Section 11.4: Equations of Lines and Planes


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1 Section 11.4: Equations of Lines and Planes Definition: The line containing the point ( 0, 0, 0 ) and parallel to the vector v = A, B, C has parametric equations = 0 + At, = 0 + Bt, = 0 + Ct, where t R is a parameter. These equations can be epressed in vector form as R(t) = 0 + At, 0 + Bt, 0 + Ct. v = A, B, C 0, 0, 0 R t = 0 + At, 0 + Bt, 0 + Ct Figure 1: Graph of the line with direction vector v passing through ( 0, 0, 0 ). Note: If A, B, and C are nonero, then the smmetric equations of the line are 0 A = 0 B = 0 C. Eample: Find parametric and smmetric equations of the line passing through the points (1, 1, 2) and (2, 1, 5). The line is defined b the vector v = 2 1, 1 ( 1), 5 2 = 1, 2, 3. Thus, parametric equations of the line are = 2 + t, = 1 + 2t, = 5 + 3t.
2 The smmetric equations of the line are 2 = 1 2 = 5 3. Definition: A vector N that is orthogonal to ever vector in a plane is called a normal vector to the plane. N v Figure 2: Illustration of a normal vector, N, to a plane. Theorem: (Equation of a Plane) An equation of the plane containing the point ( 0, 0, 0 ) with normal vector N = A, B, C is A( 0 ) + B( 0 ) + C( 0 ) = 0. Note: The equation of an plane can be epressed as A + B + C = D. This is called the standard form of the equation of a plane. Eample: Find an equation of the plane passing through the points P = ( 1, 2, 1), Q = (0, 3, 2), and R = (1, 1, 4). Two vectors in the plane are P Q = 1, 5, 1, P R = 2, 1, 5.
3 A normal vector to the plane is P Q P R = Thus, an equation of the plane is = 26, 7, 9. 26( + 1) + 7( 2) + 9( 1) = = 3. Eample: Find an equation of the plane passing through the point P = (1, 6, 4) and containing the line defined b R(t) = 1 + 2t, 2 3t, 3 t. The line passes through the point Q = (1, 2, 3) and has direction vector v = 2, 3, 1. Another vector in the plane is P Q = 0, 4, 1. A normal vector to the plane is The equation of the plane is P Q v = = 1, 2, 8. ( 1) 2( 6) + 8( 4) = = 21. Eample: Find the point at which the line defined b R(t) = 4 t, 3 + t, 2t intersects the plane defined b + 3 = 5. Substituting the parametric equations into the equation of the plane gives + 3 = 5 (4 t) (3 + t) + 3(2t) = 5 4t + 1 = 5 t = 1. Thus, the point of intersection is (3, 4, 2).
4 Definition: Two planes are parallel if the have the same normal vector (i.e. their normal vectors are parallel). Note: If two planes are not parallel, then the intersect in a line. The angle between the two planes is the angle between their normal vectors. Eample: Consider the planes defined b = 2 and = 3. (a) Find the angle between the planes. The normal vectors are N 1 = 4, 2, 1 and N 2 = 2, 1, 4. If θ is the angle between the planes, then N cos θ = 1 N 2 N 1 N 2 = 2 = Therefore, ( ) 2 θ = cos (b) Find parametric equations for the line of intersection. To find the equation of the line of intersection, we need a point on the line and a direction vector. Since the line lies in both planes, it is orthogonal to both N 1 and N 2. Thus, a direction vector for the line is N 1 N 2 = = 7, 18, 8. Since the direction vector is not horiontal, the line must intersect the plane ( = 0). Substituting = 0 into the equations for the planes gives { 4 2 = = 3. It follows that = 1 and = 1. Thus, (1, 1, 0) is a point on the line of intersection. The parametric equations are = 1 + 7t, = t, = 8t.
5 Theorem: (Distance from a Point to a Plane) The distance from ( 0, 0, 0 ) to the plane A + B + C + D = 0 is given b d = A 0 + B 0 + C 0 + D A2 + B 2 + C 2. Eample: Find the distance between the parallel planes 2 2+ = 10 and = 2. The planes are parallel since their normal vectors are parallel. Indeed, 2 N 1 = 2 2, 2, 1 = 4, 4, 2 = N 2. Find a point on one plane and then find the distance from this point to the other plane. Setting = = 0, the second plane contains (0, 0, 1). Now the distance between the planes is d = 2(0) 2(0) = 9 3 = 3. Definition: Two lines in R 3 are skew if the are not parallel and do not intersect. Parallel Intersecting Skew Figure 3: Illustration of parallel (red), intersecting (blue), and skew (green) lines. Note: Skew lines lie in parallel planes. Eample: Find the distance between the skew lines defined b L 1 : R1 (t) = 1 + t, 2 + 6t, 2t, L 2 : R2 (s) = 2 + 2s, s, 3 + 5s.
6 The lines lie in parallel planes P 1 and P 2. Thus, the distance between the lines is the distance between the planes. The common normal vector of the planes is orthogonal to the direction vectors v 1 = 1, 6, 2 and v 2 = 2, 14, 5. That is, N = = 2, 1, 2. To define the planes P 1 and P 2, we need points on the planes (on the lines L 1 and L 2 ). Setting t = 0, we find that (1, 2, 0) lies on L 1 and in P 1. Setting s = 0, we find that (2, 4, 3) lies on L 2 and in P 2. Thus, the equation of plane P 2 is 2( 2) ( 4) + 2( + 3) = = 0. The distance from the point (1, 2, 0) (in P 1 ) to the plane P 2 defined b = 0 is d = 2(1) 2 + 2(0) = 6 3 = 2.
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