Q. 71 - Calculus

Question

Let u, v and w be three vectors in ${\mathrm{ℝ}}^3$ in which the components of each vector are integers.

(a) Prove that the volume of the parallelepiped determined by u, v and w is an integer.

(b) Find examples of vectors u and v with integer components that show that the area of the parallelogram determined by u and v can be either an integer or an irrational number.

Step-by-Step Solution

Answer

(a)  The volume of the parallelepiped determined by u, v and w is an integer because it involves only addition, subtraction and multiplication. 

(b) Hence we prove that the area of the parallelogram determined by u and v can be either an integer or an irrational number.

Step 1. Given Information

Let u, v and w be three vectors in ${\mathrm{ℝ}}^3$ in which the components of each vector are integers.

(a) Prove that the volume of the parallelepiped determined by u, v and w is an integer.

(b) Find examples of vectors u and v with integer components that show that the area of the parallelogram determined by u and v can be either an integer or an irrational number.

Part (a) Step 1. Now proving that the volume of the parallelepiped determined by u, v and w is an integer.

As we know $\mathrm{Volume} \mathrm{of} \mathrm{the} \mathrm{parallelepiped}=\left|u·(v\times w)\right|$

Let $u=(-3,1,-4), v=(2,0,5), \mathrm{and} w=(1,3,13)$ that value is integer.

Now finding the value of $u·(v\times w)$

$$\begin{gathered} u·(v\times w)=\det \left[\begin{array}{ccc}-3& 1& -4\\ 2& 0& 5\\ 1& 3& 13\end{array}\right] \\ u·(v\times w)=-3\left|\begin{array}{cc}0& 5\\ 3& 13\end{array}\right|-1\left|\begin{array}{cc}2& 5\\ 1& 13\end{array}\right|-4\left|\begin{array}{cc}2& 0\\ 1& 3\end{array}\right| \\ u·(v\times w)=-3(0\times 13-5\times 3)-1(2\times 13-5\times 1)-4(2\times 3-0\times 1) \\ u·(v\times w)=-3(0-15)-1(26-5)-4(6-0) \\ u·(v\times w)=45-21-24 \\ u·(v\times w)=0 \end{gathered}$$

Part (a) Step 2. The value of u · ( v × w ) = 0 which is integer.

The volume of the parallelepiped determined by u, v and w is the absolute value of the determinant of the matrix in which the rows are the components of u, v and w. Since each component is an integer, the determinant and its absolute value are integers.

Part (b) Step 1. We have to show that the area of the parallelogram determined by u and v can be either an integer or an irrational number.

Let the $u=i \mathrm{and} v=j$

$$\begin{gathered} u\times v=\left[\begin{array}{ccc}i& j& k\\ 1& 0& 0\\ 0& 1& 0\end{array}\right] \\ u\times v=i\left|\begin{array}{cc}0& 0\\ 1& 0\end{array}\right|-j\left|\begin{array}{cc}1& 0\\ 0& 0\end{array}\right|+k\left|\begin{array}{cc}1& 0\\ 0& 1\end{array}\right| \\ u\times v=i(0-0)-j(0-0)+k(1-0 \\ u\times v=k \end{gathered}$$

The area of the parallelogram

$$\begin{gathered} \left|u\times v\right|=\sqrt{1^2} \\ \left|u\times v\right|=1 \end{gathered}$$

Part (b) Step 2. Let u = i   and   v = j + k

$$\begin{gathered} u\times v=\left[\begin{array}{ccc}i& j& k\\ 1& 0& 0\\ 0& 1& 1\end{array}\right] \\ u\times v=i\left|\begin{array}{cc}0& 0\\ 1& 1\end{array}\right|-j\left|\begin{array}{cc}1& 0\\ 0& 1\end{array}\right|+k\left|\begin{array}{cc}1& 0\\ 0& 1\end{array}\right| \\ u\times v=i(0-0)-j(1-0)+k(1-0) \\ u\times v=-j+k \end{gathered}$$

The area of the parallelogram

$$\begin{gathered} \left|u\times v\right|=\sqrt{{(-1)}^2+1^2} \\ \left|u\times v\right|=\sqrt{1+1} \\ \left|u\times v\right|=\sqrt{2} \end{gathered}$$