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In triangle ABC, AB vector is equal to a, AC vector is equal to b, when a times b is greater than 0, a times b is equal to 0, what triangle ABC is
Vector a point multiplies vector b=|a||b|cos (a^b)
|A||b|>0,
So a times b >0 can deduce cos (a^b)>0, i.e.0
In triangle ABC, the module of vector AB is equal to 4, the module of vector BC is equal to 5, the module of vector AC is equal to 7, and the inner product of vector AB and vector AC is obtained
Find the angle between AB and AC first, and use cosine formula, cos8=(AB2+AC2-BC2)/(2 x AB x AC)=5/7
Vector AB x Vector AC =|AB | x | AC | x cos8=20
For the angle between AB and AC, cosine formula can be used, cos8=(AB2+AC2-BC2)/(2 x AB x AC)=5/7
Vector AB x Vector AC =|AB | x | AC | x cos8=20
In triangle ABC AB=5 BC=7 AC=8 the value of vector AB click vector BC is?
0
In triangle ABC, vector AB=a, vector BC=b, AD is the center line on BC, G is the center of gravity of triangle ABC, then vector AG=?
AG=2/3AD=2/3(AB+BD)=2/3(AB+1/2BC)=2/3a+1/3b
AG =2/3 AD =2/3(AB+BD)=2/3(AB+1/2/2BC)=2/3a+1/3b
In the triangle ABC, AB=a, BC=b, AD is the center line of edge, G is the center of gravity of triangle ABC, find vector AG
AG=2/3*AD=2/3(AB+1/2BC)
=2/3*A+1/3*b
AG=2/3*AD=2/3(AB+1/2BC)
=2/3*A+1/3*B
In triangle ABC, AB=2, AC=3, D is the midpoint of the BC side, then AD vector times BC vector=
Vector AC - Vector AB = Vector BC
(Vector AC + Vector AB)/2= Vector AD
AD*BC =(vector AC + vector AB)*(vector AC - Vector AB)/2=(AC Square - AB Square)/2=2.5
Vector AC - Vector AB = Vector BC
(Vector AC + Vector AB)/2= Vector AD
AD*BC =(Vector AC + Vector AB)*(Vector AC - Vector AB)/2=(AC Square - AB Square)/2=2.5
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