Question

A force is specified by the vector f = [(-60)i (-80)j (-180)k] n. calculate the angles made by f with the positive x-, y-, and z-axes.

Answer 1

Use the dot product identity

`\vec a \cdot \vec b = \|\vec a\| \|\vec b\| \cos(\theta)`

Take the dot product of
`\vec f` with the unit vectors
`\vec\imath,\vec\jmath,\vec k`, each of whose magnitude is 1. Then compute the angles made with the
`x`-,
`y`-, and
`z`-axis, respectively.

Dotting with
`\vec\imath`, for instance, we have

`\vec f \cdot \vec \imath = \|\vec f\| \cos(\theta)`

`-60\,\mathrm N = \left(√((-60)^2 + (-80)^2 + (-180)^2) \,\mathrm N\right) \cos(\theta)`

`\cos(\theta) = (-60)/(20√(106)) = -\frac3{√(106)}`

`\theta = \cos^(-1)\left(-\frac3{√(106)}\right) \approx 106^\circ`

so
`\vec f` makes an angle of about 106° with the
`x`-axis.

Similarly,

`\vec f \cdot \vec \jmath = -80\,\mathrm N = (20√(106)\,\mathrm N) \cos(\theta)`

`\cos(\theta) = -\frac4{√(106)} \implies \theta \approx 113^\circ`

and

`\vec f \cdot \vec k = -180\,\mathrm N = (20√(106)\,\mathrm N) \cos(\theta)`

`\cos(\theta) = -\frac9{√(106)} \implies \theta \approx 151^\circ`