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Sec1305

# Sec1305 - Spring09 Math 231-Sec 13.5 Instructor Yanxiang...

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Spring09 Math 231-Sec 13.5 Instructor: Yanxiang Zhao Section 13.5: Equations of Lines and Planes 1 Equation of Lines 1.1 Parametric Equations of Lines A line L can be determined by the direction of L and a typical point on it. Mathe- matically, consider a line L in 3-d space whose direction is parallel to v , and a point P 0 ( x 0 , y 0 , z 0 ) sitting on L (Figure 01) . How to describe the line L analytically by using v and P 0 ( x 0 , y 0 , z 0 ) ? Lemma 1.1 Two vectors a and b are parallel if and only if there exists a constant t such that a = t b . Symbolically, a k b a = t b , where t is determined by t = | b | | a | . Let v be a vector parallel to L . Let P ( x, y, z ) be an arbitrary point on L and let r 0 and r be the position vectors of P 0 and P (namely, they have representations --→ OP 0 and -→ OP ). Then the geometric vector --→ P 0 P corresponds the algebraic vector r - r 0 . Obviously, r - r 0 is parallel to v , applying Lemma 1.1 yields r - r 0 = t v or equivalently r = r 0 + t v (1) Equation ( 1 ) is called vector equation of L , where t is the parameter. Each value of t gives the position vector r of a point on L . In other words, as t varies, the line is traced out by the tip of the vector r . If v = ( a, b, c ), and r 0 = ( x 0 , y 0 , z 0 ) , r = ( x, y, z ), then we can rewrite ( 1 ) componen- twisely as follows: x = x 0 + at, y = y 0 + bt, z = z 0 + ct (2) and equation ( 2 ) is called parametric equations of L . Example Find the vector equation and parametric equations for the line. The line through the point (6 , - 5 , 2) and parallel to the vector h 1 , 3 , - 2 / 3 i . The line through the points (1 , 3 , 2) and ( - 4 , 3 , 0). 1

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Spring09 Math 231-Sec 13.5 Instructor: Yanxiang Zhao 1.2 Symmetric Equations of Lines If we eliminate the parameter t in ( 2 ) with a, b, c are all nonzero, we can get the symmetric equations of L as follows: x - x 0 a = y - y 0 b = z - z 0 c (3) If one of a, b, c is 0, for instance if a = 0, then the 1st equation in ( 2 ) is unchanged, and by eliminating t from the 2nd and 3rd equation of ( 2 ), we get the symmetric equations of L as x = x 0 , y - y 0 b = z - z 0 c (4) This means that L lies in the plane x = x 0 . If b = 0, we have y = y 0 , x - x 0 a = z - z 0 c (5) If c = 0, we have z = z 0 , x - x 0 a = y - y 0 b (6) Example Find the symmetric equations for the line.
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