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The vector a is known to be equal to (-2.3).b=(5.x), and a⊥b. then x=? A=-15/2 b=-2 The vector a is known to be equal to (-2.3).b=(5.x), and a⊥b. then x=? A=-15/2b=-2/15c=10/3d=10/3 The vector a is known to be equal to (-2.3).b=(5.x) and a⊥b. then x=? A=-15/2 b=-2 The vector a is known to be equal to (-2.3).b=(5.x) and a⊥b. then x=? A=-15/2b=-2/15c=10/3d=10/3 The vector a is known to be equal to (-2.3).b=(5.x), and a⊥b then x=? A=-15/2 b=-2 The vector a is known to be equal to (-2.3).b=(5.x), and a⊥b then x=? A=-15/2b=-2/15c=10/3d=10/3
Parse
Axb=0
So -10+3x=0
X =10/3
(2) A=2/15
B=-15/2
C=10/3
D =3/10
What's the question?
A vector a and b have moduli 2 and 3, respectively, and |a-b|=root 19, then a, b have an angle of
|A-b|=root 19,
Square: a^2+b^2-2 a•b=19,
The moduli of vectors a and b are 2 and 3, respectively, then a^2=4, b^2=9,
Substitute for: a•b=-3,
Cos=a•b/[|a||b|]=-3/(2•3)=-1/2,
=120°, A, b is 120°.
The angle between vector a and vector b is known as θ,|vector a|=2,|vector b|=3, if |vector a+vector b|=root 19 Then θ=?
Parse
| A |=2
|B |=3
|A+b|=√(a+b)2=√a2+2ab+b2
=√4+2*2*3*Cos 9
√13+12 Cos θ=√19
So 13+12 cos θ=19
12 Cos θ=6
Cos θ=1/2
θ=60°=π/3
Urgent! Given the vector a=(2,-2), b=(0,3), then the modulus of a-b is equal to
Given the vector a=(2,-2), b=(0,3), then the modulus of a-b is equal to
A-b =(2,-5)
|A-b |=√(22+52)=√29
Module of vector a=module of vector b, module of vector a+vector b=module of root number 3 vector a-vector b, find the module of vector a-2 times vector b The module of vector a=module of vector b, the module of vector a+vector b=module of vector a-vector b of root number 3, find the module of vector a-2 times vector b The module of vector a=module of vector b, the module of vector a+vector b=module of vector a-vector b with root number 3, and find the module of vector a-2 times of vector b
Given that the modulus of a is equal to the modulus of b, then the modulus of (a+b)= the root number of the modulus of (a-b) is 3 times. Find the angle between a and b. Because |a+b|=√3 a-b|,(a+b)^2=3(a-b)^2, i.e. a^2+b^2+2ab=3a^2+3b^2-6ab, i.e.4ab=a^2+b^2, i.e. ab=|a b cosa, i.e.4|a b|*c...
6. Given vector a =(1,2), find the module of vector a 7. Given |a|=√3,|b |=-√ 6. Given vector a =(1,2), find the module of vector a 7. Given |a|=√3,|b|=-√6, ab=3 Find the angle between vector a and b ¶
6.|A|=√1 square+2 square=√5
7. Cos (a, b)=ab/|a||b|
=3/(√3·(-√6))
=-√3/2
So
Angle =150°
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