Question

Найдите длину векторов a, b, c: a (2;1), b (3;3), с (-4;0)

Answer 1

A{2;1} ;
|a| = $\sqrt{2^2+1^2}= \sqrt{4+1} = \sqrt{5}$

b{3;3} ;
|b| = $\sqrt{3^2+3^2} = \sqrt{9+9} = \sqrt{18}= \sqrt{3^2*2} =3 \sqrt{2}$

c{-4;0} ;
|c| = $\sqrt{(-4)^2+0^2}= \sqrt{16}=4$

Ответ: $\sqrt{5 } ; 3 \sqrt{2} ; 4$

Answer 2

|а|^2= 4+1=5 |а|=√5 |в|^2=9+9=18 |в|=√18=3√2 |с|^2=16+0=16 |с|=√16=4