Find the magnitude and direction angle θ of the vector.
u=4(cos102°i+sin102°j)
u=4(cos102°i+sin102°j)
= 4[cos102° , sin102°]
the direction angle is clearly 102°
magnitude = √(16(cos^2 102 + sin^2 102)
= √(16(1))
= 4
To find the magnitude and direction angle of the vector u, we first need to convert the given vector from polar form to rectangular form.
The polar form of the vector u is u = 4(cos102°i + sin102°j).
To convert it to rectangular form, we need to use the trigonometric identities:
cos(θ) = x/r
sin(θ) = y/r
Where x and y are the rectangular components of the vector and r is the magnitude of the vector.
So, we can rewrite the vector u as:
u = 4(cos102°i + sin102°j) = 4(cos(102°)i + sin(102°)j)
Now, we can equate the cosine and sine values to x/r and y/r respectively to find the rectangular components.
cos(102°) = x/4 (1)
sin(102°) = y/4 (2)
Solving equations (1) and (2) for x and y, we get:
x = 4 * cos(102°)
y = 4 * sin(102°)
Next, we can calculate the magnitude of the vector using the Pythagorean theorem:
r = sqrt(x^2 + y^2)
Substituting the values of x and y, we have:
r = sqrt((4 * cos(102°))^2 + (4 * sin(102°))^2)
Performing the calculation, we get:
r ≈ sqrt(16 * cos^2(102°) + 16 * sin^2(102°))
r ≈ sqrt(16 * (cos^2(102°) + sin^2(102°)))
r ≈ sqrt(16)
r ≈ 4
Therefore, the magnitude of the vector u is approximately 4.
To find the direction angle θ, we can use the arctangent function:
tan(θ) = y/x
Substituting the values of x and y, we have:
tan(θ) = (4 * sin(102°))/(4 * cos(102°))
tan(θ) = sin(102°)/cos(102°)
Using inverse tangent (arctan) to solve for θ, we get:
θ = arctan(sin(102°)/cos(102°))
Evaluating this expression, we have:
θ ≈ arctan(0.87630668004/(-0.48175367410))
θ ≈ -59.77°
Therefore, the magnitude of the vector u is 4 and its direction angle θ is approximately -59.77°.