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Vector a+b=(-2,-1), a-b=(4,-3), the angle between vector a, b is
Let a=(x1, y1), b=(x2, y2), x1+x2=-2, x1-x2=4, y1+y2=-1, y1-y2=-3, x1=1, x2=-3, y1=-2, y2=1, cos=(a•b)/|a b|=-(√2)/2, so the included angle is 135 degrees.
If the vector a, b satisfies |a|=1,|b|=2, and the angle between a and b is π/3, then a-b| is If the vector a, b satisfies |a|=1,|b|=2, and the angle between a and b is π/3, then |a+b| is
Solution
Ab=|a||b|cosπ/3=2×1×1/2=1
∴
|A+b|
=√(A+b)2
=√A2+2ab+b2
=√1+2×1+4
=√7
Vector A =(3 4) Vector B =(1 2) How much is the vector AB? What is the modulus of vector AB?
0
Vector A, B is (1,-1),(2,3), then A• B =() A.5 B.4 C.-2 D.-1
0
If the vector A B =(3,4), the coordinates of point A are (-2,-1), then the coordinates of point B are
Let B (x, y),
Because (x+2, y+1)=(3,4)
So x=1, y=3
B (1,3)
Given the vector set m={ a|a=(1,2)+L (3,4), L belongs to R} N={ a|a=(-2,-2)+L (4,5), L belongs to R}, then the intersection of M and N is I am dull, do not know how to explain, warm-hearted valuable, valuable ideas,
To find the intersection is to find the set of vectors contained in both sets
The vector in m can be expressed as (3x+1,4x+2)
The vector in n can be expressed as (4y-2,5y-2)
The same vector must be the same pair of ordered real numbers
So we get the equations.
3X+1=4y-2
4X+2=5y-2
Solve x, y
Then intersection is {(3x+1,4x+2)} or can also be expressed as {(4y-2,5y-2)}(replace the solved x, y with)
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