Question

find the angle between the vectors a and b,when
(iii) a = i - j and b = j + k.​

Answer 1

Resposta:   $\alpha$ = 120º

Explicação passo a passo:

i = (1,0,0)

j=(0,1,0)

k=(0,0,1)

a= i - j = (1,0,0)-(0,1,0)= (1,-1,0)

|a| = $\sqrt{(1)^{2}+(-1)^{2}+(0)^{2}} =\sqrt{2}$

b= j + k = (0,1,0)+(0,0,1)= (0,1,1)

|b| = $\sqrt{(0)^{2}+(1)^{2}+(1)^{2}} =\sqrt{2}$

produto escalar:

<a,b>= (1,-1,0) . (0,1,1) = ( ( 1 . 0 ) + ( -1 . 1 ) + ( 0 . 1 ) ) = -1

ângulo:

<a,b> = cos $\alpha$ . |a| . |b|

-1 = cos $\alpha$ . $\sqrt{2} . \sqrt{2}$

cos $\alpha$ = $\frac{-1}{\sqrt{2} . \sqrt{2} }$ = $\frac{-1}{2}$

$\alpha$ = arccos $\frac{-1}{2}$

$\alpha$ = 120º