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The answer depends on where the points A, B, C and X are. And since you have not bothered to provide that information, I cannot provide a sensible answer.
It is gibberish nonsense. There is no "B" in the Roman numeral system.
Prove that (axb)n[(bxc)x(cxa)] = [a]n(bxc)]^2 where a,b,and c are all vectors. First, multiply out the cross products. Since the cross product of two vectors is itself a vector, we'll give the cross products some names to make this a little easier to understand: (bxc)=(b2c3-b3c2)i-(b1c3-b3c1)j+(b1c2-b2c1)k = vector d (cxa)=(c2a3-c3a2)i-(c1a3-c3a1)j+(c1a2-c2a1)k = vector v (axb)=(a2b3-a3b2)i-(a1b3-a3b1)j+(a1b2-a2b1)k = vector u => (axb)n[(bxc)x(cxa)] = un[dxv] (dxv)=(d2v3-d3v2)i-(d1v3-d3v1)j+(d1v2-d2v1)k = vector w => un[dxv] = unw = u1w1 + u2w2 + u3w3 Now replace u and w with their vector coordinates (notice that the negative sign is factored into the middle terms, so the variables are switched). u1w1 + u2w2 + u3w3= (a2b3-a3b2)w1 + (a3b1-a1b3)w2 + (a1b2-a2b1)w3 = (a2b3-a3b2)(d2v3-d3v2) + (a3b1-a1b3)(d3v1-d1v3)+ (a1b2-a2b1)(d1v2-d2v1) Now we need to expand the v terms back out: (d2v3-d3v2) = d2(c1a2-c2a1) - d3(c3a1-c1a3) = d2c1a2- d2 c2a1- d3c3a1 + d3c1a3 (d3v1-d1v3) = d3(c2a3-c3a2) - d1(c1a2-c2a1) = d3c2a3 - d3c3a2- d1c1a2+ d1c2a1 (d1v2-d2v1) = d1(c3a1-c1a3) - d2(c2a3-c3a2) = d1c3a1 - d1c1a3 - d2c2a3 + d2c3a2 So: (a2b3-a3b2)(d2v3-d3v2) + (a3b1-a1b3)(d3v1-d1v3)+ (a1b2-a2b1)(d1v2-d2v1) = (a2b3-a3b2)(d2c1a2- d2 c2a1- d3c3a1 + d3c1a3) + (a3b1 - a1b3)(d3c2a3 - d3c3a2- d1c1a2+ d1c2a1)+ (a1b2-a2b1)(d1c3a1 - d1c1a3 - d2c2a3 + d2c3a2) = d2c1a2 a2b3- d2 c2a1 a2b3- d3c3a1 a2b3 + d3c1a3 a2b3- d2c1a2 a3b2+ d2 c2a1 a3b2+ d3c3a1 a3b2- d3c1a3 a3b2 + d3c2a3 a3b1 - d3c3a2 a3b1- d1c1a2 a3b1+ d1c2a1 a3b1- d3c2a3 a1b3+ d3c3a2 a1b3+ d1c1a2 a1b3- d1c2a1 a1b3+ d1c3a1 a1b2 - d1c1a3 a1b2 - d2c2a3 a1b2 + d2c3a2 a1b2- d1c3a1 a2b1+ d1c1a3 a2b1+ d2c2a3 a2b1- d2c3a2 a2b1 Some of the terms cancel out, leaving us with; = d2c1a2 a2b3 - d2 c2a1 a2b3 + d3c1a3 a2b3 - d2c1a2 a3b2 + d3c3a1 a3b2 - d3c1a3 a3b2 + d3c2a3 a3b1 - d3c3a2 a3b1 + d1c2a1 a3b1 - d3c2a3 a1b3 + d1c1a2 a1b3 - d1c2a1 a1b3 + d1c3a1 a1b2 - d1c1a3 a1b2 + d2c3a2 a1b2 - d1c3a1 a2b1 + d2c2a3 a2b1 - d2c3a2 a2b1 Now factor out d1 , d2 , and d3 = d1(c2a1 a3b1 + c1a2 a1b3 - c2a1 a1b3 + c3a1 a1b2 - c1a3 a1b2 - c3a1 a2b1) + d2(c1a2 a2b3 - c2a1 a2b3 - c1a2 a3b2 + c3a2 a1b2 + c2a3 a2b1 - c3a2 a2b1) + d3(c1a3 a2b3 + c3a1 a3b2 - c1a3 a3b2 + c2a3 a3b1 - c3a2 a3b1 - c2a3 a1b3) Now we can factor out a dot product of ( d1 + d2 + d3): = ( d1 + d2 + d3)n[(c2a1 a3b1 + c1a2 a1b3 - c2a1 a1b3 + c3a1 a1b2 - c1a3 a1b2 - c3a1 a2b1) + (c1a2 a2b3 - c2a1 a2b3 - c1a2 a3b2 + c3a2 a1b2 + c2a3 a2b1 - c3a2 a2b1) + (c1a3 a2b3 + c3a1 a3b2 - c1a3 a3b2 + c2a3 a3b1 - c3a2 a3b1 - c2a3 a1b3)] (Remember, to keep from changing the value of the equation we still need to keep the terms grouped together so that they multiply by the correct d components.) Now factor out all the "a" components within the brackets: = ( d1 + d2 + d3)n[(a1 a3{c2b1 - c1b2} + a1 a2{c1b3 - c3b1} + a1 a1{c3b2 - c2b3}) + (a1 a2{c3b2 - c2b3} + a2 a2{c1b3 - c3b1} + a2 a3{c2b1 - c1b2}) + (a1 a3{c3b2 - c2b3} + a2 a3{c1b3 - c3b1} + a3 a3{c2b1 - c1b2})] = dn[( a1 a3+ a1 a2+ a1 a1)n({c2b1- c1b2} +{c1b3 - c3b1} + {c3b2- c2b3}) + (a1 a2 + a2 a2 + a2 a3)n({c1b3- c3b1} + {c2b1- c1b2} + {c3b2- c2b3}) + ( a1 a3+ a2 a3 + a3 a3)n({c3b2- c2b3} + {c1b3- c3b1} + {c2b1- c1b2})] And we know that {c2b1- c1b2} +{c1b3 - c3b1} + {c3b2- c2b3} = (bxc), so we factor out (bxc): = dn[(bxc)n[(a1 a3+ a1 a2+ a1 a1) + (a1 a2 + a2 a2 + a2 a3) + ( a1 a3+ a2 a3 + a3 a3)] = dn[(bxc)n[a1(a3+ a2 + a1) + a2 (a1 + a2 +a3) + a3(a1+ a2 + a3)]] = dn[(bxc)n([a1 + a2 +a3]n[a1 + a2 +a3]) = dn[(bxc)n(a n a)] (from above, remember that d = (bxc) ) = (bxc)n(bxc)n a n a = [an (bxc)]^2
It is not possible to divide one rational number by another to obtain an irrational number. A rational number is of the form a/b where a and b are both integers, whereas an irrational is a number which is impossible to express in the previously mentioned way. Let A=(a/b) and B=(c/d) where A and B are both rational numbers. Consider the quotient A/B, this is the same as A(1/B). Rewrite this as (a/b)x(d/c). Assuming we all know basic arithmetic with fractions we can clearly see that the dividend is axd and the divisor is bxc, and the new expression is (axd)/(bxc). Since a, b, c, and d are all integers and the integers are closed under multiplication (two integers multiplied by each other produce another integer) our new expression as a single fraction is one integer over another and it is therefore a rational number.
The Associative property of multiplication states that the product of a set of three numbers is always the same no matter which operation is carried out first.For example Ax(BxC) = (AxB)xC and so either can be written as AxBxC.ie 3x(4x5) = 3x20 = 60and (3x4)x5 = 12x5 = 60It is important not to confuse this with the commutative (or Abelian) property which states that the order of the numbers does not matter. ie AxB = BxAMatrix multiplication, for example, is associative but NOT commutative.(a * b) * c = a * (b * c)As a result, we can write a * b * c without ambiguity.