A particle initially at the origin travels in uniform motion with velocity v=2i-j-2k. Find position venctor vector r(t) of a particle at time t. Find equatio for the plane passing through the origin and perp. to the tragectory of the particle. And find when and where the particle reaches the origin-centered unit circle.
To find the position vector r(t) of the particle at time t, we can integrate the velocity vector v with respect to time.
Given: v = 2i - j - 2k
Integrating each component with respect to time, we have:
dx/dt = 2 → x(t) = 2t + C1
dy/dt = -1 → y(t) = -t + C2
dz/dt = -2 → z(t) = -2t + C3
Since the particle is initially at the origin (x=0, y=0, z=0) at time t=0, we substitute these values into the equations above:
x(0) = 2(0) + C1 → C1 = 0
y(0) = -(0) + C2 → C2 = 0
z(0) = -2(0) + C3 → C3 = 0
Hence, the position vector r(t) of the particle is:
r(t) = (2t)i - t*j - 2t*k
To find the equation of the plane passing through the origin and perpendicular to the trajectory of the particle, we can use the normal vector of the plane, which is parallel to the velocity vector.
The normal vector N is given by N = (2i - j - 2k).
The equation of the plane passing through the origin can be written as:
N · r = 0
Substituting the values of N and r(t), we have:
(2i - j - 2k) · (2t)i - t*j - 2t*k = 0
Expanding and simplifying, we get:
(4t) - t - (4t) = 0
- t = 0
t = 0
Hence, the particle reaches the origin at time t = 0.
To find when the particle reaches the origin-centered unit circle, we need to determine the magnitude of the position vector r(t) at that time.
When t = 0:
r(0) = (2(0))i - (0)j - 2(0)k
r(0) = 0i - 0j - 0k
r(0) = 0
The magnitude of the position vector r(0) is:
| r(0) | = √(0^2 + 0^2 + 0^2)
| r(0) | = 0
Therefore, the particle reaches the origin-centered unit circle when t = 0 and its position is at the origin (0,0,0).
To find the position vector of the particle at time t, we need to integrate the velocity vector with respect to time.
Given velocity vector v = 2i - j - 2k, we can integrate each component separately to obtain the position vector:
∫v dt = ∫(2i - j - 2k) dt
Integrating each component:
∫2i dt = 2∫i dt = 2ti + C_1
∫-j dt = -∫j dt = -tj + C_2
∫-2k dt = -2∫k dt = -2tk + C_3
Putting it all together, the position vector, r(t), is given by:
r(t) = (2ti + C_1) + (-tj + C_2) + (-2tk + C_3)
Now, to find the equation for the plane passing through the origin and perpendicular to the trajectory of the particle, we need to find a vector normal to the trajectory.
For a vector to be normal to a plane, it must be perpendicular to any vector lying in the plane. Therefore, any vector parallel to the trajectory will be normal to the required plane. The direction of the trajectory vector is given by the coefficients of i, j, and k in the velocity vector, which are 2, -1, and -2 respectively.
Thus, a direction vector of the trajectory is given by v_1 = 2i - j - 2k.
Using the normal vector, the equation for the plane can be written as:
A(x - x_0) + B(y - y_0) + C(z - z_0) = 0,
where (x_0, y_0, z_0) represents the origin (0, 0, 0).
Substituting the values, we have:
2x - y - 2z = 0.
This is the equation of the plane passing through the origin and is perpendicular to the trajectory of the particle.
Finally, to find when and where the particle reaches the origin-centered unit circle, we need to find the distance from the origin for different time values.
The distance from the origin for a position vector r(t) is given by the magnitude of r(t):
| r(t) | = √(x^2 + y^2 + z^2)
Substituting the values of r(t) into the magnitude equation, we have:
| r(t) | = √((2ti + C_1)^2 + (-tj + C_2)^2 + (-2tk + C_3)^2)
To find the time and position when the particle reaches the origin-centered unit circle, we need to solve the following equation:
| r(t) | = 1
Solve this equation to find the time and corresponding position on the unit circle.