State reasons 1. Given vector AB =2, vector CD =4, the angle between vector AB and vector CD is 60°, Find (1) vector AB+vector CD (2) Vector AB + angle between vector CD and vector AB State reasons 1. Given that vector AB =2, vector CD =4, the angle between vector AB and vector CD is 60°, Find (1) vector AB+vector CD (2) Vector AB + angle between vector CD and vector AB

Let a vector be not equal to 0 vector, a vector point times b vector = a vector point times c vector, and b vector is not equal to c vector.

Use a, b, c for vectors and omit the word vector.
A·b=a·c, so a·b-a·c=0, so a·(b-c)=0(distributive law)
And b=c so b-c=0, and a=0, two vector points not equal to 0 multiplied by 0, can only be vertical, so a⊥(b-c)

What does adding two vectors equals zero mean?

These two vectors are the same size in the opposite direction

These two vectors are equal in in the opposite direction.

How does the vector |a·b|=|a b|prove this formula? I searched this formula on baidu encyclopedia but I don't understand why

By definition:, a·b=|a b|* cos (angle of a, b).
Therefore:|a·b|=|a b||cos (angle of a, b)|
Only when |coc (included angle of a, b)|=1, i.e. a, b collinear,
|A·b|=|a b|.
So generally, you can't say,|a.b|=|a.b|.

By definition:, a·b=|a b|* cos (angle of a, b).
Therefore:|a·b|=|a b||cos (included angle of a, b)|
Only when |coc (included angle of a, b)|=1, i.e. a, b collinear,
|A·b|=|a b|.
Therefore, in general, we can not say,|a.b|=|a. b|.

Vector proof Given that AM is the center line of BC in triangle ABC, prove it by vector AM Square =1/2(AB Square + AC Square - BM Square Vector proof It is known that AM is the center line of BC in triangle ABC, by vector AM Square =1/2(AB Square + AC Square - BM Square

The following is proof: AM=(AB+AC)/2; so AM2=(AB2+AC2)/4+AB*AC/2, but in order to become the side length, AB*AC should be transformed into the formula about the square. Therefore, AB*AC=(AM+MB)(AM+MC)=(AM+MB)(AM-MB)=AM2-MB2. Substitute AM2=(AB2+AC2)/4+(AM2-MB2)/2 shift term, and multiply it by 2 to obtain the solution.

A proof of a vector. Let vector groups α1,α2,α3,α4,α5 be linearly independent β1=α1+α2 β2=α2+α3 β3=α3+α1 β4=α4+α5 β5=α5+α1 Proving that β1,β2,β3,β4,β5 are linearly independent

Let A =(α1,α2,α3,α4,α5)
B =(β1,β2,β3,β4,β5)
β1=α1+α2 β2=α2+α3 β3=α3+α4 β4=α4+α5 β5=α5+α1
Then B=AK
K=
〔1 0 0 0 1
1 1 0 0 0
0 1 1 0 0
0 0 1 1 0
0 0 0 1 1〕
Because |K| is not equal to 0
So R (B)= R (A)
Because alpha 1, alpha 2, alpha 3, alpha 4, alpha 5 are linearly independent
So R (A)=5, so R (B)=5
Thus β1,β2,β3,β4,β5 are linearly independent

Let a vector be not equal to 0 vector, a vector point times b vector = a vector point times c vector, and b vector is not equal to c vector.

What does adding two vectors equals zero mean?

How does the vector |a·b|=|a b|prove this formula? I searched this formula on baidu encyclopedia but I don't understand why

Vector proof Given that AM is the center line of BC in triangle ABC, prove it by vector AM Square =1/2(AB Square + AC Square - BM Square Vector proof It is known that AM is the center line of BC in triangle ABC, by vector AM Square =1/2(AB Square + AC Square - BM Square

A proof of a vector. Let vector groups α1,α2,α3,α4,α5 be linearly independent β1=α1+α2 β2=α2+α3 β3=α3+α1 β4=α4+α5 β5=α5+α1 Proving that β1,β2,β3,β4,β5 are linearly independent

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