How do you find two unit vectors that make an angle of 60° with v = ‹3, 4›?

Solution:

The vector \( \underset{v}{\rightarrow} \) can be written as (v₁, v₂).

Let the required unit vector \( \underset{u}{\rightarrow} \) with coordinates (x,y) and magnitude 1(unit vector).

The magnitude of 1 implies (x)² + (y)² = 1 ----------->(1)

We know that v₁ = 3, v₂ = 4. Therefore the magnitude of the vector is √[(3)² + (4)²] = 5

Given that the angle between \( \underset{v}{\rightarrow} \) and \( \underset{u}{\rightarrow} \) is 𝛑/3 we take the two vectors’ dot product:

\( \underset{u}{\rightarrow} \).\( \underset{v}{\rightarrow} \) = I u I.I v I Cos(𝛑/3)

(x,y).(v₁, v₂) = (√(x)² + (y)²).( √(3)² + (4)²)Cos(𝛑/3)

(x,y).(3,4) = 1.5.(1/2)

3x + 4y = 5/2

3x = 5/2 - 4y

x = (1/3)(5/2 - 4y) ----------->(2)

Substituting (2) in (1)

[(1/3)(5/2 - 4y)]² + (y)² = 1

(1/9)(25/4 + 16y² -20y) + y² = 1

25/4 + 16y² -20y + 9 y² = 9

25y² - 20y + 25/4 - 9 = 0

25y² - 20y - 11/4 = 0

100y² - 80y - 11 = 0

The roots of the equation are given by the quadratic formula:

y =(-b ± √b² - 4ac)/2a

Where b = -80; a = 100; c = 11

y= (80 ± √6400 + 4400)/200

= 80/200 ± √10800/200

= 2/5 ± (10/200)√3 × 4 x 3 x 3

= 2/5 ± (1/10) (3√3)

Hence the y = 2/5 + 3√3/10 and 2/5 - 3√3/10

If y = 2/5 + 3√3/10

Then x = (1/3)(5/2 - 4(2/5 + 3√3/10)) = 3/10 - (2/5)√3

If y = 2/5 - 3√3/10

Then x = (1/3)(5/2 - 4(2/5 - 3√3/10)) = 3/10 + (2/5)√3

The two unit vectors are :

[3/10 - (2/5)√3, 2/5 + 3√3/10] and [3/10 + (2/5)√3, 2/5 - 3√3/10)

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How do you find two unit vectors that make an angle of 60° with v = ‹3, 4›?

Summary:

The two unit vectors that make an angle of 60° with v = ‹3, 4› are [(3/10 - (2/5)√3), (2/5 + 3√3/10)] and [(3/10 + (2/5)√3), (2/5 - 3√3/10)]. The two vectors are unit vectors, can be verified by squaring x and y coordinates and adding them. Their sum is 1.