Chapter 6, Problem 53CR

To determine

To calculate:

Unit vector and verify the result has magnitude of $1$ .

Answer to Problem 53CR

Unit vector in the direction of the given vector is $\frac{8}{17}i-\frac{15}{17}j$ .

Explanation of Solution

Given information:

Vector: $v=8i-15j$

Calculation:

Consider the vector $v=8i-15j$

The objective is to find the unit vector in the direction of the given vector and verify that the result has a magnitude of $1$ .

A unit vector is a vector that has a magnitude of 1 unit. The unit vector in the direction of $v$ is,

  $\begin{array}{l}\frac{v}{\Vert v\Vert }=\frac{8i-15j}{\Vert 8i-15j\Vert }\\ =\frac{8i-15j}{\sqrt{{\left(8\right)}^2+{\left(-15\right)}^2}}\\ =\frac{8i-15j}{\sqrt{64+225}}\\ =\frac{8i-15j}{\sqrt{289}}\\ =\frac{8i-15j}{17}\\ =\frac{8}{17}i-\frac{15}{17}j\end{array}$

Thus, the vector in the direction of $v$ is $\frac{8}{17}i-\frac{15}{17}j$ .

The result obtained is verified by showing that the vector obtained has a magnitude $1$ unit.

  $\begin{array}{l}\Vert \frac{8}{17}i-\frac{15}{17}j\Vert =\sqrt{{\left(\frac{8}{17}\right)}^2+{\left(\frac{-15}{17}\right)}^2}\\ =\sqrt{\frac{64}{289}+\frac{225}{289}}\\ =\sqrt{\frac{64+225}{289}}\\ =\sqrt{\frac{289}{289}}\\ =1\end{array}$

Thus, the result is verified.