Angle A and Angle R are in the 1st quadrant.
If you draw these 2 vectors, the line from the end of vector A to the end of vector R is Vector B.
The direction of the line is 45° above the negative x axis. The angle of vector B is 135° counter clockwise from the positive x axis.
Refer your answer to the questions on the problem below.
Given that: A = -2i + 5j and B = 3i - j
Magnitude = square root of sum the square of coefficients of i and j.
A = (-2^2 + 5^2)^0.5
B = (3^2 + -1^2)^0.5
http://hyperphysics.phy-astr.gsu.edu/hba…
See web site above for scalar product equation
Angle between the 2 vectors = Inverse tangent of (∆j ÷ ∆i)
Inverse tan ((-1 – -2) ÷ (3 – 5)
Angle = -26.6°
Angle of vector = Inverse tangent of (j ÷ i)
Angle A = Inverse tan (5/-2)
Angle B = Inverse tan (-1/3)
Find the magnitude of Vector A.
Find the magnitude of Vector B.
The magnitude of the scalar product vectors A•B is?
The angle between the two vectors A and B is?
Find the angle α of vector A?
Find the angle α of vector B?
What is the magnitude of Σx in the product of A x B?
What is the magnitude of Σy in the product of A x B?
What is the magnitude of Σz in the product of A x B?
x – i, y = j, z = k
There is no k, so there is no z component!
What is the magnitude of the product of A x B?
http://hyperphysics.phy-astr.gsu.edu/hba…
see web site above for cross product equation