Chapter 11.3, Problem 2E

To determine

To fill: The blank in the statement “ $u\times u=$$\text{\_\_\_\_\_\_}$ ”.

Answer to Problem 2E

The complete statement is “ $u\times u=0$ .”

Explanation of Solution

Given information:

The statement “ $u\times u=$$\text{\_\_\_\_\_\_}$ ”.

Consider the provided statement “ $u\times u=$$\text{\_\_\_\_\_\_}$ ”.

Recall that if there are two three-dimensional vectors in a space, say, $a=a_{1}i+a_{2}j+a_{3}k$ and $b=b_{1}i+b_{2}j+b_{3}k$ , then the cross product of two vectors is defined as ,

  $\begin{array}{c}a\times b=\begin{array}{ccc}i& j& k\\ a_1& a_2& a_3\\ b_1& b_2& b_3\end{array}\\ \text{= }\left(a_2{b}_3-a_3{b}_2\right)i-\left(a_1{b}_3-a_3{b}_1\right)j+\left(a_1{b}_2-a_2{b}_1\right)k\end{array}$

The cross product of a vector with itself is always zero.

Let $a=\text{i}+2\text{j}-\text{k}$ be any vector in the space, then the cross product of $a$ and $a$ will be calculated as,

  $\begin{array}{c}a\times a=\begin{array}{ccc}i& j& k\\ 1& 2& -1\\ 1& 2& -1\end{array}\\ \text{= }\left(\left[2\times \left(-1\right)\right]-\left[2\times \left(-1\right)\right]\right)i-\left(\left[1\times \left(-1\right)\right]-\left[1\times \left(-1\right)\right]\right)j+\left(1\times 2-1\times 2\right)k\\ \text{=}\left(-2-\left(-2\right)\right)i-\left(-1-\left(-1\right)\right)j+\left(2-2\right)k\\ \text{=}\left(-2+2\right)i-\left(-1+1\right)j+0k\end{array}$

Simplify, it further as,

  $\begin{array}{c}a\times a=\left(-2+2\right)i-\left(-1+1\right)j+0k\\ \text{=}0i-0j+0k\\ \text{=0}\end{array}$

Thus, the complete statement is $u\times u=0$ .