Respuesta :
Answer:
Explanation:
Given
[tex]\vec{A}=-2\hat{i}+3\hat{j}+4\hat{k}[/tex]
[tex]\vec{B}=3\hat{i}+1\hat{j}-3\hat{k}[/tex]
Cross of two Product gives a vector which is perpendicular to the both vectors
[tex]\vec{A}\times \vec{B}=\begin{vmatrix}i &j &k \\ -2 &3 &4 \\ 3 &1 &-3 \end{vmatrix}[/tex]
[tex]\vec{A}\times \vec{B}=\hat{i}(-9-4)-\hat{j}(6-12)+\hat{k}(-2-9)[/tex]
[tex]\vec{A}\times \vec{B}=\vec{r}=-13\hat{i}+6\hat{j}-11\hat{k}[/tex]
Now unit vector in this direction is given by
[tex]\hat{n}=\frac{\vec{r}}{|\vec{r}|}=\frac{-13\hat{i}+6\hat{j}-11\hat{k}}{\sqrt{13^2+6^2+11^2}}[/tex]
[tex]\hat{n}=\frac{-13\hat{i}+6\hat{j}-11\hat{k}}{\sqrt{326}}[/tex]
