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Why is the vector multiplication negative? This is a coordinate multiplication problem Isn't vector multiplication a length? How can there be negative numbers?
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Is the coordinate point multiplication of the vector the corresponding coordinate multiplication
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What does the vector coordinate multiply? Why is it a number?
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How do two coordinate vectors multiply?
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Parallel vectors with equal modulus are equal vectors
No. A non-zero vector is parallel (collinear) to its opposite vector and moduli are equal, but they are not equal
Equal vector A vector whose moduli are equal and whose directions are the same
No. A non-zero vector is parallel (collinear) to its opposite vector and moduli are equal, but they are not equal
Equal vector A vector whose modules are equal and whose directions are the same
No. A non-zero vector is parallel (collinear) to its opposite vector and the modules are equal, but they are not equal
Equal vector A vector whose modules are equal and whose directions are the same
It is known that the two vectors a, b, c are equal in angle, and the module of vector a is equal to 1, the module of vector b is equal to 2, and the module of vector c is equal to 3. 1. Find the module of vector a+b+c. 2. Find the angle between vector a+b+c and vector a. The vectors a, b, c are known to be equal angles, and the module of vector a is equal to 1, the module of vector b is equal to 2, and the module of vector c is equal to 3. 1. Find the module of vector a+b+c. 2. Find the angle between vector a+b+c and vector a. The vectors a, b, c are known to have equal angles, and the module of vector a is equal to 1, the module of vector b is equal to 2, and the module of vector c is equal to 3. 1. Find the module of vector a+b+c. 2. Find the angle between vector a+b+c and vector a.
Let a be (1,0), b be (-2cos [60],2sin [60]), i.e.(-1,2√3) c be (-3cos [60],-3sin [60]), i.e.(-1(-1.5,-3√3/2) a+b+c=(1-1-1.5,0+2√3-3√3/2)=(-3/2,√3/2) modulo:√(9/4+3/4)=√32.
RELATED INFORMATIONS
- 1. It is known that the angles of the vectors a, b, c are equal, and the moduli of a, b, c are 1,2,3 respectively. It is known that the angles of the vectors a, b, c are equal, and the modules of a, b, c are 1,2,3, respectively.
- 2. Is the module of an equal vector equal
- 3. Vector a, b, c has the same starting point,|a|=|b|=2, ab=-1,(a-c, b-c)=60 degrees,|a-c|=Root 7, then the included angle of vector a, c is The vector a, b, c has the same starting point,|a|=|b|=2, ab=-1,(a-c, b-c)=60 degrees,|a-c|=Root 7, then the included angle of vector a, c is The vector a, b, c the same starting point,|a|=|b|=2, ab=-1,(a-c, b-c)=60 degrees,|a-c|=Root 7, then the included angle of vector a, c is
- 4. Given that the point G is the center of gravity of the triangle ABC, make a straight line through G on both sides of AB and AC and intersect at M and N respectively, and vector AM=x, vector AN=y vector AC, Find the value of 1/x+1/y Given that the point G is the center of gravity of the triangle ABC, the straight line passing through G intersects the two sides of AB and AC at M and N respectively, and the vector AM=x, the vector AN=y vector AC, Find the value of 1/x+1/y
- 5. If vector AD=X vector AB, vector AE=Y vector AC, and XY=0, Try to find the value of 1/x+1/y. Make any straight line DE, points A and G through G, and make a straight line intersection BC at M. Let vector AB=a, vector AC=b, then vector AG=2/3 vector AM=2/3[1/2(a+b)]=1/3(a+b), so vector GD=vector AD-vector AG=(x-1/3) a-1/3b vector ED=vector AD-vector AE=xa-xb, because vector GD is collinear with vector ED, vector GD=λ vector ED, So (x-1/3) a-1/3b=xλa-yλb, so there are equations: x-1/3=λx,1/3=λy, eliminate λ to get (x-1/3)/1/3=x/y, i.e.1/x+1/x=3. My question is how to get the equations from (x-1/3) a-1/3b=xλa-yλb, so there are equations: x-1/3=λx,1/3=λy,
- 6. Given that the three sides of triangle ABC are abc and a+b=4 ab=7/2c=3, judge whether triangle ABC is a right triangle, and explain that
- 7. In the known triangle ABC, AB=a, AC=b, when ab <0 or ab=0, try to judge the shape of △ABC (related to the number product of plane vector)
- 8. Given vector a=(cosx, sinx), b=(root number 2, root number 2), if a·b=8/5, and ∏/4 find the value of sin2x (1+tanx)/1-tanx
- 9. Given vector a=(1,2), a·b=5,|a-b|=2√5, then |b| equals
- 10. Vector cross multiplication The result of multiplication by "·" is a scalar quantity; A·B=|A||B|cosW (A, B have vector scale, it is inconvenient to print out. W is the angle of two vectors). The result of multiplication by "×" is a vector perpendicular to the plane of the original vector. A×B=|A||B|sinW. Why? Vector cross multiplication The result of multiplication by "·" is a scalar quantity; A·B=|A||B|cosW (A, B have vector scale, so it is inconvenient to print out. W is the angle of two vectors). The result of multiplication by "×" is a vector perpendicular to the plane of the original vector. A×B=|A||B|sinW. Why?
- 11. Known A =(1,1), and A and A+2 B in the same direction, A• B.
- 12. If a and b are mutually negative vectors, then a-b=, a+b=
- 13. What is a slip vector and what is a free vector
- 14. What is the cross multiplication formula of three vectors? For example, A vector cross by B vector cross by C vector
- 15. What is the meaning of multiplying two vectors in a plane? When two scalars are multiplied, we know how many and what are added. And what does it mean to multiply two vectors? What's the point of production? What is the meaning of multiplying two vectors in a plane? When two scalars are multiplied, we know how many and what are added. And what does it mean to multiply two vectors? What's the point of being productive? What is the meaning of multiplying two vectors in a plane? When two scalars are multiplied, we know how many and what are added. And what does the multiplication of two vectors mean? What's the point of production?
- 16. How to multiply two vectors?
- 17. λ,μ Are real numbers, if λa=μb, then a and b are not collinear. When λ=μ=0, a and b can be arbitrary vectors, satisfying λa=μb. How to understand this sentence
- 18. A vector is equal to (n+2, n^2-cosa^2), b vector is equal to (m,2/m+sina), where n, m, a are real numbers. If vector a=2 vector b, calculate the value range of n/m Cosa^2 is the square of cosa A vector is equal to (n+2, n^2-cosa^2), b vector is equal to (m,2/m+sina), where n, m, a are real numbers. If vector a=2 vector b, find the value range of n/m Cosa^2 is the square of cosa
- 19. Given vector a, b satisfies |a|=|b|=1, real number m, n satisfies m^2+n^2=1, then the value range of |ma+nb| is |Ma+nb ma nb 2(m^2+n^2)
- 20. Let e1, e2 be two non-collinear vectors, vector AB=2e1+ke2, vector CB=e1+3e2, vector CD=2e1-e2, if A, B, D are collinear, find the value of k