Chapter 11.3, Problem 15E

To determine

To find $u\times v$ and show that it is orthogonal to both $u\text{and}v$ .

Answer to Problem 15E

The cross product is $u\times v=-2i+3j+3k$ .

Explanation of Solution

Given information:

The vectors are $u=\langle 3,-2,4\rangle ,v=\langle 0,1,-1\rangle$ .

Cross product:

Let two vectors are $a=a_{1}i+a_{2}j+a_{3}k\text{}\text{}\text{and}b=b_{1}i+b_{2}j+b_{3}k$ ,

Cross product is,

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

Dot product:

Let two vectors $a=a_{1}i+a_{2}j+a_{3}k\text{}\text{}\text{and}b=b_{1}i+b_{2}j+b_{3}k$ are orthogonal if and only if,

  $\begin{array}{l}a\cdot b=\left(a_{1}i+a_{2}j+a_{3}k\right)\text{}\text{}\cdot \left(b_{1}i+b_{2}j+b_{3}k\right)\\ a\cdot b=a_1{b}_1+a_2{b}_2+a_3{b}_3\end{array}$

If $a\cdot b=0$ , the both vectors are orthogonal.

Calculation:

  $u=\langle 3,-2,4\rangle ,v=\langle 0,1,-1\rangle$ write as,

  $u=3i-2j+4k\text{}\text{}\text{and}v=0i+j-k$

So, cross product is,

  $\begin{array}{c}u\times v=\left|\begin{array}{ccc}i& j& k\\ 3& -2& 4\\ 0& 1& -1\end{array}\right|\\ u\times v=\left|\begin{array}{cc}-2& 4\\ 1& -1\end{array}\right|i-\left|\begin{array}{cc}3& 4\\ 0& -1\end{array}\right|j+\left|\begin{array}{cc}3& -2\\ 0& 1\end{array}\right|k\\ u\times v=\left(2-4\right)i-\left(-3-0\right)j+\left(3+0\right)k\\ u\times v=-2i+3j+3k\end{array}$

So, cross product is $u\times v=-2i+3j+3k$ .

Now, the dot product of $\left(u\times v\right)$ and $u$ is,

  $\begin{array}{l}\left(u\times v\right)\cdot u=\left(-2i+3j+3k\right)\cdot \left(3i-2j+4k\right)\\ \left(u\times v\right)\cdot u=-2\left(3\right)+3\left(-2\right)+3\left(4\right)\left[\text{from}\text{dot}\text{product}\text{concept}\right]\\ \left(u\times v\right)\cdot u=-6-6+12\\ \left(u\times v\right)\cdot u=0\end{array}$

the dot product of $\left(u\times v\right)$ and $u$ is 0.

So, vector $\left(u\times v\right)$ and $u$ are orthogonal.

Now, the dot product of $\left(u\times v\right)$ and $v$ is,

  $\begin{array}{l}\left(u\times v\right)\cdot v=\left(-2i+3j+3k\right)\cdot \left(0i+j-k\right)\\ \left(u\times v\right)\cdot v=-2\left(0\right)+3\left(1\right)+3\left(-1\right)\\ \left(u\times v\right)\cdot v=0+3-3\\ \left(u\times v\right)\cdot v=0\end{array}$

the dot product of $\left(u\times v\right)$ and $v$ is 0.

So, vector $\left(u\times v\right)$ and $v$ are orthogonal.