Question

Find \(|u \times v|\) and determine whether \(u \times v\) is directed into the page or out of the page.

Short answer

\(|{\rm{u}} \times {\rm{v}}|\) is \(96\sqrt 3 \) and \(u \times v\) is directed into the page.

Step-by-step solution

Formula used

If\(\theta \)is the angle between vectors\({\bf{u}}\)and\({\bf{v}}\), then the cross product is,

\(|u \times v| = |u||v|\sin \theta \ldots \ldots \ldots (1)\)

Here,

|u| is the magnitude of vector\({\bf{u}}\)and

|v| is the magnitude of vector\({\bf{v}}\).

Right-hand rule:

The fingers of the right hand curl in the direction of rotation from\({\bf{u}}\)to\({\bf{v}}\), then the thumb points are in the direction of\(u \times v\). The rotation angle should be less than\({180^^\circ }\).

Find the \(|u \times v|\)

As given, \(|{\rm{u}}| = 12,|{\rm{v}}| = 16\) and \(\theta = {120^^\circ }\)

Find actual angle \((\theta )\)as shown in figure 1

\(\begin{array}{l}\theta = {180^^\circ } - {120^^\circ }\\\theta = {60^^\circ }\end{array}\)

Substitute 12 for |u|, 16 for |v| and \({60^^\circ }\) for \(\theta \) in equation (1),

\(\begin{array}{l}|{\rm{u}} \times {\rm{v}}| = (12)(16)\sin \left( {{{60}^^\circ }} \right)\\|{\rm{u}} \times {\rm{v}}| = (12)(16)\left( {\frac{{\sqrt 3 }}{2}} \right)\\|{\rm{u}} \times {\rm{v}}| = (12)(8)(\sqrt 3 )\\|{\rm{u}} \times {\rm{v}}| = 96\sqrt 3 \end{array}\)

Thus, the \(|{\rm{u}} \times {\rm{v}}|\) is \(96\sqrt 3 \).

Since the rotation angle is \({60^^\circ }\), which is from \(v\) to \(u\), the resultant \(u \times v\) is directed into the page (by the right hand rule).

Thus, the resultant \(u \times v\) is directed into the page.