Chapter 6.1, Problem 26E

Solution

(a)

To find: The unit vector in the direction of the given vector in component form.

The unit vector is equal to $〈-\frac{3}{\sqrt{13}},\frac{2}{\sqrt{13}}〉$ .

Given information:

The given vector is $\overrightarrow{u}=〈-3,2〉$ .

Concept used:

The unit vector is $\widehat{a}=\frac{\overrightarrow{a}}{\left|a\right|}$ .

Calculation:

The given vector is $\overrightarrow{u}=〈-3,2〉$ .

Apply the above concept and find the unit vector below:

  $\begin{array}{c}\widehat{u}=\frac{〈-3,2〉}{\sqrt{{\left(-3\right)}^2+{\left(2\right)}^2}}\\ =\frac{〈-3,2〉}{\sqrt{9+4}}\\ =\frac{〈-3,2〉}{\sqrt{13}}\\ =〈-\frac{3}{\sqrt{13}},\frac{2}{\sqrt{13}}〉\end{array}$

Therefore, the unit vector is $〈-\frac{3}{\sqrt{13}},\frac{2}{\sqrt{13}}〉$ .

(b).

To find: The unit vector in the direction of the given vector in standard unit vectors $i$ and $j$ .

The unit vector is equal to $\frac{-3i+2j}{\sqrt{13}}$ .

Given information:

The given vector is $\overrightarrow{u}=〈-3,2〉$ .

Concept used:

The unit vector is $\widehat{a}=\frac{\overrightarrow{a}}{\left|a\right|}$ .

Calculation:

The given vector is $\overrightarrow{u}=〈-3,2〉$ .

Rewrite the given vector below:

  $\overrightarrow{u}=-3i+2j$

Apply the above concept and find the unit vector below:

  $\begin{array}{c}\widehat{u}=\frac{-3i+2j}{\sqrt{{\left(-3\right)}^2+{\left(2\right)}^2}}\\ =\frac{-3i+2j}{\sqrt{9+4}}\\ =\frac{-3i+2j}{\sqrt{13}}\end{array}$

Therefore, the unit vector is $\frac{-3i+2j}{\sqrt{13}}$ .