Vector A is directed along the positive y-axis and has a magnitude of sqrt 3.0 units. Vector B is directed along the negative x-axis and has magnitude 1.0 units. What are the magnitude and direction of A + B?

Since A and B are perpendicular, use the Pythagorean theorem for the magnitude.

Mag = sqrt[(sqrt3)^2+1] = sqrt 4 = 2

The sum vector A + B is arctan 1/sqrt3 = 30 degrees west of north.

Thank you. Could you explain to me why we use arctan?? I just want to have a better understanding of why?

Vector A is directed along the positive y-axis and has magnitude sqrt 3.0 units. Vector B is directed along the negative x-axis and has a magnitude 1.0 units. (a) what are the magnitude and direction of A-B? (b) what are the x and y components of B-A?

To find the magnitude and direction of A + B, we need to add the two vectors.

First, let's find the components of the vectors A and B:

Vector A is directed along the positive y-axis, so its x-component is 0 and its y-component is √3.0 units.

Vector B is directed along the negative x-axis, so its x-component is -1.0 units and its y-component is 0.

To find the components of vector A + B, we simply add the corresponding components of A and B:

The x-component of A + B = x-component of A + x-component of B = 0 + (-1.0) = -1.0 units.

The y-component of A + B = y-component of A + y-component of B = √3.0 + 0 = √3.0 units.

Now, we can find the magnitude of A + B using the Pythagorean theorem:

Magnitude of A + B = √(x-component of A + B)^2 + (y-component of A + B)^2
= √((-1.0)^2 + (√3.0)^2)
= √(1.0 + 3.0)
= √4.0
= 2.0 units.

The magnitude of A + B is 2.0 units.

To find the direction of A + B, we can use the tangent function:
tan(θ) = (y-component of A + B) / (x-component of A + B)
tan(θ) = (√3.0) / (-1.0)
θ = arctan((-√3.0) / 1.0)
θ ≈ -60°.

Therefore, the direction of A + B is approximately -60°.