Given |a|=2√13, b=(-2,3) and a⊥b, find the coordinates of vector a

Given |a|=2√13, b=(-2,3) and a⊥b, find the coordinates of vector a

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The product of the number of plane vectors is a number, but a·b=x1x2+y1y2, which is a vector, does this conflict with the lallblcos number? The product of the number of plane vectors is a number, but a·b=x1x2+y1y2, which is a vector, does this conflict with lallblcos being the number?

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Let vector a=(x1, y1), b=(x2, y2), the product of a and b=x1x2+y1y2?

The size of the quantity product is the horizontal multiplication plus the vertical multiplication;
The results are correct;

In the coordinate operation of space vector, what conditions can satisfy the intersection of two vectors?

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The Operation of Space Vector All operations of the space vector, such as vector A minus the module of vector B, vector A plus the module of vector B, vector A multiplied by the module of vector B... Single operation is acceptable. It is better to have modulo operation, such as │A│minus │B│module |A+B|=sqr (|A||B||+2AB) | A |= sqr (a*a) |B |= sqr (b*b) The Operation of Space Vector All operations of the space vector, such as vector A minus the module of vector B, vector A plus the module of vector B, vector A multiplied by the module of vector B... Single operation is fine, and it is better to have modulo operation, such as │A│minus │B│module |A+B|=sqr (|A||B||+2AB) | A |= sqr (a*a) |B |= sqr (b*b)

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Space vector coordinate operation Given A (-1,2) B (2,8) and, vector DA=negative third vector BA, find the coordinates of points C, D and vector CD

Let D (x, y)
Because A (-1,2) B (2,8)
So vector DA=(-1-x,2-y), vector BA=(-3,-6)
Because vector DA = negative third vector BA
So -1-x=1,2-y=2
Solution x=-2, y=0
So coordinates of D are (-2,0)

Let D (x, y)
Because A (-1,2) B (2,8)
So vector DA=(-1-x,2-y), vector BA=(-3,-6)
Because vector DA = negative third vector BA
So -1-x=1,2-y=2
Solution: x=-2, y=0
So coordinates of D are (-2,0)