Question

Use the definition of the vector product and the definitions of the unit vectors$\overrightarrow{\mathbf{i}}\mathbf{,}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{j}}\mathbf{,}\mathit{a}\mathit{n}\mathit{d}\stackrel{\mathbf{\rightharpoonup }}{\mathbf{k}}$to prove Equations 11.7. You may assume the x axis points to the right, the y axis up, and the z axis horizontally toward you (not away from you). This choice is said to make the coordinate system a right-handed system.

Short answer

$\widehat{i}\times \widehat{j}=1\widehat{k}$

Step-by-step solution

Define vector product

In physics and astronomy, the vector product and the scalar product are the two methods of multiplying vectors that are used the most. Taking the product of the magnitude of the vector times the sine of the angle between them yields the magnitude of the vector product of two vectors.

To find the vector product

By vector product of the two vectors are we get

$\overrightarrow{A}\times \overrightarrow{B}=ABsin\theta$

The angle between the vectors is

To Prove that

$\widehat{i}\times \widehat{i}=\widehat{j}\times \widehat{j}=\widehat{k}\times \widehat{k}=\mathrm{0....}\left(1\right)$

$\widehat{i}\times \widehat{j}=-\widehat{j}\times \widehat{i}=\widehat{k}\mathrm{....}\left(2\right)$

$\widehat{j}\times \widehat{k}=-\widehat{k}\times \widehat{j}=\widehat{i}\mathrm{.....}\left(3\right)$

$\widehat{k}\times \widehat{i}=-\widehat{i}\times \widehat{k}=\widehat{j}\mathrm{.....}\left(4\right)$

The two vectors$\widehat{i}$ are$\widehat{i}$ and are parallel, the angle between them be zero

From the first equation as$\widehat{i}\times \widehat{i}=1.1sin{0}^o$

$\widehat{i}\times \widehat{j}=1\times 1sin{90}^o$

$=1\widehat{k}$