Finding parallel and perpendicular components of a force

[Poto23]

Homework Statement

Consider a plane with the outer normal unit vector of (1/√2, -1√2, 0). There is a force of (2,1,5) acting on the surface. What is the component of force normal and parallel to the plane? What is the maximum parallel component to the plane?

I would like to check my answers for the first two, but do not know how to find the maximum parallel component. I'm guessing since my direction is at a 45 degree angel, the maximum parallel component of the force should be given at the 45 degree angle?

Homework Equations

Dot product and Pythagorean theorem
Force = 2^2 +1^2 + 5^2 = √30

The Attempt at a Solution

I found the normal force to be (1/2, -1/2, 0) through doing the dot product of force and the unit vector, and then multiplying the scalar product by the unit vector again.
For the parallel component, I found the magnitude of the Force and the magnitude of the normal Force, and used Pythagorean theorem. I got √Force^2 - 1^2. = 29.98(1/√2, 1/√2, 0).

Thank you

 

[Poto23]

bump 10char

 

[HallsofIvy]

Homework Statement

Consider a plane with the outer normal unit vector of (1/√2, -1√2, 0). There is a force of (2,1,5) acting on the surface. What is the component of force normal and parallel to the plane? What is the maximum parallel component to the plane?

I would like to check my answers for the first two, but do not know how to find the maximum parallel component. I'm guessing since my direction is at a 45 degree angel, the maximum parallel component of the force should be given at the 45 degree angle?

Homework Equations

Dot product and Pythagorean theorem
Force = 2^2 +1^2 + 5^2 = √30

The Attempt at a Solution

I found the normal force to be (1/2, -1/2, 0) through doing the dot product of force and the unit vector, and then multiplying the scalar product by the unit vector again.

For the parallel component, I found the magnitude of the Force and the magnitude of the normal Force, and used Pythagorean theorem. I got √Force^2 - 1^2. = 29.98(1/√2, 1/√2, 0).

Thank you

Your normal vector is correct. To find the parallel vector, just use the fact that the normal and parallel vector must add to the original vector: (x, y, z)+ (1/2, -1/2, 0)= (2, 1, 5).

 

[Poto23]

I will wait next time. Thank you for letting me know.