Question

16) sabendo que o angulo entre os vetores u(2,1,-1) e v(1,-1,m+2) é pi sobre 3, determinar m.

Answer 1

$\boxed{\boxed{cos(\theta)= \frac{U.V}{|U|*|V|} }}$

Ф= angulo entre os vetores = π/3 (60º)
U.V = produto escalar entre os vetores

$U=(2,1,-1)\\|U|= \sqrt{2^2+1^2+(-1)^2} = \sqrt{6} \\\\\\V=(1,-1,m+2)\\|V|= \sqrt{1^2+(-1)^2+(m+2)^2} \\|V|= \sqrt{2+m^2+4m+4} \\\\|V| = \sqrt{m^2+4m+6}$

então

$cos( \frac{\pi}{3} )= \frac{(2,1,-1).(1,-1,m+2)}{ \sqrt{6}* \sqrt{m^2+4m+6}} \\\\ \frac{1}{2}= \frac{(2*1)+(1*(-1)) + (-1*(m+2))}{ \sqrt{6*(m^2+4m+6)} } \\\\ \frac{1}{2}= \frac{2-1-m-2}{6m^2+24m+36} \\\\ \sqrt{6m^2+24m+36}=2*(-1-m)\\\\6m^2+24m+36=2^2*(-1-m)^2\\\\6m^2+24m+36= 4m^2+8m+4\\\\ 2m^2+16m+32=0\\\\m^2+8m+16=0\\\\m^2+2*4*m+4^2=0\\(m+4)^2=0\\\\\boxed{\boxed{m=-4}}$