### Home > CALC > Chapter 11 > Lesson 11.1.1 > Problem11-10

11-10.

If vector $\vec { a } = \langle - 2,5 )$, $\vec { b } = \langle 6,0 \rangle$, and $\vec { c } = \langle \frac { 3 } { 2 } , 8 )$, find:

1. $2 \vec { c } - \vec { a }$

$2\big<\frac{3}{2},8\big>-\big<-2,5\big>$

$\big<3,16\big>-\big<-2,5\big>$

1. $- \vec { b } + \vec { c }$

$-\big<6,0\big>+\Big<\frac{3}{2},8\Big>$

1. $|| \vec { a } + \vec { b } ||$

Express $\vec{a}+\vec{b}$ as a single vector.

The double bars indicate "the length of". Use the Pythagorean Theorem to calculate this length.

1. $\frac { \vec { b } } { \| \vec { b } \| }$

$||\vec{b}||=6$