Chapter 9, Problem 45RE

To determine

To find: The unit vector which is orthogonal to both $v$ and $w$ .

Answer to Problem 45RE

The unit vector which is orthogonal to both $v$ and $w$ is $3i+9j+9k$ .

Explanation of Solution

Given information:

The first vector is

  $v=3i+j-2k$

The second vector is

  $w=-3i+2j-k$

Calculation:

The cross product of the two vectors is

  $v\times w=\left(b_1{c}_2-b_2{c}_1\right)i-\left(a_1{c}_2-a_2{c}_2\right)j+\left(a_1{b}_2-a_2{b}_1\right)k$

Here, the values of $a_1=3$ , $b_1=1$ , $c_1=-2$ , $a_2=-3$ , $b_2=2$ and $c_2=-1$ .

Substitute the values in the above expression.

  $\begin{array}{c}v\times w=\left(1\left(-1\right)-2\left(-2\right)\right)i-\left(3\left(-1\right)-\left(-3\right)\left(-2\right)\right)j+\left(3\left(2\right)-\left(-3\right)1\right)k\\ =\left(-1+4\right)i-\left(-3-6\right)j+\left(6+3\right)k\\ =3i+9j+9k\end{array}$

The vector $3i+9j+9k$ is orthogonal to both $v$ and $w$ .

The two vectors are orthogonal if their dot product is zero.

Calculate the dot product of $v\cdot \left(v\times w\right)$ .

  $\begin{array}{c}v\cdot \left(v\times w\right)=\left(3i+j-2k\right).\left(3i+9j+9k\right)\\ =9+9-18\\ =0\end{array}$

Calculate the dot product of $w\cdot \left(v\times w\right)$ .

  $\begin{array}{c}w\cdot \left(v\times w\right)=\left(-3i+2j-k\right).\left(3i+9j+9k\right)\\ =-9+18-9\\ =0\end{array}$

Therefore, the unit vector which is orthogonal to both $v$ and $w$ is $3i+9j+9k$ .