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2313-1-4 - S ECTION 1.4 1.4 L INES AND P LANES PART I 9 L...

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S ECTION 1.4 L INES AND P LANES , P ART I 9 1.4. L INES AND P LANES , P ART I In two dimensions, there are numerous ways of presenting the equation of a line, such as point-slope form , y y 0 m x x 0 . In this form we are specifying a point on the line, x 0 , y 0 , and the slope of the line, m . In three dimensions, one way to specify a line is with a point, x 0 , y 0 , z 0 , on the line, and the direction of the line, which we specify with a vector m m 1 , m 2 , m 3 . L m P 0 x 0 , y 0 , z 0 P x, y, z You should think of an equation for such a line as a “point-testing procedure”. If the equation holds for a point x, y, z , then x, y, z lies on the line, and otherwise x, y, z does not lie on the line. From this viewpoint, the point P x, y, z lies on the line L if and only if the vector P 0 P is parallel to the direction vector m . Remembering that two vectors are parallel if and only if they are scalar multiples, we see that P lies on the line if and only if P 0 P m t for some scalar t . Expanding P 0 P , we see that x, y, z lies on the line if and only if x x 0 , y y 0 , z z 0 m t, or x, y, z m t x 0 , y 0 , z 0 . This is called a vector-valued function , or simply a vector function , because it takes one num- ber as input, t , and returns a vector, x, y, z . You should think of a vector function as tracing out a curve consisting of all the points its vectors point to from the origin. Vector Equation for a Line. The vector equation for the line through the point x 0 , y 0 , z 0 parallel to the vector m is r t m t x 0 , y 0 , z 0 , where t . By solving the vector equation for a line for x , y , and z , we obtain another form.
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10 C HAPTER 1 V ECTORS AND V ECTOR -V ALUED F UNCTIONS Parametric Equations for a Line. The parametric equations for the line through the point x 0 , y 0 , z 0 parallel to the vector m m 1 , m 2 , m 3 are x m 1 t x 0 , y m 2 t y 0 , z m 3 t z 0 , where t . Notice that unlike in the 2 d case, where the slope-intercept form y mx b uniquely de- termines a line, in 3 d neither the vector nor the parametric equations uniquely determine a line. It is always possible to give different equations for the line by using a different “base point” P 0 x 0 , y 0 , z 0 . Example 1. Give a vector equation for the line through the point 4 , 0 , 2 parallel to the vector 2 , 1 , 1 . Solution. We have been given everything we need. The vector equation r t 2 , 1 , 1 t 4 , 0 , 2 specifies this line. Example 2. Give a vector equation for the line through the points P 3 , 2 , 2 and Q 0 , 1 , 3 . Solution. This line is parallel to the vector PQ , PQ 0 3 , 1 2 , 3 2 , 3 , 1 , 1 , so the line is given by r t 3 , 1 , 1 t 3 , 2 , 2 . (Note that here we used the point P as our base point, but we could instead have chosen Q , or indeed, any other point on the line.) Example 3. Determine whether the point P 3 , 1 , 4 lies on the line r t 0 , 2 , 2 t 3 , 5 , 9 .
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S ECTION 1.4 L INES AND P LANES , P ART I 11 Solution. This point lies on the line if and only if we can find a value of t satisfying 3 , 1 , 4 0 , 2 , 2 t 3 , 5 , 9 .
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