Section 12.1 Three-Dimensional Coordinate System

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1 Chapter 1 Vectors and Geometry of Space Section 1.1 Three-Dimensional Coordinate System Suppose P1 ( x1, y1, z 1) and P ( x, y, z ) are the given points. Find the distance D. The distance D = ( x x ) + ( y y ) + ( z z ) Important equations in 3-D to remember: 1. ax + by + cz = d represents a plane. x = a is a surface parallel to yz-plane 3. y = b is a surface parallel to zx-plane 4. z = c is a surface parallel to xy-plane 5. y = x is a vertical plane that intersects xy-plane in the line y = x 6. ( x a) ( y b) ( z c) d + + = is a sphere center at (,, ) and radius d 7. ( x a) + ( y b) + ( z c) = d, z c is a hemisphere center at ( a, b, c ) 1. Find the equation of a sphere center at (1,, 1) and radius 1. Solution: = ( x 1) ( y ) ( z 1) 1. Determine whether the points lie on the straight line a) A(5,1,3), B(7,9 1), C(1, 15,11) Solution: Check that AB = 1, BC = 6 1, AC = 4 1 and AB + AC = BC, The points are on a line. b) K(0,3, 4), L(1,, ), C(3,0,1). Like in a) you can show the points are not on the same line. 3. Find the center and radius of the sphere given by Solution: Complete the square as ( x 3) ( y ) ( z 1) 5 (3,,1) and radius = 11 x y z x y z = and then center is at 3 4. Describe in words the region of R represented by the equations or inequalities a) y = 5 b) x = 5 c) x > 4 d) y 0 e) 0 z 6 f) y = z g) i) x y z z + + < 3 j) x x + y + z > 1 h) xyz = 0 + y = 1 k) x + z 9

2 Section 1. Vectors Parallelogram law: If we place two vectors u, v so that they start at a same point, then u + v lies along the diagonal of the parallelogram with u, v vectors as sides. For two vectors u =< a, b, c > and v =< x, y, z > the vector represented and defined by a = AB =< x a, y b, z c > and a = BA =< a x, b y, c z > The length or magnitude of a vector: a = ( x a) + ( y b) + ( z c) 1. Given a =< 3, 1 >, b =< 5,3 >. Find a + b, a b, 3a b and a + b, a b, 3a b. Find a vector that has same direction as the vector <,4,5 > and magnitude 6. 1 Solution: find unit vector u = <,4,5 >, the vector we are looking for is 3 5 w = 6u = <, 4,5 > 5 Section 1.3 The Dot Product of Vectors For two vectors u =< a, b, c >, v =< x, y, z > the dot product is defined as u v = ax + by + cz If θ is the angle between the vectors u, v, then u v = u v cosθ Case 1. θ = 0 vectors are parallel in the same direction and u v = u v Case. θ = π vectors are parallel in the opposite direction and u v = u v π Case 3. θ = vectors are perpendicular (Orthogonal) and u v = 0 Direction angles and direction cosine: If α, β, γ are the angles of a vector with the coordinate axes x, y and z respectively then those are called the direction angles. And cos α, cos β, cosγ are direction cosines and cos α + cos β + cos γ = 1 Question: Show that α β γ cos + cos + cos = 1 Projections: Prove that scalar projection of b onto a is a i b compab = a and

3 Vector projection of b onto a a b i a is projab = (See your text book page # 811) a a 1. Find the dot product between two given vectors. 1 a) u =, 4 and v =< 8, 3 > b) u = 4i 3k and v = i + 3 j 4k. Given that u = 4, and v = 10, θ = 10 0 find u v 1 3. Find the angle between the vectors u =, 4 and v =< 8, 3 > 4. Show that u =<, 6, 4 > and v =< 3, 9, 6 > are parallel 5. Find a unit vector that is orthogonal to both i + j and i + k Solution: Suppose a =< a, b, c > is the unit vector. Now a + b + c = 1 and < a, b, c > i < 1,1,0 >= 0 and < a, b, c > i < 1, 0,1 >= 0. Solve for a, b, and c for the unit 1 vector a =< a, b, c >= ± < 1, 1, 1 > Find direction cosines and direction angles of the vector a =<,3, 6 > Solution: cosα = = α = arccos( / 7), cos β = 3 3 = β = arccos(3/ 7) and cosγ = 6 6 = γ = arccos( 6 / 7) If two direction angles are given α = π / 4, β = π / 3, find γ. 8. Determine the scalar and vector projection of b =< 4,1 > onto a =< 1, > 9. A crate is hauled 8 cm up a ramp under a constant force of 300 N applied at an angle of 30 degrees to the ramp. Find the work done. Solution: Work done W = 0 F D = F D cosθ = 300(8) cos 30 Nm = J Nm stands for Newton-meter and 1 Nm = 1 Joules, which is the unit of work. 10. A force is given by a vector F = <3, 4, 5> and moves a particle from the point P(,1, 0) to the point Q(3,, 1). Find the work done.

4 Solution: F D =< 3, 4, 5 > < 1,1,1 > units = 1 units of work, where D is the distance vector from P to Q. 11. A woman exerts a horizontal force of 5 lb on a crate as she pushes it up a ramp that is 0 ft long and inclined at an angle of 15 degrees above the horizontal. Find the work done on the box. Answer: About 483 ft-lb Section 1.4 The Cross Product of Vectors For two vectors u =< a, b, c >, v =< d, e, f > the cross product is defined as i j k u v = = i( bf ec) j( af dc) + k( ae bd) d e f You need to review determinants from any algebra book. Theorem: The vector u v is orthogonal to both u and v Theorem: If θ is the angle between two vectors then u v = u v sin θ, 0 θ π Corollary: Two nonzero vectors u and v are parallel if u v = 0 Note: The length of the cross product u v is equal to the area of the parallelogram determined by the vectors. Scalar Triple Product (STP): For three vectors u, v and w the scalar triple product is defined as u ( v w ) Volume of a parallelepiped is given by V = u ( v w), which is the magnitude of a scalar triple product. Note: If u ( v w ) = 0 the vectors are coplanar Properties: 1. a ( b c) = c ( a b) = b ( c a). u v = v u 3. i i = 0, i j = k, k i = j and so on 4. a ( b c) = ( a c) b) ( a b) c 1. Show that a =< 1, 4, 7 >, b =<, 1, 4 > Solution: One needs to verify that a ( b c) = 0 and c =< 0, 9,18 > are coplanar

5 . Find a vector perpendicular to the plane that that passes thru P(1,4,6), Q(,5, 1) and R(1, 1,1) Solution: Find PQ PR =< 40, 15,15 > 3. Find the area of a triangle with vertices P(1,4,6), Q(,5, 1) and R(1, 1,1) 5 8 Solution: Area A = 1/ PQ PR = 4. Find a b for a =< 1, 4, 7 >, b =<, 1, 4 > and show that the cross product is orthogonal to both a and b 5. Find two unit vectors orthogonal to both < 1,1,1 > and <,0,1 > 6. Show that a ( b a) = 0 a c b c 7. Show that ( a b) ( c d ) = a d b d Solution: ( a b) ( c d ) = ( a b) v = a ( b v) = a ( b c d ) = a [( b d ) c ( b c) d ] = ( a c)( b d) ( b c)( a d ) Section 1.5 Equations of lines and planes Lines: For a given direction vectors v =< a, b, c >, the vectors r =< x, y, z >, and r 0 =< x0, y0, z0 > lies on the line L r = r + tv 0 < x, y, z >=< x, y, z > + t < a, b, c > =< x + at, y + bt, z + ct > The parametric equation of the line L is x = x0 + at, y = y0 + bt, z = z0 + ct Also we can write the symmetric form (eliminating t) x x0 y y0 z z0 = = where a, b, and c are called direction numbers or direction ratios. The line segment from r 0 =< x0, y0, z0 > to r1 =< x1, y1, z1 > is the vector given by r( t) = (1 t) r 0 + tr1, 0 t 1 For two points p0 = ( x0, y0, z0) and p1 = ( x1, y1, z1) on L has the symmetric equation x x0 y y0 z z0 = = x x y y z z

6 Planes: Vector equation of a plane is defined as n ( r r 0) = 0, where n is the unit normal vector to the plane, containing r, and r 0. A plane passing thru a point p0 = ( x0, y0, z0) with normal vector n =< a, b, c > has scalar equation a( x x0) + b( y y0) + c( z z0) = 0 which also can be written as ax + by + cz + d = 0. The distance of a point p1 = ( x1, y1, z1) from the plane ax1 + by1 + cz1 + d ax + by + cz + d = 0 is defined as D = a + b + c Angle between two planes ax + by + cz + d = 0 and a1x + b1 y + c1z + d1 = 0 is given by cosθ n n n n 1 =, where n1,, 1 =< > and n =< a1, b1, c1 > A. Lines 1. Find a vector equation and parametric equations for the line that passes thru (5, 1, 3) and is parallel to v =< 1, 4, > Vector equation < x, y, z >=< 5,1,3 > + t < 1, 4, > =< 5 + t,1+ 4 t,3 t >= (5 + t) i + (1 + 4 t) j + (3 t) k Parametric equation x = 5 + t, y = 1+ 4 t, z = 3 t. Find a symmetric equation and parametric equations for the line that passes thru (, 4, 3) and (3, 1, 1) The symmetric equation x x0 y y0 z z0 = = x x y y z z x y 4 z 3 = = 1 3 The parametric form is x = + t, y = 4 3 t, z = 3 t 3. In example, find intersection of the line with xy-plane. On the xy-plane z = 0. Then x = + t, y = 4 3 t, z = 0 = 3 t t = 3/ We have the point (7/, -1/, 0)

7 B. Planes 4. Find an equation of a plane through (, 4, -1) with a normal vector n =<,3,4 > The plane has equation n ( r r 0) = 0 <,3, 4 > < x, y 4, z + 1 >= 0 ( x ) + 3( y 4) + 4( z + 1) = 0 5. Find the equation of a plane thru P(1, 3, ), Q(3, -1, 6) and R(5,, 0) Derive vectors PQ =<, 4, 4 >, PR =< 4, 1, >, and n = PQ PR. Now you can consider the point P(1, 3, ) and the normal vector to find your plane 1( x 1) + 0( y 3) + 14( z ) = 0 6. Find the angle between two given planes x + y + z = 1 and x y + 3z = 1. Notice that we have n 1 =< 1,1,1 > n1 n and n =< 1,, 3 >. Now find cosθ = n n 1 7. Find the symmetric equations of the line of intersection L of two planes x + y + z = 1 and x y + 3z = 1. Suppose n and n 1 are the normal vectors to the given planes. Then n 1 =< 1,1,1 > and n =< 1,, 3 >. The line L has direction vector v = n1 n =< 5,, 3 >. Let us find a point common to both the planes letting z = 0, which could be (1, 0, 0). Thus we have the x 1 y z equation of L in symmetric form, = = 5 3 Section 1.6 Cylinder and quadric surfaces A cylinder is a surface that consists of all lines (called rulings) that are parallel to a given line and passes thru a given plane curve. A quadric surface is the graph of a second-degree equation in three variables x, y and z. The most general equation of a quadric surface is Ax + By + Cz + Dxy + Eyz + Fzx + x + Hy + Iz + J = 0 General forms: Look at page number 386 on your text for the diagrams x y z = 1is an ellipsoid. For a = b = c, the ellipsoid is a sphere z x y. = + is an elliptic paraboloid. For a = b it is circular paraboloid. c a b

8 z x y = is a hyperbolic paraboloid. c a b x y z + = 1is a hyperboloid of one sheet x y z + = 1 is a hyperboloid of two sheets x y z + = 0 is acone Homework problems: 8. since z is missing in x y = 1, we consider x y = 1with z = k, is a hyperbola on the z = k plane. The surface is hyperbolic cylinder. 1. Find the traces of 4y = x + z in the planes x = k, y = k, and z = k. When x = k: 4y = k + z is a parabola, y = k: 4k = x + z is a circle and z = k: 4y = x + k is also a parabola Thus the surface is a circular paraboloid with axis in the y axis and vertex (0, 0, 0) x y z. 9x + 4y + z = = 1 is an ellipsoid with intercepts 1/ 9 1/ 4 1 ( ± 1/ 3,0,0), (0, ± 1/,0), (0,0, ± 1) 34. Reduce the equation 4y + z x 16y 4z + 0 = 0 to one of the standard forms a classify the surface and make a rough sketch. x ( y ) ( z ) Solution: We find the form = + is an elliptic paraboloid vertex at (0,, ) and axis is the horizontal line y =, z =.