Question

A watermelon seed has the following coordinates: x = -6.4 m, y = 5.0 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 (3.0 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°].

Answer 1

Part 1. When seed is at coordinate x = -6.4 m & y = 5.0 m, z = 0 m, So

r = position vector about origin = -6.4 i + 5.0 j + 0 k

Magnitude of position vector will be:

|r| = sqrt((-6.4)^2 + 5.0^2 + 0^2) = 8.12 m

Part 2.

Direction of position vector will be:

Direction = arctan (y/x) = arctan (5.0/(-6.4)) = -38.0 deg

Since we need direction relative to +ve x-axis, So

Direction = 180 - 38.0 = 142.0 deg

Part 3.

Now when seed is moved to position r1 = (3.0, 0, 0) m = 3.0 i

then displacement vector of seed will be:

dr = r1 - r

dr = 3.0 i - (-6.4 i + 5.0 j)

dr = 9.4 i - 5.0 j

Now Magnitude of displacement vector will be:

|dr| = sqrt((9.4)^2 + (-5.0)^2 + 0^2) = 10.65 m

Part 4.

Direction of displacement vector will be:

Direction = arctan (y/x) = arctan (-5.0/9.4) = -28.0 deg

Since we need direction relative to +ve x-axis, and in range of (-180, 180) deg, So

Direction = -28.0 deg

Let me know if you've any query.