# 5 solution the lines are not parallel because the

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Unformatted text preview: nd therefore do not lie in the same plane). SOLUTION The lines are not parallel because the corresponding vectors 1, 3, 5 10 5 x y 1 and 2, 1, 4 are not parallel. (Their components are not proportional.) If L 1 and L 2 had a point of intersection, there would be values of t and s such that 1 2s 3t 3 4 FIGURE 5 t 2 _5 t s 3 4s But if we solve the ﬁrst two equations, we get t 151 and s 8 , and these values don’t 5 satisfy the third equation. Therefore, there are no values of t and s that satisfy the three equations. Thus, L 1 and L 2 do not intersect. Hence, L 1 and L 2 are skew lines. Planes Although a line in space is determined by a point and a direction, a plane in space is more difﬁcult to describe. A single vector parallel to a plane is not enough to convey the “direction” of the plane, but a vector perpendicular to the plane does completely specify its direction. Thus, a plane in space is determined by a point P0 x 0 , y0 , z0 in the plane and a vector n that is orthogonal to the plane. This orthogonal vector n is called a normal vector. Let P x, y, z be an arbitrary point in the plane, and let r0 and r be the position vectors of P0 and P. Then the vector r r0 is represented by P0 P . (See Figure 6.) The norA mal vector n is orthogonal to every vector in the given plane. In particular, n is orthogonal to r r0 and so we have z n P (x, y, z) r 0 5 r-r ¸ n r r0 0 which can be rewritten as r¸ P¸(x¸, y¸, z¸) x 6 nr n r0 y FIGURE 6 Either Equation 5 or Equation 6 is called a vector equation of the plane. 5E-13(pp 858-867) 862 ❙❙❙❙ 1/18/06 11:28 AM Page 862 CHAPTER 13 VECTORS AND THE GEOMETRY OF SPACE r0 To obtain a scalar equation for the plane, we write n x 0 , y0 , z0 . Then the vector equation (5) becomes a, b, c a, b, c , r x0, y x y0 , z z0 0 y0 cz z0 x, y, z , and 0 or ax 7 x0 by Equation 7 is the scalar equation of the plane through P0 x 0 , y0 , z0 with normal vector n a, b, c . EXAMPLE 4 Find an equation of the plane through the point 2, 4, z SOLUTION Putting a...
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## This note was uploaded on 02/04/2010 for the course M 56435 taught by Professor Hamrick during the Fall '09 term at University of Texas at Austin.

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