Chapter 6.1, Problem 27E

Solution

(a).

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

The unit vector is equal to $〈-\frac{4}{\sqrt{41}},-\frac{5}{\sqrt{41}}〉$ .

Given information:

The given vector is $\overrightarrow{u}=〈-4,-5〉$ .

Concept used:

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

Calculation:

The given vector is $\overrightarrow{u}=〈-4,-5〉$ .

Apply the above concept and find the unit vector below:

  $\begin{array}{c}\widehat{u}=\frac{〈-4,-5〉}{\sqrt{{\left(-4\right)}^2+{\left(-5\right)}^2}}\\ =\frac{〈-4,-5〉}{\sqrt{16+25}}\\ =\frac{〈-4,-5〉}{\sqrt{41}}\\ =〈-\frac{4}{\sqrt{41}},-\frac{5}{\sqrt{41}}〉\end{array}$

Therefore, the unit vector is $〈-\frac{4}{\sqrt{41}},-\frac{5}{\sqrt{41}}〉$ .

(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 $\frac{-4i-5j}{\sqrt{41}}$ .

Given information:

The given vector is $\overrightarrow{u}=〈-4,-5〉$ .

Concept used:

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

Calculation:

The given vector is $\overrightarrow{u}=〈-4,-5〉$ .

Rewrite the given vector below:

  $\overrightarrow{u}=-4i-5j$

Apply the above concept and find the unit vector below:

  $\begin{array}{c}\widehat{u}=\frac{-4i-5j}{\sqrt{{\left(-4\right)}^2+{\left(-5\right)}^2}}\\ =\frac{-4i-5j}{\sqrt{16+25}}\\ =\frac{-4i-5j}{\sqrt{41}}\end{array}$

Therefore, the unit vector is $\frac{-4i-5j}{\sqrt{41}}$ .