Question

Given that vector ?⃗ is 5.00 m at 37⁰ North of East, find vector ?⃗⃗ such that their sum is directed East and has a magnitude of 3.00 m. Give the result with respect to compass like how ?⃗ is specified, and with the angle less than or equal to 45⁰

Answer 1

Suppose given that Vector is R and it makes angle with +x-axis, then it's components are given by:

Rx = R*cos

Ry = R*sin

Using above rule:

Vector A = 5.00 m at 37 deg North of East, So

Ax = A*cos = 5.00*cos 37 deg = 4.00

Ay = A*sin = 5.00*sin 37 deg = 3.00

Now given vector B is such that Sum of A and B is directed towards east, So

R = A + B

Where, R = 3.00 m towards east, So

Rx = R*cos = 3.00*cos 0 deg = 3.00

Ry = R*sin = 3.00*sin 0 deg = 0

Now Since

R = A + B

B = R - A

B = (Rx - Ax) i + (Ry - Ay) j

Using above values:

B = (3.00 - 4.00) i + (0 - 3.00) j

B = -1.00 i - 3.00 j

Magnitude of vector B will be:

|B| = sqrt ((-1.00)^2 + (-3.00)^2) = 3.16 m

Direction of vector B will be:

Since vector B is in 3rd quadrant, So angle will be about South of West

Direction = arctan (|By|/|Bx|) = arctan (3.00/1.00) = 71.57 deg

Direction = 71.57 deg South of West = 90 - 71.57 = 18.43 deg West of South

So,

Vector B = 3.16 m at 18.43 deg West of South

Let me know if you've any query.