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Published byKaylee Bradford Modified over 4 years ago

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IB Revision 4 How do Cartesian graphs work? (i.e. y =....) e.g. y = 3x + 2 x can be any number xy 15 28 -3-7 ½3.5

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IB Revision 4 Consider the equation initial point direction vector (2,-1) t12½-2 direction vector t can be any number

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IB Revision 4 e.g. find the vector equation of the line through A and B. A (2,-5) B (-3, 8) Direction vector = (or ) Initial point = A (or B) So

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IB Revision 4 e.g. find the vector equation of the line through: 1. 2. A (5, -3) and B (-4, 2) 3.

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IB Revision 4 e.g. find the vector equation of the line through (-1, 0) and parallel to y = 4x - 2

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IB Revision 4 Does (5, -36) lie on ? A (2,-5) B (-3, 8) i.e. is there some value of t which will generate the point (5, -36)? If it does: t cant be both things at once – so the point is not on the line.

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IB Revision 4 Does (3, 2) lie on Does (1, -3) lie on

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IB Revision 4 Parametric form – each dimension (x, y, z) given in terms of a parameter λ e.g. r = i + 2j – 4k + λ(6i – 7j + 3k)

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IB Revision 4 Find the parametric equations of the line passing through A (-2, 1, 3) and B (1, -1, 4)

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IB Revision 4 Cartesian coordinates from parametric x = 3μ + 1 y = 2μ – 1 z = 4 + μ

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IB Revision 4 Cartesian coordinates from parametric x = 3μ + 1 y = 2μ – 1 z = 4 + μ

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IB Revision 4 Cartesian coordinates from parametric x = 3μ + 1 y = 2μ – 1 z = 4 + μ

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IB Revision 4 Cartesian coordinates from parametric x = 3μ + 1 y = 2μ – 1 z = 4 + μ

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IB Revision 4 Cartesian coordinates from parametric x = 3μ + 1 y = 2μ – 1 z = 4 + μ

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IB Revision 4 Cartesian coordinates from parametric x = 3μ + 1 y = 2μ – 1 z = 4 + μ

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IB Revision 4 A line is parallel to the vector 2i – j + 2k and passes through (2, -3, 5). Find: the vector equationthe parametric equation the Cartesian equation

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IB Revision 4 Are these lines parallel? Consider the direction vectors. Therefore, are parallel.

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IB Revision 4 Are these lines parallel? so not parallel. Consider parametric form: r1:r2:r1:r2: x = λx = 2 y = - λy = 1 + 3μ z = 1 - 3λz = 5 μ Do they intersect? If so x = x y = yz = z λ = 2-λ =1 + 3μ 1 - 3λ = 5μ -2 = 1 + 3μ -1 = μ 1 – 6 = -5 Equation systems work intersect, at: x = λ = 2 y = - λ= -2 (2, -2, -5) z = 1 - 3λ= -5 Solve the simultaneous equation set. If you can find values for λ and μ that work, then they intersect. If, for example, this line was 1 – 6 = 10, the equation set is inconsistent and cannot be solved, so the lines do not intersect. (are skew) Use either value in the respective equation to find the point of intersection.

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