# Engineering

posted by
**Aaron**
.

A force of 26 lb. acts through a point (4; 5; -7) ft. and is equally inclined

to the positive ends of the orthogonal axes. What is the moment of this

force about the origin

So i'm assuming vector r is 4i+5j-7k?

And using the fact that cos^2α + cos^2β + cos^2γ = 1

therefore l=m=n=√1/3

I understand the force vector is equivalent to (26l + 26m + 26n) which equates to: (26/√3 i + 26/√3 j + 26/√3 k)

Thus the moment of force about the origin is (4i+5j+7k) X (26/√3 i + 26/√3 j + 26/√3 k)

This just seems very wrong to me as the cross products would yield very ugly numbers.

I have close to zero confidence in my answer