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12.8 Algebraic Vectors & Parametric Equations

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In 12-7, we focused on the geometric aspect of vectors. 12-8 focuses on the algebraic properties. Note: the book will use (, ) for a point and a vector! Be careful! We will use , Unit vector: vector with magnitude 1 y x (x, y) is the norm/ magnitude (horizontal)(vertical) is the unit vector w/ same direction as Component Form: Polar Form:

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We can determine a vector if we know its initial (x 1, y 1 ) and terminal points (x 2, y 2 ). Ex 1) Given v with initial point (2, 3) & terminal point (7, 9), determine: a)component form b)polar form c)unit vector in same direction as

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Vector operations: vector sum: vector difference: scalar multiplication: Ex 2)

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In this picture, Q(x, y) is a point and P(a, b) is a point. P(a, b) Q(x, y) If you wanted to get to Q(x, y) from P(a, b), we could add a vector (x, y) = (a, b) + vector Since we don’t know the size of the vector, we can multiply by a scalar to get to the point., the direction vector will be given to you, or you can find it by subtracting the 2 nd point – 1 st point that they give you. so, (x, y) = (a, b) + t c, d (x, y) = (a, b) + tc, td (x, y) = (a + tc, b + td) this leads to x = a + tc and y = b + td these two eqtns are called parametric equations with parameter t of the line

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Ex 3) Determine a direction vector of the line containing the two points P(5, 8) and Q(11, 2). Then find the equation of the line & a pair of parametric equations of the line. direction vector = Q – P = (11 – 5, 2 – 8) = 6, –6 vector equation of line: (x, y) = (5, 8) + t 6, –6 parametric: x = 5 + 6t and y = 8 – 6t We can also represent other graphs (not just lines) in parametric. Ex 4) Graph the curve with parametric equations x = 3cost and y = 5sint. Find an equation of the curve that contains no other variables but x & y. 1 Ellipse square put together

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Homework #1208 Pg 651 #1, 3, 6, 9, 11, 15–18, 20, 21, 24, 29, 31, 33, 35

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