12.5

Lines and Planes Review: v aaa222   va1,, a 2 a 3  a 1 i  a 2 j  a 3 k 1 2 3

u v u1 v 1  u 2 v 2  u 3 v 3  u v cos u u u u and v are orthogonal if and only if u v 0

uv uv compvu  projvuv  v vv

u v uv23  uv 32 i  uv 13  uv 31 j  uv 12  uv 21  k u v is orthogonal to both u and v . u v u v sin u and v are paralallel if and only if u v 0

u varea of the parallelogram determined by u and v Volume of parallelepiped a  b  c

Vector equation of a line L: r r0 t v rv0 point on the line, direction  abc , ,

Parametric (scalar) equation of a line L: xx0  aty,  y 0  btz ,  z 0  ct Planes In order to find the equation of a plane, we need : this vector is called A a point on the plane P0 x 0 , y 0 , z 0  the normal vector B a vector that is orthogonal to the plane n  abc , , to the plane

n r  r0   0

vector equation of the plane

n r  r0  0 

a x x0  b y  y 0  c z  z 0   0 scalar (or component) equation of the plane r0OP 0 x 0,, y 0 z 0 r OP x,, y z n  abc,, ax by  cz  ax0  by 0  cz 0

P0 P r -,, r 0  x  x 0 y  y 0 z  z 0 Plane: ax by  cz  d Problem : Determine the equation of the plane that passes through PQR(1,2,3),  (3,2,1) and  (  1,  2,2) For a point in the plane we can choose e.g. P = (1,2,3)

For a vector normal to the plane we can choose the cross product of two vectors in the plane:

For two vectors in the plane we can choose e.g. u PQ 2,0,  2 and v=PR   2,  4,  1

i j k uv 2 0  2  08 i 42 j  80  k  8i  6 j  8 k 2  4  1

n 4 i  3 j  4 k is normal to the plane 4(x 1)  3( y  2)  4( z  3)  0 4x 3 y  4 z  10 Two distinct planes in 3-space either are parallel or intersect in a line.

z

y

x

Question: How do I decide if two planes are parallel?

Answer: The normal must be parallel. Are the planes 2x y  z  1 and  4x  2 y  2 z   3 parallel? Problem : Find the line of intersection of the two planes x2 y  z  0 and 2 x  3 y  2 z  0

2x 4 y  2 z  0 2x 3 y  2 z  0 40xy yx 4

Take the first equation: x2 y  z  0 and plug in y  4 x x24x  z 07  z  x x can be anything and we have y and z as functions of x . Let xt then y 4 t and z 7 t L: x t , y  4 t , z  7 t If two planes intersect, then you can determine the angle between them. between planes between their normal vectors

n1 n2  nn cos  12 nn12

Example: Find the angle between the planes x20 y  z  6    54 2x 3 y  2 z  0 102

nn12 1,  2,1 and  2,3,  2

nn126 and 17 nn12  2  6  2   6 Distance between a point and a plane: Q D The point is given by Q b

The plane is given by a point P0 and a normal n n

Project the vector b PQ0 onto the normal direction n :

||bn P Distance D  | compb | 0 n n Example: ||bn Find the distance from Q (1,2,0) to the plane 2 x  y  1 D  P  (0,1,0) n 0 n  2,1,0 b = PQ0  1,1,0 | 1,1,0 . 2,1,0 | 3 D  4 1 5 Distance between a point and a line:

The point is given by S The line is given by a point P and a direction v

||PS  v Distance D | PS | sin( )  | v |

Example: ||PS  v Find the distance from S  (1,0,1) to the line containing D 

PP01 (1,1,1) and  (2,  1,1) | v | i j k 1 PS0 0, 1,0 v PP01 1,  2,0 PS0 v 0  1 0  k D  1 2 0 5 12.6

Quadric surfaces Review conic sections in the plane: x intercept: x a xy22 Equation of an ellipse: 1 y intercept: y b ab22 circle: ab 2 Equation of a parabola: y ax

xy22 x intercept: x a Equation of a hyperbola: 1 ab22 no y intercept! x2 y2 z 2 Ellipsoid:   1 sphere: abc a2 b2 c2

Slice the surface with planes parallel to the yz - plane:

y2 z 2 k 2 Set x constant: x k , then   1  b2 c 2 a 2

All traces are ellipses a  k  a Hyperboloid of one sheet:

x2 y 2 k 2 x2 y 2 z 2 zk :  1  any k !   1 a2 b 2 c 2 abc2 2 2 Horizontal traces are ellipses

x2 z 2 k 2 yk :  1  a2 c 2 b 2

Vertical traces are hyperbolas Hyperboloid of two sheets: zk x2 y 2 z 2    1 2 2 2 222 x y k a b c   1 only kc ! a2 b 2 c 2 Horizontal traces are ellipses

Vertical traces are hyperbolas (Elliptical) Paraboloid: Horizontal traces are ellipses

z x22 y  Vertical traces are parabolas c a22 b (Hyperbolic) Paraboloid: Horizontal traces are hyperbolas

z x22 y  Vertical traces are parabolas c a22 b Cone Horizontal traces are ellipses z2 x 2 y 2 2 2 2 c a b Vertical traces are hyperbolas x k hyperbolas k  0 y k hyperbolas k0 lines z y z x  ,   c b c a Cylinders: xy22 yx22 y ax2 1 1 ab22 ab22 VII IV

II III VI I VIII V Draw a picture of the following surfaces:

x y22 z a) x 2 y22 3 z or  elliptic paraboloid 6 3 2 opens in the positive x  direction

yz22 b) 4 x y22  4 z  0 or x  41

hyperbolic paraboloid Draw a picture of the surface : 4x2 y 2  4 z 2  4 y  24 z  36  0

4x2 y 2  4 y  ___  4 z 2  6 z  ___9    36  ___4  ___36

4x2  y  222  4 z  3  4

22 x2  yz23     1 1 4 1

ellipsoid center:  0,2,3

notice e.g. y  0 Summary:

x2 y2 z 2   1 Ellipsoid a2 b2 c2 x2 y 2 z 2   1 Hyperboloid of one sheet abc2 2 2 all variables present x2 y 2 z 2    1 Hyperboloid of two sheets a222 b c all variables squared z2 x 2 y 2  Cone  elliptic c2 a 2 b 2

z x22 y  Paraboloid  elliptic c a22 b z x22 y one variable Paraboloid  hyperbolic 22 c a b not squared

one variable not present cylinder opening in the direction of the missing variable 22 22 xy xz 2 1 Elliptic cylinder 1 Hyperbolic cylinder z ax Parabolic cylinder ab22 ab22