已知点A(2,-1)B(-1,3)C(-2,-5)①求向量AB+向量AC的坐标②求向量AB-向量AC

2024-12-28 06:22:14
推荐回答(2个)
回答1:

1

AB=(-1,3)-(2,-1)=(-3,4),AC=(-2,-5)-(2,-1)=(-4,-4)
AB+AC=(-3,4)+(-4,-4)=(-7,0)
2
AB-AC=(-3,4)-(-4,-4)=(1,8)
3
AB·AC=(-3,4)·(-4,-4)=12-16=-4
4
AB·AC=|AB|*|AC|*cos(∠BAC)=5*4sqrt(2)*cos(∠BAC)
故:cos(∠BAC)=AB·AC/(20sqrt(2))=-4/(20sqrt(2))=-sqrt(2)/10
即:∠BAC=π-arccos(sqrt(2)/10)

回答2:

①AB=(-1-2)x+(3-(-1))y=-3x+4y
AC=(-2-2)x+(-5-(-1))y=-4x-4y
AB+AC4x-4=(-3x+4y)+(4x-4y)=-7x+0y=-7x 为起点在原点,长度为7,方向为x轴的负方向的向量,座标为(-7,0);
②AB-AC=(-3x+4y)-(4x-4y)=x+8y,座标为(1,8);