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Vector 3a+b is perpendicular to 7a-5b, and vector a-4b is perpendicular to vector 7a-2b, then the included angle of vectors a and b RT
Vertical so there is:(a+5b)∙(7a-5b)=0, i.e.7a^2+16a∙b-15b^2=0,∙(7a-b)=0, i.e.7a^2-30a∙b+8b^2=0,2ab=a^2=b^2, so cos=(|b|/|a|)/2=1/2, so=π/3...
Given vector AB=2a-b, vector CB=a-2b, vector CD=3a-2b, find vector AD, vector AC
Vector AD=vector AB+vector BC+vector CD=2a-b-(a-2b)+3a-2b=4a-b
Vector AC=vector AB+vector BC=2a-b-(a-2b)=a+b
Vector AD =vector AB+vector BC+vector CD=2a-b-(a-2b)+3a-2b=4a-b
Vector AC =vector AB+vector BC=2a-b-(a-2b)=a+b
Simplify vector AB-CB-AD-DC The Reduction Vector AB-CB-AD-DC
-CB=+BC
In the same way
Expressed as
AB+BC+DA+CD=AC+CD+DA=AD+DA=0
-CB=+BC
In the same way
Formulated as
AB+BC+DA+CD=AC+CD+DA=AD+DA=0
If AB=(-3a), CD=5a, and AD=CB, what is the shape of the quadrilateral ABCD? (AB,CD,AD,CB,a are all vectors)
Isosceles trapezoid
AB=-3/5CD indicates that AB//CD and AB is not equal to CD ABCD can not be parallelogram
AD=CB
So ABCD can only be isosceles trapezoid.
4A-2b=(-2,2√3), c=(1,√3), a·c=3,|b|=4. Find the angle between b and c (abc is a vector)
4A-2b=(-2,2√3), c=(1,√3), then yes
(4A-2b)*c=(-2*1+2√3 3)=4,
4Ac-2bc =4, and ac =3,
4*3-2Bc=4,
Bc =4.
|B |=4.
| C |=√(1+3)=2.
Cos (b, c)=bc/|b c|=4/(4*2)=1/2= cos60.
The angle between b and c is:60 degrees.
The vector abc is known as three vectors in the same plane, where a=(-3,4)|b|=2.5 and (a+2b) is perpendicular to (2a-b). Such as title
Given,|a|=√(9+16)=5,
Because a+2b is perpendicular to 2a-b,(a+2b)*(2a-b)=0,
I.e.2a^2+3a*b-2b^2=0,
So a*b=(2b^2-2a^2)/3=-12.5,
Therefore, cosα=a*b/(|a b|)=-1.5/(5*2.5)=-1,
Then α=180°.
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