Chapter 11.3, Problem 17E

To determine

The value of the expression $u\text{×}v$ and show that the product obtained is orthogonal to both u and v.

Answer to Problem 17E

The value of the expression $u\text{×}v$ is $i-2j-2k$ and both the vectors u and v are orthogonal to $u\text{×}v$.

Explanation of Solution

Given information:

The value of the vectors in component form $u=\langle 2,-3,4\rangle$ and $v=\langle 0,-1,1\rangle$ .

Formula used:

If there are two three-dimensional vectors in a space given in the form of their components, say, $a=\langle a_1,a_2,a_3\rangle$ and $b=\langle b_1,b_2,b_3\rangle$ , then the cross product of two vectors is defined as ,

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

Any two non-zero vectors are said to be orthogonal to each other if and only if their dot product is zero, i.e. the angle between the two vectors is ${90}^{\circ }$ . Dot product of any two vectors, say, $x=x_{1}i+x_{2}j+x_{3}k$ and $y=y_{1}i+y_{2}j+y_{3}k$ is expressed as,

  $\begin{array}{c}x\cdot y=\left(x_{1}i+x_{2}j+x_{3}k\right)\cdot \left(y_{1}i+y_{2}j+y_{3}k\right)\\ =x_1{y}_1+x_2{y}_2+x_3{y}_3\end{array}$

Calculation:

Consider the given value of the vectors $u=\langle 2,-3,4\rangle$ and $v=\langle 0,-1,1\rangle$ .

Recall that if there are two three-dimensional vectors in a space given in the form of their components, say, $a=\langle a_1,a_2,a_3\rangle$ and $b=\langle b_1,b_2,b_3\rangle$ , then the cross product of two vectors is defined as ,

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

Apply it,

So, the cross product of u and v vectors will be calculated as,

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

Simplify it further as,

  $\begin{array}{c}u\text{×}v=\left(-3+4\right)i-\left(2\right)j+\left(-2\right)k\\ \text{=}i-2j-2k\end{array}$

Now, recall that any two non-zero vectors are said to be orthogonal to each other if and only if their dot product is zero, i.e. the angle between the two vectors is ${90}^{\circ }$ . Dot product of any two vectors, say, $x=x_{1}i+x_{2}j+x_{3}k$ and $y=y_{1}i+y_{2}j+y_{3}k$ is expressed as,

  $\begin{array}{c}x\cdot y=\left(x_{1}i+x_{2}j+x_{3}k\right)\cdot \left(y_{1}i+y_{2}j+y_{3}k\right)\\ =x_1{y}_1+x_2{y}_2+x_3{y}_3\end{array}$

So, to prove that the product obtained is orthogonal to both u and v, we will calculate the dot product of u and v with $u\text{×}v$ .

So, the dot product of u and $u\text{×}v$ will be calculated as,

  $\begin{array}{c}u\cdot \left(u\text{×}v\right)=\left(\text{2}i-3j\text{+4}k\right)\cdot \left(i-2j-2k\right)\\ =\left[2\cdot 1\right]+\left[\left(-3\right)\cdot \left(-2\right)\right]+\left[4\cdot \left(-2\right)\right]\\ =2+6-8\\ =0\end{array}$

Since, dot product of u and $u\text{×}v$ is zero, so, u is orthogonal to $u\text{×}v$ .

Now, dot product of v and $u\text{×}v$ will be calculated as,

  $\begin{array}{c}v\cdot \left(u\text{×}v\right)=\left(\text{0}i-1j\text{+1}k\right)\cdot \left(i-2j-2k\right)\\ =\left[0\cdot 1\right]+\left[\left(-1\right)\cdot \left(-2\right)\right]+\left[1\cdot \left(-2\right)\right]\\ =0+2-2\\ =0\end{array}$

Since, dot product of v and $u\text{×}v$ is zero, so, v is also orthogonal to $u\text{×}v$

  $\text{u×v}$ .

Hence, the value of the expression $u\text{×}v$ is $i-2j-2k$ and it is also proved that both u and v are orthogonal to $u\text{×}v$ .