Finding Unit Vectors in Same Position

Question

For each different vector, find a unit vector in the same direction.

 

​a.​u1 = [ −5, 3]

​b.​u2 = ​with point M = (2, −8) and point N = (−5, 4).

​c.​u3 is the vector with magnitude 6 and direction angle of 225° .

Answer 1

a) u₁ = [-5, 3]

First, find the magnitude:

|u₁| = √[(-5)² + 3²]

= √(25 + 9)

= √34

Now, divide each component by √34.

Unit vector = <-5/√34, 3/√34>

b) Vector from M = (2, -8) to N = (-5, 4)

First, find the vector from M to N:

u₂ = N - M

= [-5 - 2, 4 - (-8)]

= [-7, 12]

Now, find its magnitude:

|u₂| = √[(-7)² + 12²]

= √(49 + 144)

= √193

Therefore, the unit vector is:

Unit vector = <-7/√193, 12/√193>

c) Magnitude 6 and direction angle 225°

A unit vector with direction angle θ is:

[cos θ, sin θ]

Let's substitute θ = 225º.

Since 225º is in Quadrant III:

cos 225º = -√2/2

Therefore,

Unit vector = <-√2/2, -√2/2>

Note that the given magnitude of 6 does not affect the unit vector because every unit vector has a magnitude of 1.

Answer 2

A unit vector is a vector with a magnitude of 1. To turn any vector into a unit vector in the same direction involves finding the magnitude, then dividing each component of the vector by the magnitude.

a) u1=<-5,3> the magnitude = √[ (-5)²+(3)² ] = √ (25+9) = √34

so the unit vector in the same direction is < -5/√34, 3/√34 >

b) u2 = < 4-(-8) , (-5) - 2 > = <12, -7> the magnitude is √[ (12)²+(-7)² ] √(144+49) = √193

so the unit vector in the same direction as <12, -7> is

<12/√193, -7/√193 >

c) u3 is actually deceptively simple. They gave you the direction already, so the unit vector in the

same direction is a vector with magnitude of 1, and a direction angle of 225 degrees. If they

want this in x-y-component form, then we would need to compute: 1 ·cos225 = x-component and

1·sin225 = y-component or < -√2/2, -√2/2 >