Question

Determine um vetor u ortogonal a v=(1,1,0) tal que ||u||=raiz de 2 e que a medida do ângulo em graus entre u e vetor (1,-1,0) seja 45º.

Answer 1

angulo entre vetores A e B

$\boxed{\boxed{cos(\theta)= \frac{\vec A \cdot \vec B}{||\vec A||*||\vec B||} }}$

dados:
$U=(x,y,z)\\\\||U||= \sqrt{x^2+y^2+z^2}= \sqrt{2}\\\\\\V=(1,1,0)\\\\||V||= \sqrt{1^1+1^1+0^2}=\sqrt{2}\\\\\\P=(1,-1,0)\\\\||P||= \sqrt{1^1+(-1)~2+0^2}=\sqrt{2}$

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U é ortogonal ao vetor V..então o angulo entre eles é 90°

$cos(90)= \frac{(x,y,z)*(1,1,0)}{\sqrt{2}*\sqrt{2}} \\\\0= \frac{(x*1)+(y*1)+(z*0)}{2}\\\\\boxed{0=x+y}$

o vetor U forma um angulo de 45° com o vetor P
$cos(45)= \frac{(x,y,z)*(1,-1,0)}{\sqrt{2}*\sqrt{2}} \\\\ \frac{\sqrt{2}}{2}= \frac{(x*1)+(y*(-1))+(z*0)}{2} \\\\ \boxed{\sqrt{2}=x-y }$

agora temos
$\Bmatrix{x+y=0\\\\x-y=\sqrt{2}\end$

somando as duas equaçoes
$2x= \sqrt{2}\\\\\boxed{x= \frac{ \sqrt{2} }{2} }$

logo o valor de y sera
$y=- \frac{\sqrt{2}}{2}$

calculando o z
$||U||= \sqrt{x^2+y^2+z^2}= \sqrt{2}\\\\ \sqrt{(\frac{\sqrt{2}}{2} )^2+(-\frac{\sqrt{2}}{2} )^2+z^2}=\sqrt{2}\\\\ \frac{2}{4}+ \frac{2}{4}+z^2=2\\\\z^2=2-1\\\\\boxed{z=1 }$

Answer 2

Vetor v = (x,y,z) ortogonal (1,1,0)

(x,y,z) . (1,1,0) = 0

x + y = 0

y = -x

v = (x,-x,z)

Norma u ser √2

√x²+(-x)²+z² = √2

2x² + z² = 2

Ângulo u e (1,-1,0) ser 45°

cos 45° = (x,-x,z).(1,-1,0)/||u||.||(1,-1,0)||

√2/2 = 2x/√2.√2

x = √2/2

Logo

2 . 2/4 + z² = 2

z² = 1

z = ±1

u = (√2/2, -√2/2, 1) ou u = (√2/2, -√2/2, -1)