Vector midpoint proof In the known isosceles triangle ABC,∠C=90°, D is the midpoint of CB, E is the point on AB, and AE=2EB.

Vector midpoint proof In the known isosceles triangle ABC,∠C=90°, D is the midpoint of CB, E is the point on AB, and AE=2EB.

Let CA=a, CB=b (the above arrows are omitted), then we can calculate BA=a-b, AD=1/2b-a, CE=b+1/3(a-b)=1/3a+2/3b; calculate AD·CE=(1/2b-a)·(1/3a+2/3b)=1/3b-1/3a-1/2ab; angle C is right angle, ab=0, CA=CB, b=a, so AD·C...

ON VECTORS IN COLLEGE MATHEMATICAL PROOF Let A and B be square matrices of order n and the determinant of A be equal to 2. 2 Proves that if n-dimensional unit vectors e1, e2... en can be linearly represented by a set of dimensional vectors a1, a2... an, then the set of vectors a1, a2... an is linearly independent

1.|A|=0, so A is reversible, and its inverse matrix is denoted as P.
Thus BA=EBA=(PA) BA=P (AB) A
AB is similar to BA by definition of similarity
2. If the n-dimensional unit vectors e1, e2... en can be linearly represented by the set of dimensional vectors a1, a2... an, then the two sets of vectors are known to be equivalent so as to have the same rank.
I don't know. Is that okay?

Prove:|b vector-a vector |a vector |-|b vector | Prove:|b vector-a vector a vector |-|b vector | Prove:|b vector-a vector | a vector |-|b vector |

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For any two vectors a, b, there is ||a|-|b a-b a b|

Let θ=first left:||a|-|b a-b|from |a-b|2-||a|-|b||2=(a2-2|a||b|cos b2)-(a2-2|a||b b2)=2|a||b|b|(1-cosθ)≥0, then right:|a-b a|-|b||:

Proving that (λa) b=λ(ab)=a (λb) Such as ~ Proving that (λa) b=λ(ab)=a (λb) For example ~

If the angle between a and b is x, then λa and b, a and λb are also x
The values are λabcosx (where a, b are the moduli of vectors a, b)

If the angle between a and b is x, then λa and b, a and λb are also x
Then all values are λabcosx (where a, b are the moduli of vectors a, b)

As shown in the figure, in the isosceles trapezoidal ABCD, E is the midpoint of CD, EF⊥AB is F, and if AB=6, EF=5, the area of the trapezoidal ABCD is ______. As shown in the figure, in the isosceles trapezoid ABCD, E is the midpoint of CD, EF⊥AB is F, and if AB=6, EF=5, the area of the trapezoid ABCD is ______.

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