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The torque vector τ has magnitude
|τ| = |F| |r| sin(θ)
where θ is the angle made by the force vector F and position vector r.
Recall the dot product identity,
F • r = |F| |r| cos(θ)
and use to find the measure of the angle in question.
F = (-3 i + 3 j + 4 k) N
r = (-2 i + 4 j) m
F • r = (6 + 12 + 0) N•m = 18 N•m
|F| = √((-3 N)² + (3 N)² + (4 N)²) = √34 N
|r| = √((-2 m)² + (4 m)²) = √20 m
So, we have
cos(θ) = (18 N•m) / ((√20 m) (√34 N)) = 9/√170
⇒ θ = arccos(9/√170) ≈ 46.3°
Now just find |τ| :
|τ| = (√34 N) (√20 m) sin(arccos(9/√170))
|τ| = (√680 N•m) sin(arccos(9/√170))
|τ| = (√680 N•m) √(89/170)
|τ| = 2√89 N•m ≈ 18.9 N•m