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AXC and BXC form a linear pair True or false?
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.
What is BXC in Roman numerals?
It is gibberish nonsense. There is no "B" in the Roman numeral system.
Axb Dot bxc x cxa equals a Dot bxc squared How can I prove this using Mathmatica X stands for cross product and dot is or the dot product?
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
Is it possible to divide one rational number by another to obtain an irrational number as the quotient?
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.
What is the associative property of multiplication?
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.
