Final answer:
The magnitude of vector A is approximately 8.94 units. The direction of vector A is approximately -26.57° with respect to the y-axis.
Step-by-step explanation:
To find the magnitude of a vector, you can use the formula:
|A| = sqrt(Ax^2 + Ay^2 + Az^2)
For vector A = -8j + 4k, the x-component (Ax) is 0, so we only need to calculate the y-component (Ay) and z-component (Az).
Ay = -8
Az = 4
Substituting these values into the formula, we get:
|A| = sqrt(0^2 + (-8)^2 + 4^2) = sqrt(0 + 64 + 16) = sqrt(80) = 8.94
The magnitude of vector A is approximately 8.94 units.
The direction of a vector can be represented as an angle with respect to a reference axis. To find the direction of vector A, we need to find the angles it makes with the x, y, and z axes.
Since Ax = 0, the angle with the x-axis is undefined.
The angle with the y-axis (θy) can be found using:
θy = arctan(Ay/Ax)
θy = arctan(-8/0) = undefined
The angle with the z-axis (θz) can be found using:
θz = arctan(Az/Ay)
θz = arctan(4/-8) = -26.57°
Therefore, the direction of vector A is approximately -26.57° with respect to the y-axis.