Chapter 11.3, Problem 7E

To determine

The value of the expression $v\times v$ .

Answer to Problem 7E

The value of the expression $v\times v$ is $0$ .

Explanation of Solution

Given information:

The value of the vectors $u=3i-j+4k$ and $v=2i+2j-k$ .

Formula used:

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 ,

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

Calculation:

Consider the given value of the vectors $u=3i-j+4k$ and $v=2i+2j-k$ .

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 ,

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

So, the value of expression $v\times v$ will be calculated as,

  $\begin{array}{c}v\times v=\begin{array}{ccc}i& j& k\\ 2& 2& -1\\ 2& 2& -1\end{array}\\ \text{= }\left(\left[2\times \left(-1\right)\right]-\left[2\times \left(-1\right)\right]\right)i-\left(\left[2\times \left(-1\right)\right]-\left[2\times \left(-1\right)\right]\right)j+\left(\left[2\times 2\right]-\left[2\times 2\right]\right)k\\ \text{=}\left(-2-\left(-2\right)\right)i-\left(-2-\left(-2\right)\right)j+\left(4-4\right)k\\ \text{=}\left(-2+2\right)i-\left(-2+2\right)j+\left(0\right)k\end{array}$

Simplify it further as,

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

Hence, the value of the expression $v\times v$ is 0.