Intersections of Three Planes MCV4U: Calculus & Vectors There - PDF document

i n t e r s e c t i o n s o f l i n e s a n d p l a n e s i n t e r s e c t i o n s o f l i n e s a n d p l a n e s Intersections of Three Planes MCV4U: Calculus & Vectors There are many more ways in which three planes may intersect (or not) than two planes. First consider the cases where all three normals are collinear. • All three planes are parallel and distinct (inconsistent) Intersections of Three Planes • Two planes are coincident, and the third is parallel (inconsistent) J. Garvin • All three planes are coincident (infinite solutions) J. Garvin — Intersections of Three Planes Slide 1/15 Slide 2/15 i n t e r s e c t i o n s o f l i n e s a n d p l a n e s i n t e r s e c t i o n s o f l i n e s a n d p l a n e s Intersections of Three Planes Intersections of Three Planes Example Example Determine any points of intersection of the planes Determine any points of intersection of the planes π 1 : 3 x − 2 y + 1 z + 5 = 0, π 2 : 6 x − 4 y + 2 z + 10 = 0 and π 1 : 2 x − y + 5 z + 4 = 0, π 2 : 4 x − 2 y + 10 z + 15 = 0 and π 3 : − 6 x + 3 y − 15 z + 7 = 0. π 3 : 15 x − 10 y + 5 z + 25 = 0. The three normals are � n 1 = (3 , − 2 , 1), � n 2 = (6 , − 4 , 2) and The three normals are � n 1 = (2 , − 1 , 5), � n 2 = (4 , − 2 , 10) and n 3 = (15 , − 10 , 5). � n 3 = ( − 6 , 3 , − 15). � The equations for π 2 and π 3 are multiples of that of π 1 (by 2 n 2 = 2 � � n 1 , but the equation for π 2 is not twice that of π 1 . and by 5). Similarly, � n 3 = − 3 � n 1 , but the equation for π 3 is not triple Therefore, the planes are all coincident. that of π 1 . As before, let s and t be parameters and set y = s and Therefore, the plans are parallel and distinct, and there are z = t . Then x = 2 3 s + 1 3 t − 5. no points of intersection. These form the parametric equations of the plane that contains all solutions. J. Garvin — Intersections of Three Planes J. Garvin — Intersections of Three Planes Slide 3/15 Slide 4/15 i n t e r s e c t i o n s o f l i n e s a n d p l a n e s i n t e r s e c t i o n s o f l i n e s a n d p l a n e s Intersections of Three Planes Intersections of Three Planes Next, consider the cases where only two normals are collinear. Finally, consider the cases where none of the normals are collinear. • Two planes are coincident, and the third cuts the others (intersection is a line) • Normals are coplanar, planes intersect in pairs (inconsistent) • Two planes are parallel, and the third cuts the others • Normals are coplanar, planes intersect each other (inconsistent) (intersection is a line) • Normals not coplanar (intersection is a point) In the case of the first scenario, solve as earlier using the intersection of two planes. J. Garvin — Intersections of Three Planes J. Garvin — Intersections of Three Planes Slide 5/15 Slide 6/15

i n t e r s e c t i o n s o f l i n e s a n d p l a n e s i n t e r s e c t i o n s o f l i n e s a n d p l a n e s Intersections of Three Planes Intersections of Three Planes Example We need to solve a linear system of three equations with three unknowns. Determine any points of intersection of the planes π 1 : x − y + z + 2 = 0, π 2 : 2 x − y − 2 z + 9 = 0 and x − y + z + 2 = 0 π 3 : 3 x + y − z + 2 = 0. 2 x − y − 2 z + 9 = 0 3 x + y − z + 2 = 0 By inspection, none of the normals are collinear. Check if the normals are coplanar using the triple scalar product. Choose one variable that is easy to eliminate in two different pairs of equations, such as y . ( � n 1 × � n 2 ) · � n 3 = ((1 , − 1 , 1) × (2 , − 1 , − 2)) · (3 , 1 , − 1) Using equations 1 and 2, eliminate y . = (3 , 4 , 1) · (3 , 1 , − 1) x − y + z + 2 = 0 = 12 2 x − y − 2 z + 9 = 0 − Since the triple scalar product is non-zero, the normals are − x + 3 z − 7 = 0 not coplanar and the planes intersect at a single point. J. Garvin — Intersections of Three Planes J. Garvin — Intersections of Three Planes Slide 7/15 Slide 8/15 i n t e r s e c t i o n s o f l i n e s a n d p l a n e s i n t e r s e c t i o n s o f l i n e s a n d p l a n e s Intersections of Three Planes Intersections of Three Planes Using equations 2 and 3, eliminate y . Solve for z . 2 x − y − 2 z + 9 = 0 5( − 1) − 3 z + 11 = 0 + 3 x + y − z + 2 = 0 z = 2 5 x − 3 z + 11 = 0 Use the equation of any of the three planes to solve for y . Now we have a system of two equations with two unknowns, − 1 − y + 2 + 2 = 0 and we can solve for x . y = 3 − x + 3 z − 7 = 0 The point of intersection is at ( − 1 , 3 , 2). + 5 x − 3 z + 11 = 0 4 x + 4 = 0 x = − 1 J. Garvin — Intersections of Three Planes J. Garvin — Intersections of Three Planes Slide 9/15 Slide 10/15 i n t e r s e c t i o n s o f l i n e s a n d p l a n e s i n t e r s e c t i o n s o f l i n e s a n d p l a n e s Intersections of Three Planes Intersections of Three Planes Example Set up a linear system of three equations with three unknowns. Determine any points of intersection of the planes π 1 : x − 5 y + 2 z − 10 = 0, π 2 : x + 7 y − 2 z + 6 = 0 and x − 5 y + 2 z − 10 = 0 π 3 : 8 x + 5 y + z − 20 = 0. x + 7 y − 2 z + 6 = 0 Again, none of the normals are collinear. 8 x + 5 y + z − 20 = 0 Check if the normals are coplanar using the triple scalar product. Using equations 1 and 2, eliminate z . ( � n 1 × � n 2 ) · � n 3 = ((1 , − 5 , 2) × (1 , 7 , − 2)) · (8 , 5 , 1) x − 5 y + 2 z − 10 = 0 = ( − 4 , 4 , 12) · (8 , 5 , 1) + x + 7 y − 2 z + 6 = 0 = 0 2 x + 2 y − 4 = 0 Since the triple scalar product is zero, the normals are coplanar and the planes either intersect in a line or in pairs. J. Garvin — Intersections of Three Planes J. Garvin — Intersections of Three Planes Slide 11/15 Slide 12/15

i n t e r s e c t i o n s o f l i n e s a n d p l a n e s i n t e r s e c t i o n s o f l i n e s a n d p l a n e s Intersections of Three Planes Intersections of Three Planes Using equations 2 and 3, eliminate z . Multiplying the first equation by 17 and the second by 2, we obtain the following system. x + 7 y − 2 z + 6 = 0 34 x + 34 y − 68 = 0 + 16 x + 10 y + 2 z − 40 = 0 34 x + 34 y − 68 = 0 17 x + 17 y − 34 = 0 This system is true for any values of x and y , so the planes Now we have a system of two equations with two unknowns. intersect in a line. Let x = t and solve for y and z . 2 x + 2 y − 4 = 0 2 t + 2 y − 4 = 0 17 x + 17 y − 34 = 0 y = 2 − t t + 7(2 − t ) − 2 z + 6 = 0 z = 10 − 3 t The planes intersect in a line with parametric equations x = t , y = 2 − t and z = 10 − 3 t . J. Garvin — Intersections of Three Planes J. Garvin — Intersections of Three Planes Slide 13/15 Slide 14/15 i n t e r s e c t i o n s o f l i n e s a n d p l a n e s Questions? J. Garvin — Intersections of Three Planes Slide 15/15