9y^2 + 25y^2} = 1$

Taking the square of both sides

$16y^2 + 9y^2 + 25y^2 = 1$

=> $50y^2 = 1$

=> $y = \pm \frac{1}{\sqrt{50}}$

=> $y = \pm \frac{1}{5 \sqrt{2}}$

Rationalizing

=> $y = \pm \frac{\sqrt{2}}{10}$

Given that the first coordinate is positive

$y = - \frac{\sqrt{2}}{10}$

Hence the unit vector is

$u =$

=> $u = < \frac{2\sqrt{2}}{10} , \frac{3\sqrt{2}}{10} , - \frac{\sqrt{2}}{2}>$