1. In equilateral triangle ABC, the angle between vector AB and vector BC is a.90 degree b.60 degree c.120 degree d.45 degree (why choose C). 2. In triangle ABC, if vector ab × vector BC > 0, then the triangle is? 3. Given that unit vectors E1 and E2 are perpendicular to each other, then (a + 2b) × (3a-b) =? 3 there is a mistake. It should be (3e1-2e2) × (E1 + E2) =?

1. In equilateral triangle ABC, the angle between vector AB and vector BC is a.90 degree b.60 degree c.120 degree d.45 degree (why choose C). 2. In triangle ABC, if vector ab × vector BC > 0, then the triangle is? 3. Given that unit vectors E1 and E2 are perpendicular to each other, then (a + 2b) × (3a-b) =? 3 there is a mistake. It should be (3e1-2e2) × (E1 + E2) =?

1. Vector AB and vector BC are originally different starting points, but if you translate them to the same starting point and look at them again, it's not 120 degrees. 2. Vector multiplication also needs to draw a triangle at the same starting point. See vector AB multiplied by vector BC = the module length of vector AB to form the module length of vector BC, and then multiplied by the cos value of the angle formed by abbc and the starting point. Because the module length is definitely greater than 0, the cosine value is greater than 0, so it's the acute angle of the first quadrant, because it's complementary, So one of the angles in the triangle is obtuse, so obtuse triangle 3 didn't understand your question