Chapter 11.3, Problem 34E

To determine

To calculate: For the set of vectors $u=7i-14j+5k$ and $v=14i+28j-15k$ a unit vector that is orthogonal to both the vectors.

Answer to Problem 34E

A unit vector that is orthogonal to both the vectorsis $\frac{\sqrt{429}}{9009}\left(70i+175j+392k\right)$ .

Explanation of Solution

Given information:

The set of vectors $u=7i-14j+5k$ and $v=14i+28j-15k$ .

Formula used:

Let $a\times b$ represents the vector x such that $x=x_{1}i+x_{2}j+x_{3}k$ then the magnitude of the vector $x$ is given by $\Vert x\Vert =\sqrt{x_1{}^2+x_2{}^2+x_3{}^3}$ .

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 ,

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

A unit vector that is orthogonal to both $a=\langle a_1,a_2,a_3\rangle$ and $b=\langle b_1,b_2,b_3\rangle$ is, $\frac{a\times b}{\Vert a\times b\Vert }$ .

Calculation:

Consider the set of vectors $u=7i-14j+5k$ and $v=14i+28j-15k$ .

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 vectors $u=7i-14j+5k$ and $v=14i+28j-15k$ is,

  $\begin{array}{c}u\times v=\begin{array}{ccc}i& j& k\\ 7& -14& 5\\ 14& 28& -15\end{array}\\ \text{=}\left(210-140\right)i-\left(-105-70\right)j+\left(196+196\right)k\\ \text{=}70i+175j+392k\end{array}$

Let $a\times b$ represents the vector $x$ such that $x=x_{1}i+x_{2}j+x_{3}k$ then the magnitude of the vector $x$ is given by $\Vert x\Vert =\sqrt{x_1{}^2+x_2{}^2+x_3{}^3}$ .

Magnitude of $u\times v$ is,

  $\begin{array}{c}\Vert u\times v\Vert =\sqrt{{\left(70\right)}^2+{\left(175\right)}^2+{\left(392\right)}^2}\\ =21\sqrt{429}\end{array}$

Recall that a unit vector that is orthogonal to both $a=\langle a_1,a_2,a_3\rangle$ and $b=\langle b_1,b_2,b_3\rangle$ is, $\frac{a\times b}{\Vert a\times b\Vert }$ .

Apply it, so, a unit vector that is orthogonal to both $u=7i-14j+5k$ and $v=14i+28j-15k$ is,

  $\begin{array}{c}\frac{u\times v}{\Vert u\times v\Vert }=\frac{70i+175j+392k}{21\sqrt{429}}\\ =\frac{70i+175j+392k}{21\sqrt{429}}\cdot \frac{\sqrt{429}}{\sqrt{429}}\\ =\frac{\sqrt{429}}{9009}\left(70i+175j+392k\right)\end{array}$

Thus, a unit vector that is orthogonal to both the vectorsis $\frac{\sqrt{429}}{9009}\left(70i+175j+392k\right)$ .