Question

Find the area of the parallelogram determined by the vectors u(3,2,1) and v= (1,2,3)

Answer 1

Answer:

$4\sqrt6$ 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

$\vec{u}=3\hat{i}+2\hat{j}+\hat{k}$

$\vec{v}=\hat{i}+2\hat{j}+3\hat{k}$

We know that area of parallelogram determined by two vectors a and b

$\mid {a\times b}\mid =\begin{vmatrix}1&j&k\\x_1&x_2&x_3\\x_4&x_5&x_6\end{vmatrix}$

Using this formula

$\mi{u\times v}\mid=\begin{vmatrix}i&j&k\\3&2&1\\1&2&3\end{vmatrix}$

$\mid{u\times v}\mid=\mid{\hat{i}(6-2)-\hat{j}(9-1)+\hat{k}(6-2)\mid$

$\mid{u\times v}\mid=\mid{4i-6j+4k}\mid=\sqrt{4^2+(-6)^2+4^2}=4\sqrt6$

Area of parallelogram=$4\sqrt6$ square units