Vector algebra proves that "given a × B = C × D, a × C = B × D, prove that a - D and B - C are collinear."

Vector algebra proves that "given a × B = C × D, a × C = B × D, prove that a - D and B - C are collinear."

a*b-a*c=a*(b-c)
c*d-b*d=d*(c-b)=-d*(b-c)=d*(b-c)
Because a * B-A * C = C * D-B * D
So a * (B-C) = D * (B-C)
So (A-D) * (B-C) = 0
So A-D and B-C are collinear

Given that a, B, C and D are vectors, it is proved that (a × b) · (C × d) = (a · C) (B · d) - (a · d) (B · C)

(a×b)·(c×d)=(a×b,c,d)=(a×b×c,d)=[(a·c)b-(b·c)a]·d=(a·c)(b·d)-(a·d)(b·c)
Where (·, ·, ·) denotes mixed product, and the third equal sign uses double outer product formula

In the following vector, the vector perpendicular to vector a = (7,9) is () A. (- 7, - 9) B. (- 7,9) C. (9,7) d. (9, - 7)

We haven't learned it. Think it should be d

Let non-zero vectors a, B, C, D satisfy D vector = (a vector. C vector) * (a vector * B vector) * C vector, and prove that a vector is perpendicular to D vector

There must be something wrong with the title:
If d = (a · C) (a · b) C, a and D cannot be perpendicular
If d = (a · C) (a × b) × C, a and D cannot be perpendicular
It should be: D = (A.C) B - (A.B) C
a·d=a·((a·c)b-(a·b)c)
=(a·c)(a·b)-(a·b)(a·c)=0
That is: a ⊥ D

If the roots of two real numbers of the quadratic equation Y2 + my + n = 0 with respect to y are opposite to each other, then ()
A. M = 0 and N ≥ 0b. N = 0 and m ≥ 0C. M = 0 and N ≤ 0d. N = 0 and m ≤ 0

∵ the two real number roots of the quadratic equation Y2 + my + n = 0 with respect to y are opposite to each other, ∵ △ m2-4n ≥ 0, ∵ n ≤ 0, so C

If the equation (X-2) 2 = A-4 has real roots, then the value range of a is ()
A. a＞4B. a≥4C. a＞2D. a≥2

According to the meaning of the question A-4 ≥ 0 solution: a ≥ 4, so choose B

Quick solution: when m takes any value, the quadratic equation MX2 + m2x-1 = x2 + X has no linear term
When m takes what value, there is no linear term in the quadratic equation MX2 + m2x-1 = x2 + X with respect to X. if so, a reward will be offered

mx²+m²x-1=x²+x
They are: (m-1) x & # 178; + (M & # 178; - 1) X-1 = 0
∵ not once
∴m²-1=0
‖ M = 1 or M = - 1
And the equation is a quadratic equation of one variable about X
∴m-1≠0
∴m≠1
∴m=-1

Given the quadratic equation of one variable 8x2 - (2m + 1) x + M-7 = 0, the values of M are obtained according to the following conditions: (1) two reciprocal numbers; (2) two opposite numbers; (3) one is zero; (4) one is 1

Let two of the original equations be reciprocal of X1 and X2 (1) ∵ two of them are reciprocal of each other, the product of them is 1x1 · x2 = m − 78 = 1, and the solution is m = 15, (2) ∵ two of them are opposite to each other, ∵ X1 + X2 = 2m + 18 = 0, ∵ M = - 12, (3) when one of them is zero, ∵ M-7 = 0, ∵ M = 7, (4) when one of them is 1, ∵ 8-2m-1 + M-7 = 0, and the solution is m = 0

It is known that the quadratic equation x2-2 (m-1) x + M2 = 0 with respect to X. if two of the equations are reciprocal to each other, then M=______ ．

∵ the univariate quadratic equation x2-2 (m-1) x + M2 = 0 with respect to X. if two of the equations are reciprocal to each other, let one of the equations be a, then the other one is 1a, ∵ M2 = 1 and △ = 4 (m-1) 2-4m2 ≥ 0. The solution is m = ± 1m ≤ & nbsp; 12, so the value of M is - 1

When are two of the quadratic equations of one variable opposite or reciprocal to each other

When the coefficient of the first term is 0, the two of the quadratic equations of one variable are opposite to each other
When the coefficients of quadratic term and constant term are the same, the two of quadratic equation of one variable are reciprocal to each other