find two unit vectors that make an angle of 60° with v = 4, 3

The two unit vectors that makes an angle of  [tex]60^{0}[/tex] with [tex]v=4,3[/tex] are (-0.3928, 0.9196) and (0.9928, -0.1196).

Given that the angle made by unit vectors is [tex]60^{0}[/tex] with [tex]v=4,3[/tex].

The magnitude value of the v is [tex]\sqrt{x^{2} +y^{2} } =\sqrt{3^{2} +4^{2} }[/tex].

Consider that the unit vector as [tex]a= (x,y)[/tex], where its magnitude is [tex]|a|=x^{2} +y^{2} =1[/tex] ... (1)

Taking the dot product for the unit vector and v, the equation becomes,

[tex]a \cdot v=|a||v| \cos \emptyset \\[/tex]

Substitute the known values, we get

[tex](x, y) \cdot(3,4)=1 \cdot \sqrt{3^2+4^2 } \cos 60^{\circ} \\[/tex]

[tex]3 x+4 y=1 \cdot \sqrt{(9+16) }(\frac{1}{2} ) \\[/tex]

[tex]3 x+4 y=\frac{5}{2}[/tex]

[tex]& 4 y=\frac{5}{2} -3 x \\[/tex]

[tex]& y=\frac{1}{4} (2.5-3 x)[/tex] ... (2)

Substituting (2) in (1) to find the value of x, then the equation becomes,

[tex]x^2+((2.5-3 x) / 4)^2=1 \\[/tex]

[tex]16 x^2+6.25+9 x^2-15 x=16 \\[/tex]

[tex]25 x^2-15 x-9.75=0[/tex]

By using the quadratic formula [tex]x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a} \\[/tex], we have [tex]a=25, b=-15,[/tex] and [tex]c=-9.75[/tex].

[tex]x=\frac{-(-15) \pm \sqrt{(-15)^2-4(25)(-9.75)}}{2(25)} \\[/tex]

[tex]x=\frac{15 \pm \sqrt{1200}}{50} \\[/tex]

The first root is given by,

[tex]x=\frac{15+34.64}{50} \\[/tex]

[tex]x=\frac{49.64}{50} \\[/tex]

[tex]x=0.9928 \\[/tex]

The second root is given by,

[tex]& x=\frac{15-34.64}{50} \\[/tex]

[tex]& x=\frac{-19.64}{50} \\[/tex]

[tex]& x=-0.3928[/tex]

Thus, the values of x are 0.9928 and -0.3928.

Substitute the x-values in equation (2), and we get

[tex]y = \frac{(2.5 - 3(0.9928))}{4}[/tex]

[tex]y = -0.1196[/tex]

[tex]y = \frac{(2.5 - 3(-0.3928))}{4}[/tex]

[tex]y = 0.9196[/tex]

Thus, the values of y are -0.1196 and 0.9196.

Therefore, the two unit vectors that makes an angle of  [tex]60^{0}[/tex] with [tex]v=4,3[/tex] are (-0.3928, 0.9196) and (0.9928, -0.1196).

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To find two unit vectors that make an angle of 60° with v = 4, 3, we can use the formula. The two unit vectors that make an angle of 60° with v = 4, 3 are approximately u1 = (0.259, 0.966) and u2 = (0.965, 0.259).

u = (cosθ, sinθ)

where θ is the angle between u and v.

First, we need to find the angle between v and the x-axis. We can use trigonometry to do this:

tanθ = (3/4)
θ = tan^-1(3/4)
θ ≈ 36.87°

Next, we can find two unit vectors that make an angle of 60° with v by adding or subtracting 60° from θ and plugging it into the formula for u:

u1 = (cos(θ + 60°), sin(θ + 60°))
u1 = (cos(96.87°), sin(96.87°))
u1 ≈ (0.259, 0.966)

u2 = (cos(θ - 60°), sin(θ - 60°))
u2 = (cos(16.87°), sin(16.87°))
u2 ≈ (0.965, 0.259)

So the two unit vectors that make an angle of 60° with v = 4, 3 are approximately u1 = (0.259, 0.966) and u2 = (0.965, 0.259).

To find two unit vectors that make an angle of 60° with v = (4, 3), we can use the rotation matrix method.

First, normalize vector v to get a unit vector:
||v|| = sqrt(4² + 3²) = 5
u = (4/5, 3/5)

Now, rotate u by ±60° to obtain two new unit vectors. A 2D rotation matrix is given by:
R(θ) = [cos(θ)  -sin(θ)]
           [sin(θ)   cos(θ)]

For θ = 60°:
R(60) = [cos(60°)  -sin(60°)]
              [sin(60°)   cos(60°)]

For θ = -60°:
R(-60) = [cos(-60°)  -sin(-60°)]
               [sin(-60°)   cos(-60°)]

Multiply R(60) and R(-60) by the unit vector u, respectively, to obtain the two desired unit vectors:
u1 = R(60) * u
u2 = R(-60) * u

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