Dados os vetores u=(5,x,-2) , v=(x,3,2) e os pontos A(-1,5,-2) e B(3,2,4), determinar o valor de x tal que u.(v+BA)=10.
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vamos determinar o vetor BA
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∵
u.(v + BA) = 10
![\\ (5, x , -2).[(x,3,2)+(-4,3,-6)] = 10
\\
\\ (5, x , -2).(x-4,3+3,2-6)=10
\\
\\ (5, x , -2).(x-4, 6, -4) = 10
\\
\\ 5(x-4)+x(6)-2(-4)=10
\\
\\ 5x-20+6x+8=10
\\
\\ 11x-12=10
\\
\\ 11x=10+12
\\
\\ 11x=22
\\
\\ x = \frac{22}{11}](https://tex.z-dn.net/?f=+%5C%5C+%285%2C+x+%2C+-2%29.%5B%28x%2C3%2C2%29%2B%28-4%2C3%2C-6%29%5D+%3D+10%0A+%5C%5C+%0A+%5C%5C+%285%2C+x+%2C+-2%29.%28x-4%2C3%2B3%2C2-6%29%3D10%0A+%5C%5C+%0A+%5C%5C+%285%2C+x+%2C+-2%29.%28x-4%2C+6%2C+-4%29+%3D+10%0A+%5C%5C+%0A+%5C%5C+5%28x-4%29%2Bx%286%29-2%28-4%29%3D10%0A+%5C%5C+%0A+%5C%5C+5x-20%2B6x%2B8%3D10%0A+%5C%5C+%0A+%5C%5C+11x-12%3D10%0A+%5C%5C+%0A+%5C%5C+11x%3D10%2B12%0A+%5C%5C+%0A+%5C%5C+11x%3D22%0A+%5C%5C+%0A+%5C%5C+x+%3D++%5Cfrac%7B22%7D%7B11%7D+ "\\ (5, x , -2).[(x,3,2)+(-4,3,-6)] = 10
\\
\\ (5, x , -2).(x-4,3+3,2-6)=10
\\
\\ (5, x , -2).(x-4, 6, -4) = 10
\\
\\ 5(x-4)+x(6)-2(-4)=10
\\
\\ 5x-20+6x+8=10
\\
\\ 11x-12=10
\\
\\ 11x=10+12
\\
\\ 11x=22
\\
\\ x = \frac{22}{11}")
x = 2
------------------------------
∵
u.(v + BA) = 10
x = 2
joelsonccruz:
obrigado
Por nada :-)
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