For the vector a, b, c and the real number λ, is the true proposition () in the following propositions possible to analyze these four in detail? A. If a*b=0, then a=0 or b=0 B. If λa=0, then λ=0 or a=0 C. If a^2= b^2, then a = b or a =-b D. If a*b=a*c, then b=c For the vector a, b, c, and real number λ, could the following real proposition () analyze these four in detail? A. If A*B=0, then A=0 or B=0 B. If λa=0, then λ=0 or a=0 C. If A2= B2, then A = B or A = B D. If a*b=a*c, then b=c

For the vector a, b, c and the real number λ, is the true proposition () in the following propositions possible to analyze these four in detail? A. If a*b=0, then a=0 or b=0 B. If λa=0, then λ=0 or a=0 C. If a^2= b^2, then a = b or a =-b D. If a*b=a*c, then b=c For the vector a, b, c, and real number λ, could the following real proposition () analyze these four in detail? A. If A*B=0, then A=0 or B=0 B. If λa=0, then λ=0 or a=0 C. If A2= B2, then A = B or A = B D. If a*b=a*c, then b=c

B

Let a, b, c be arbitrary nonzero plane vectors and not collinear with each other, then how to prove the following two false propositions? 1(A·b) c-(c·a) b=0;... Let a, b, c be arbitrary nonzero plane vectors and not collinear with each other, then how to prove the following two false propositions? 1(A·b) c-(c·a) b=0;2(b·c) a-(c·a) b is not perpendicular to c?

Certificate:
(1)(A·b) c-(c·a) b is a vector, but c, b are not collinear
So it can't be a 0 vector
So proposition 1 is false
(2)[(B·c) a-(c·a) b ]·c
=(B·c) a·c-(c·a) b·c
=0
So 2 false

Certificate:
(1)(A·b) c-(c·a) b is a vector, but c, b are not collinear
So it can't be a 0 vector
So proposition 1 is false
(2)[(B·c) a-(c·a) b ]·c
=(B·c) a·c-(c·a) b·c
=0
So 2 False

How to Orthogonize Two Vectors

If the eigenvalues are not equal, only the corresponding eigenvectors need to be unitized. The reason is that the eigenvectors of the real symmetric with different eigenvalues are orthogonalized.

Let a=(1,2,2), b=(2, a,3), and a be orthogonal to b, then a=

1*2+2*A+(-2)*3=0, so a=2(orthogonal i.e. vertical, the sum of each product is zero)

On the Problem of Orthogonal Vector 1. Find two unit vectors orthogonal to and. 2. If the points Q and R are on the straight line L and the point P is not on L, prove that the distance d from the point P to the straight line L is D= a*b / a where a=QRb=QP The distance d from point P to straight line L is actually the projection of b on a. If so, d should be equal to a.b/ a However, the proof in this question is d= a*b / a. Under what conditions are the two formulas used respectively? I've already tried the first floor method. It's wrong, but thank you. I just forgot to say that the correct answer to the first question is and. The answer on the third floor is correct. But there is a sentence that I still don't understand." The vector orthogonal to both vectors is a=k (2,-1), and the unitized vector is ±1/√6(2,1,-1)". I use English textbooks abroad. I do n' t quite understand some professional vocabulary in this sentence. What is unitization and how to find the final answer? And why is it wrong to say on the first floor that a*b and b*a are perpendicular to a and b? Thank you so much.

1. Let (x, y, z) be orthogonal to both vectors, then
X-y+z=0
4Y +4z =0
Solution x=-2z, y=-z
Therefore, the vector orthogonal to both vectors is a=k (2,1,-1), and the unitized vector is
±1/√6(2,1,-1)
2." The distance d from point P to straight line L is actually the projection size of b on a "is wrong. Think about the length of the projection.

Orthogonality of linear algebraic vector Vector a1=(-1.1.1) T a2=(1.0.1) T. Find a vector a3 so that a3 and a1, a2 are orthogonal.

0