Unit Vector Perp. to a: Solving Problem

[WeiLoong]

Homework Statement

if a =3i-4j
find a unit vector perpendicular a

Homework Equations

Vector

The Attempt at a Solution

<a> = <3 , -4>
<n> = <x, y> : x² + y² = 1
<a>•<n> = 3x-4y = 0

y = (3/4)x
x² + (9/16)x² = 1
25x² = 16
x = -4/5, 4/5
y = -3/5, 3/5

There are two unit vectors normal to a which have opposite directions:
n =(4i + 3j)/5
and
n = -(4i+3j)/5
This is the answer that given by my textbook. Can somebody explain to me the 2nd step <n> = <x, y> : x² + y² = 1
I do not understand why it use x^2+y^2=1 as this is circle equation.

 

[jedishrfu]

Its saying to define a vector n with components x,y such that x^2 + y^2 =1 in other words that n is a unit vector with components x and y.

Dotting it with the known vector gives you the other equation so that you now have two equations in x and y and thus can solve for x and y

 

[HallsofIvy]

Homework Statement

if a =3i-4j
find a unit vector perpendicular a

Homework Equations

Vector

The Attempt at a Solution

<a> = <3 , -4>
<n> = <x, y> : x² + y² = 1
<a>•<n> = 3x-4y = 0

y = (3/4)x
x² + (9/16)x² = 1
25x² = 16
x = -4/5, 4/5
y = -3/5, 3/5

There are two unit vectors normal to a which have opposite directions:
n =(4i + 3j)/5
and
n = -(4i+3j)/5
This is the answer that given by my textbook. Can somebody explain to me the 2nd step <n> = <x, y> : x² + y² = 1
I do not understand why it use x^2+y^2=1 as this is circle equation.

Yes, it is a circle centered at the origin with radius 1. So any unit vector with "tail" at the origin has its "tip" on that circle. Any unit vector <x, y> satisfies x² + y² = 1.