Answer:
[tex]4\sqrt6[/tex] square units
Step-by-step explanation:
We are given that two vectors
u=<3,2,1>
v=<1,2,3>
We have to find the area of parallelogram determined by the given vectors
[tex]\vec{u}=3\hat{i}+2\hat{j}+\hat{k}[/tex]
[tex]\vec{v}=\hat{i}+2\hat{j}+3\hat{k}[/tex]
We know that area of parallelogram determined by two vectors a and b
[tex]\mid {a\times b}\mid =\begin{vmatrix}1&j&k\\x_1&x_2&x_3\\x_4&x_5&x_6\end{vmatrix}[/tex]
Using this formula
[tex]\mi{u\times v}\mid=\begin{vmatrix}i&j&k\\3&2&1\\1&2&3\end{vmatrix}[/tex]
[tex]\mid{u\times v}\mid=\mid{\hat{i}(6-2)-\hat{j}(9-1)+\hat{k}(6-2)\mid[/tex]
[tex]\mid{u\times v}\mid=\mid{4i-6j+4k}\mid=\sqrt{4^2+(-6)^2+4^2}=4\sqrt6[/tex]
Area of parallelogram=[tex]4\sqrt6[/tex] square units
