Svar 2.2:8
FörberedandeFysik
(Skillnad mellan versioner)
			  			                                                      
		          
			 (Ny sida: <math>2</math>'''b'''–'''c'''<math>+3</math>'''d'''<math>=2(3,-1)–(-2,4)+3(1,2)=(11,0)</math>;   <math>e=\frac{(11,0)}{\sqrt{11^2+0^2}}=(1,0)=e_x</math>)  | 
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| Rad 1: | Rad 1: | ||
| - | <math>2  | + | <math> 2\mathbf{b}\, –\, \mathbf{c} + 3\mathbf{d} = 2(3, -1) – (-2, 4) + 3(1, 2) = (11, 0); </math>  | 
| - | <math>e=\frac{(11,0)}{\sqrt{11^2+0^2}}=(1,0)=e_x</math>  | + | |
| + | <math>\displaystyle \mathbf{e} = \frac{(11,0)}{\sqrt{11^2+0^2}} = (1,0) = \mathbf{e_x} </math>  | ||
Nuvarande version
\displaystyle 2\mathbf{b}\, –\, \mathbf{c} + 3\mathbf{d} = 2(3, -1) – (-2, 4) + 3(1, 2) = (11, 0);
\displaystyle \displaystyle \mathbf{e} = \frac{(11,0)}{\sqrt{11^2+0^2}} = (1,0) = \mathbf{e_x} 
