pesso resposta da pergunta a seguir

a) (-2,3).(4,3) = 1

a) O valor de AB. U será:

\(\begin{align} & A=(3,-2) \\ & B(1,1) \\ & u=(4,3) \\ & \\ & AB\cdot u=\left( B-A \right)u \\ & AB\cdot u=\left( (1-3,1+2) \right)(4,3) \\ & AB\cdot u=\left( -2,3 \right)(4,3) \\ & AB\cdot u=-8+9 \\ & AB\cdot u=1 \\ \end{align}\ \)

b) Agora para cos(ABû) temos:

\(\begin{align} & \cos (AB\hat{\ }u)=\cos \theta \\ & \cos \theta =\frac{AB\cdot u}{|AB|\cdot |u|} \\ & \cos \theta =\frac{1}{|\sqrt{4+9}|\cdot |\sqrt{16+9}|} \\ & \cos \theta =\frac{1}{|\sqrt{13}|\cdot |5|} \\ & \,\cos \theta =\frac{1}{5\sqrt{13}} \\ & \theta =87{}^\text{o} \\ \end{align}\ \)

c) Para encontrarmos as coordenadas, realizaremos os cálculos abaixo:

\(\begin{align} & AB|u\hat{\ }AB|=10 \\ & AB\times u\cdot AB=10 \\ & AB\times u=10 \\ & \\ & \left[ \begin{matrix} -2 & 3 & 1 \\ 4 & 3 & 1 \\ i & j & 1 \\ \end{matrix} \right]\begin{matrix} -2 & 3 \\ 4 & 3 \\ i & j \\ \end{matrix}=10 \\ & -6+3i-4j-3i+2j-12=10 \\ & -18-2j=10 \\ & -2j=28 \\ & j=14 \\ & \\ & P(0,14) \\ \end{align}\ \)