Chapter 9.1, Problem 32E

To determine

The value of $2\text{u, -}3\text{v, u + v, 3u - 4v}$ .

Answer to Problem 32E

The desired value is $\begin{array}{l}2\text{u}=\langle \text{-4,10}\rangle \\ -3\text{v}=\langle \text{-6,24}\rangle \\ \text{u + v}=\langle \text{0,-3}\rangle \\ \text{3u - 4v}=\langle 2,-17\rangle \end{array}$ .

Explanation of Solution

Given :

  $\begin{array}{l}\text{u = }\langle \text{-2,5}\rangle \\ \text{v = }\langle \text{2,-8}\rangle \end{array}$

Calculation:

Formula used:

If.

  $\begin{array}{l}\text{u = }\langle {\text{a}}_{\text{1}}{\text{,b}}_{\text{1}}\rangle \\ \text{v = }\langle {\text{a}}_{\text{2}}{\text{,b}}_{\text{2}}\rangle \\ \text{u + v = }\langle {\text{a}}_{\text{1}}{\text{+ a}}_{\text{2}}{\text{,b}}_{\text{1}}{\text{+ b}}_{\text{2}}\rangle \\ \text{u - v =}\langle {\text{a}}_{\text{1}}{\text{- a}}_{\text{2}}{\text{,b}}_{\text{1}}{\text{- b}}_{\text{2}}\rangle \end{array}$

  $\text{cu = c}\langle {\text{a}}_{\text{1}}{\text{,b}}_{\text{1}}\rangle {\text{ = ca}}_1+{\text{cb}}_{\text{1}};\text{c}\in \text{R}$

  $\text{cv = c}\langle {\text{a}}_{\text{2}}{\text{,b}}_{\text{2}}\rangle {\text{ = ca}}_2+{\text{cb}}_{\text{2}};\text{c}\in \text{R}$

So,

  $\begin{array}{l}2\text{u}=2\langle \text{-2,5}\rangle =\langle \text{-4,10}\rangle \\ -3\text{v}=\text{ -3}\langle \text{2,-8}\rangle =\langle \text{-6,24}\rangle \\ \text{u + v }\\ \text{=}\langle \text{-2,5}\rangle +\langle \text{2,-8}\rangle \\ =\langle \text{0,-3}\rangle \\ \text{3u - 4v }\\ \text{= 3}\langle \text{-2,5}\rangle -4\langle \text{2,-8}\rangle \\ =\langle \text{-6,15}\rangle -\langle \text{-8,32}\rangle \\ =\langle 2,-17\rangle \end{array}$