A watermelon seed has the following coordinates: x = -8.7 m, y = 1.6 m, and z = 0 m. Find its position vector as (a) a magnitude and (b) an angle relative to the positive direction of the x axis. If the seed is moved to the xyz coordinates ( 3.4 m, 0 m, 0 m), what is its displacement as (c) a magnitude and (d) an angle relative to the positive direction of the x axis? Put the angles in the range (-180°, 180°].
To find the position vector of the watermelon seed, we can use the given coordinates (x, y, z).
(a) Magnitude of the position vector:
The magnitude of the position vector is given by the formula:
|𝑟| = √(𝑥^2 + 𝑦^2 + 𝑧^2)
Substituting the given values:
|𝑟| = √((-8.7)^2 + (1.6)^2 + (0)^2)
Calculating this, we find:
|𝑟| ≈ √(75.69 + 2.56) ≈ √78.25 ≈ 8.85 meters
Therefore, the magnitude of the position vector is approximately 8.85 meters.
(b) Angle relative to the positive direction of the x-axis:
The angle can be found using the inverse tangent function:
𝜃 = tan^(-1)(𝑦/𝑥)
Substituting the given values:
𝜃 = tan^(-1)(1.6/(-8.7))
Calculating this, we find:
𝜃 ≈ tan^(-1)(-0.183908046) ≈ -10.5°
The angle is approximately -10.5° relative to the positive direction of the x-axis.
(c) Magnitude of displacement:
The displacement is the change in position vector when the seed is moved to the new coordinates. We subtract the initial position vector from the final position vector to obtain the displacement vector.
𝑑 = 𝑟(final) - 𝑟(initial)
Given initial coordinates: (x_initial, y_initial, z_initial) = (-8.7 m, 1.6 m, 0 m)
Given final coordinates: (x_final, y_final, z_final) = (3.4 m, 0 m, 0 m)
𝑑 = (𝑥_final - 𝑥_initial)𝑖̂ + (𝑦_final - 𝑦_initial)𝑗̂ + (𝑧_final - 𝑧_initial)𝑘̂
Substituting the given values, we have:
𝑑 = (3.4 - (-8.7))𝑖̂ + (0 - 1.6)𝑗̂ + (0 - 0)𝑘̂
𝑑 = 12.1𝑖̂ - 1.6𝑗̂
Therefore, the displacement vector is 𝑑 = 12.1𝑖̂ - 1.6𝑗̂.
(d) Angle relative to the positive direction of the x-axis:
To find the angle, we can again use the inverse tangent function:
𝜃 = tan^(-1)(𝑦_displacement / 𝑥_displacement)
Substituting the values from the displacement vector:
𝜃 = tan^(-1)(-1.6 / 12.1)
Calculating this, we find:
𝜃 ≈ tan^(-1)(-0.132231405) ≈ -7.44°
The angle is approximately -7.44° relative to the positive direction of the x-axis.
To find the position vector of the watermelon seed, we can combine the given coordinates into a 3D vector.
(a) Magnitude of the position vector:
The magnitude of a vector can be found using the Pythagorean theorem. In this case, the position vector is a 3D vector. Let's calculate the magnitude using the distance formula:
magnitude = sqrt(x^2 + y^2 + z^2)
magnitude = sqrt((-8.7)^2 + (1.6)^2 + (0)^2)
magnitude = sqrt(75.69 + 2.56 + 0)
magnitude = sqrt(78.25)
magnitude ≈ 8.84 m (rounded to two decimal places)
So, the magnitude of the position vector is approximately 8.84 m.
(b) Angle relative to the positive direction of the x axis:
To find the angle relative to the positive direction of the x axis, we can use trigonometry. The angle θ can be calculated using the inverse tangent function:
θ = arctan(y / x)
θ = arctan(1.6 / -8.7)
θ ≈ -10.54° (rounded to two decimal places)
The angle relative to the positive direction of the x axis is approximately -10.54°.
(c) Magnitude of the displacement:
To find the displacement, we need to subtract the initial position vector from the final position vector and calculate the magnitude of the resulting vector.
Displacement vector = Final position vector - Initial position vector
displacement_x = 3.4 m - (-8.7 m) = 12.1 m
displacement_y = 0 m - 1.6 m = -1.6 m
displacement_z = 0 m - 0 m = 0 m
magnitude_displacement = sqrt(displacement_x^2 + displacement_y^2 + displacement_z^2)
magnitude_displacement = sqrt((12.1)^2 + (-1.6)^2 + (0)^2)
magnitude_displacement = sqrt(146.41 + 2.56 + 0)
magnitude_displacement = sqrt(148.97)
magnitude_displacement ≈ 12.20 m (rounded to two decimal places)
The magnitude of the displacement vector is approximately 12.20 m.
(d) Angle relative to the positive direction of the x axis:
Similar to part (b), we can calculate the angle relative to the positive direction of the x axis for the displacement vector.
θ_displacement = arctan(displacement_y / displacement_x)
θ_displacement = arctan(-1.6 / 12.1)
θ_displacement ≈ -7.56° (rounded to two decimal places)
The angle relative to the positive direction of the x axis for the displacement vector is approximately -7.56°.
Note: The angles are given in the requested range (-180°, 180°] where -180° is included but 180° is not, i.e., the angles are in the range (-180°, 180°].
a. 6.087
b. 104.265
c. 8.13449
d. 46.49