Find the vector's magnitude and direction.

Part A
A⃗ = 4 i^− 8 j^

A =?
θA =?

To find the magnitude (A) and direction (θA) of the vector A⃗ = 4i^ - 8j^, we can use the Pythagorean theorem and trigonometry.

Part A: Finding the Magnitude (A)
The magnitude of a vector is the length or size of the vector. In a Cartesian coordinate system, the magnitude of a vector with components (x,y) can be found using the Pythagorean theorem:

A = sqrt(x^2 + y^2)

In this case, the x-component is 4 and the y-component is -8. Plugging these values into the formula, we get:

A = sqrt((4)^2 + (-8)^2)
= sqrt(16 + 64)
= sqrt(80)

So, the magnitude of vector A⃗ is sqrt(80), or approximately 8.94.

Part B: Finding the Direction (θA)
The direction of a vector is usually given as an angle measured counterclockwise from the positive x-axis. To find this angle (θA), we can use trigonometry.

θA = arctan(y/x)

In this case, the y-component is -8 and the x-component is 4. Plugging these values into the formula, we get:

θA = arctan((-8)/4)
= arctan(-2)

Since the arctan function gives us an angle in radians, we can convert it to degrees by multiplying by (180/π):

θA = arctan(-2) * (180/π)
≈ -63.43°

Therefore, the direction of vector A⃗ is approximately -63.43° counterclockwise from the positive x-axis.