Chapter 9, Problem 85RE

To determine

To calculate:The unit vector in the same direction as $v$ and then in the opposite direction of $v$ . Where the vectors are $v=3i+j-2k$ and $w=-3i+2j-k$ .

Answer to Problem 85RE

Theunit vector in the direction of vector $v$ and in the opposite direction of $v$ :

  $\begin{array}{l}{u}_1=\frac{3}{\sqrt{14}}i+\frac{1}{\sqrt{14}}j-\frac{2}{\sqrt{14}}k\\ u_2=-\frac{3}{\sqrt{14}}i-\frac{1}{\sqrt{14}}j+\frac{2}{\sqrt{14}}k\end{array}$ .

Explanation of Solution

Given information:

The vectors:

  $v=3i+j-2k$ and $w=-3i+2j-k$

Formula used:

Formula of the unit vector on the direction of vector.

  $u_1=\frac{v}{\Vert v\Vert }$

Formula of unit vector on the direction opposite to the given vector.

  $u_1=-\frac{v}{\Vert v\Vert }$

Calculation:

Consider the vectors:

  $v=3i+j-2k$ and $w=-3i+2j-k$

First find the unit vector in the direction as $v$ of the vector,

Apply the formula of the unit vector in the direction as $v$ of the vector,

  $u_1=\frac{v}{\Vert v\Vert }$

Apply the values in the formula,

First calculate the magnitude of $\Vert v\Vert$ .

  $\begin{array}{c}\Vert v\Vert =\sqrt{{\left(3\right)}^2+{\left(1\right)}^2+{\left(-2\right)}^2}\\ =\sqrt{9+1+4}\\ =\sqrt{14}\end{array}$

Now apply the value of magnitude of $\Vert v\Vert$ in the formula:

  $\begin{array}{c}{u}_1=\frac{v}{\Vert v\Vert }\\ =\frac{3i+j-2k}{\sqrt{14}}\\ =\frac{3}{\sqrt{14}}i+\frac{1}{\sqrt{14}}j-\frac{2}{\sqrt{14}}k\end{array}$

Observe that, above result is the result of unit vector in the direction as v.

Now find the direction opposite to the vector $v$ .

Apply the formula here for direction opposite to the vector $v$ .

  $u_1=-\frac{v}{\Vert v\Vert }$

Now rewrite the expression in the formula:

As the value of unit vector we already found above then,

  $\begin{array}{c}{u}_1=-\frac{v}{\Vert v\Vert }\\ =\frac{3}{\sqrt{14}}i+\frac{1}{\sqrt{14}}j-\frac{2}{\sqrt{14}}k\end{array}$

Now for opposite direction take minus sign to the unit vector and got the result for the opposite direction of the vector $v$ .

  $\frac{3}{\sqrt{14}}i+\frac{1}{\sqrt{14}}j-\frac{2}{\sqrt{14}}k=-\frac{3}{\sqrt{14}}i-\frac{1}{\sqrt{14}}j+\frac{2}{\sqrt{14}}k$

Thus, The unit vector in the direction of vector $v$ and in the opposite direction of $v$ :

  $\begin{array}{l}{u}_1=\frac{3}{\sqrt{14}}i+\frac{1}{\sqrt{14}}j-\frac{2}{\sqrt{14}}k\\ u_2=-\frac{3}{\sqrt{14}}i-\frac{1}{\sqrt{14}}j+\frac{2}{\sqrt{14}}k\end{array}$ .