The given vectors quantities can be described by their properties of both
magnitude and direction.
- a. The drawing of the vector extending from point (0, 0) to (190, 0) on the coordinate plane is attached.
- b. The velocity vector extending from (0, 0) to (108.76, 50.714) on the coordinate plane is attached.
- c. The displacement vector extending from (0, 0) to (30·√2, 30·√2) is attached.
Reasons:
a. The magnitude of the vector = 190 N
The direction in which the vector acts = East
Therefore, in vector form, we have;
= 190 × cos(0)·i + 190 × sin(0)·j = 190·i
The vector can be represented by an horizontal line, 190 units long
Coordinate points on the vector = (0, 0) and (190, 0)
The drawing of the vector with the above points using MS Excel is attached.
b. Magnitude of the velocity vector = 120 km/hr. 25° North of east
Solution;
The vector form of the velocity is; = 120 × cos(25)·i + 120×sin(25)·j, which gives;
= 120 × cos(25)·i + 120×sin(25)·j ≈ 108.76·i + 50.714·j
≈ 108.76·i + 50.714·j
Therefore, points that define the vector are; (0, 0) and (108.76, 50.714)
The drawing of the vector is attached
c. The magnitude of the vector = 60 m
The direction of the vector is southwest = West 45° south
The vector form of the displacement is = 60 × cos(45°)·i + 60 × sin(45°)·j
Which gives;
= 30·√2·i + 30·√2·j
Points on the vector are therefore; (0, 0), and (30·√2, 30·√2)
The drawing of the vector is attached
Learn more about vectors here:
