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Lecture1b.pdf

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Department
Mathematics
Course
MATH 104
Professor
Steve Spencer
Semester
Summer

Description
Lecture 1b 2 Vector Equation of a Line in R (pages 5-6) The equation x 2 (0:5)x 1 1 de▯nes the x c2mponent of a point on the line in terms the x1component. But when we graph the line, we think less of this equation, and more of the following two facts: the x2-intercept is 1 and the slope is 0.5. And so, when graphing, we may start at the point (0;1), and then move \up one, right two" to ▯nd a second point on the line (the point (2;2)), and then connecting those points gives us our line. In fact, though, we could have used any multiple of \up one, right two" to ▯nd a second point on the line. So, we can think of the lin2 x = (0:1)x +1 as starting at the point (0;1), and adding all scalar multiples of the directions \up one, right two". Well, if we decide to think of vectors as directions in▯tead▯of points, then the direction 2 \up one, right two" corresponds to the vector . Then we can turn the ▯ 1▯ ▯ ▯ ▯ ▯ x1 0 2 equation x2= (0:5)x 1 1 into the equation x = 1 + t 1 for all 2 ▯ ▯ 0 t 2 R. Notice that picking t = 0 gives us our starting point o1 . ~ De▯nition: A line through ~ with direction vector d is the set f~ + td j t 2 Rg which has the vector equation ~ x = ~ + td ; t 2 R Note that in the case when ~ = 0, we end up with the equation x = td, which ~ is simply all scalar multiples of d. This ends up being the equation of the line through the origin that goes through d. Another thing to note is that since the direction vector d corresponds to the slope of a line, we see that two lines are parallel if and only if their direction vectors are NON-ZERO MULTIPLES of each other. Sometimes it is useful to w▯ite t▯e co▯pon▯nts o▯ a ▯ector equation separately. x 0 2 For
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