Respuesta :
let us consider
[tex]\vec{A} = sint\hat{i}+cost\dot{j}+ t\hat{k}\ ,\vec{B} =cost\hat{i} + sint\hat{j}[/tex]
now we have to evaluate
[tex]\vec{A}\times \vec{B}[/tex]
To proof =
[tex]\vec{A}\times \vec{B}=\begin{vmatrix}i &j &k \\ sint&cost &t \\ cost&sint &0 \end{vmatrix}[/tex]
now solving the determinant form
we get
=[tex]\hat{i}\left ( -sint \right )-\hat{j}\left ( -tcost \right )+\hat{k}\left ( sin^{2}t - cos^{2}t\right )[/tex]
by using the formula
[tex]sin^{2}t + cos^{2}t = 1[/tex]
[tex]cos^{2}t = sin^{2} t - 1[/tex]
put this value in the above equation
we get
=[tex]\hat{i}\left ( -sint \right )+\hat{j}\left ( tcost \right )+\hat{k}\left ( 2sin^{2}t - 1\right )[/tex]
Hence proved
