MATH 100 WORKSHEET 1.1 & 1. Vectors in the Plane Find the vector v where u =, 1 and w = 1, given the equation v = u w. Solution. v = u w =, 1 1, =, 1 +, 4 =, 1 4 = 0, 5 Find the magnitude of v = 4, 3 Solution. v = 4 + 3 = 16 + 9 = 5 = 5 Forces with magnitudes of 500 pounds and 00 pounds act on a machine part at angles of 30 o and 45 o with the x-axis respectively. Find the resultant force vector. Solution. We first write out the vectors for the two forces, which we label F 1 and F. Then we can add the two vectors together to get the resultant force vector. F 1 = 500 cos (30 o î + 500 sin (30 o ĵ = 500 3 î + 500 ĵ 1 F = 00 cos ( 45 o î + 00 sin ( 45 o ĵ = 00 î + 00 ĵ F tot = [ 50 3 + 100 ] î + [ 50 100 ] ĵ
Vectors in Space We introduce the z-direction and look at points and vectors in 3 dimensions. The triangle given by the vertices A(5, 3, 4, B(7, 1, 3, C(3, 5, 3 is translated three units to the right along the y-axis. Determine the coordinates of the translated triangle. Solution. Translating the triangle along the y-axis only effects the y-components of the points. Thus our new vertices for the triangle are A t (5, 6, 4 B t (7, 4, 3 C t (3, 8, 3. Complete the square to write the equation in the form of a standard sphere. Solution. ( x + y + z + 9x y + 10z + 19 = 0 x + 9x + ( x + 9x + 81/4 + ( y y + ( + + ( y y + 1 + z + 10z + ( z + 10z + 5 (x + 9/ + (y 1 + (z + 5 = 109 4 Thus the circle has a center of ( 9 109, 1, 5 and a radius of r =. = 0 + + + = 0 + 81/4 + 1 + 5 Use vectors to determine whether the points A(1, 1, 5, B(0, 1, 6, C(3, 1, 3 lie in a straight line. Solution. To determine if the three given points lie in the same line, compare any two vectors that use all three points. If the two vectors are parallel, then the points are co-linear, that is, they lie in the same line. AB = 0 1, 1 ( 1, 6 5 = 1, 0, 1 AC = 3 1, 1 ( 1, 3 5 =, 0, we observe that AC = AB and since they are multiples, the two vectors are parallel. Thus we have that the points are colinear. Find a unit vector in the direction of u = 3,, 5 and a unit vector in the direction opposite of u. Solution. The unit vector in the direction of u is 3,, 5 3,, 5 =. 9 + 4 + 5 38 The unit vector in the direction opposite of u is 3,, 5 = 9 + 4 + 5 3,, 38.
MATH 100 WORKSHEET 1.3 Dot Product The dot product between two vectors u = u 1, u, u 3 and v = v 1, v, v 3 is defined as u v = u 1 v 1 + u v + u 3 v 3 = u v cos θ. Find the cosine of the angle between the vectors u = 1, 1, 1 and v =, 1, 1. Solution. Applying the formula above we find that cos θ = u v u v = 1 + 1 1 + 1 ( 1 1 + 1 + 1 + 1 + ( 1 = 3 6 = 3 3 = 3 Find the project of u = 0, 4, 1 onto v = 0,, 3. Find the vector projection of u onto v. Solution. The vector projection of u onto v is found by determining how much of the vector u lies in the direction of v and multiplying that scalar by the unit direction vector in the direction of v. proj v u = = ( ( u v v u v v v v v = 0 0 + 4 + 1 3 ( 0 + + 3 0,, 3 = 11 0,, 3 13
A toy wagon is pulled by exerting a force of 15 pounds on the handle that makes a 30 o angle with the horizontal. Find the work done in pulling the wagon 50 feet. Solution. Using the equation for Work and the information we are given we find that W = F PQ or F PQ cos θ = (15(50 cos (30 o 3 = 15 (50 = 375 3 ft-lbs
MATH 100 WORKSHEET 1.4 Cross Product The cross product of the vectors u = u 1, u, u 3 and v = v 1, v, v 3 is defined as î ĵ ˆk u v = u 1 u u 3 v 1 v v 3 = u u 3 v v 3 î u 1 u 3 v 1 v 3 ĵ + u 1 u v 1 v ˆk u v = u v sin θ Find u v given u = î + ĵ + ˆk and v = î + ĵ ˆk. Solution. Applying the cross product formula given above we find that î ĵ ˆk u v = 1 1 1 = ( 1 1 î ( 1 ĵ + (1 ˆk 1 1 Thus we have that u v = î + 3ĵ ˆk or, 3, 1. Find the area of the triangle with the vertices A(, 3, 4, B(0, 1,, C( 1,, 0. Solution. Determining two vectors from two sides of our triangle we have that AB = 0, 1 ( 3, 4 =, 4, and CB = 0 ( 1, 1, 0 = 1, 1,. Using that the area of a triangle equals 1/ base times height and geometry, we have the formula Area = 1 AB CB First we determine the cross product giving us AB CB î ĵ ˆk = 4 = (8 î ( 4 ( ĵ + ( 4 ˆk 1 1 Thus we need to take half of the magnitude of the vector 6,, which gives us Area = 1 6,, = 1 36 + 4 + 4 = 1 1 44 = 4 11 = 11 Thus the area of the triangle is 11 units.
Find the triple scalar product u ( v w given u =, 1,, v = 1, 1, 1, and w = 0,, Solution. To quickly determine the triple scalar product, one slightly modifies the matrix used to find the cross product of ( v w. By replacing the first row in the matrix (î ĵ ˆk with the components of the vector u, one can calculate the triple scalar product in one step. u ( v w = Thus we have that 1 1 1 1 0 = ( ( ( 0 (1 + ( 0 ( = 0 4 = 6 u ( v w = 6. A bolt is tightened by applying a 30-N force to a 40 cm wrench at an angle of 45 o. Find the magnitude of the torque about the center of the bolt. Solution. Using the formula for torque and the information we have given we find that τ = PQ F or PQ F sin θ = (.40(30 sin (45 o =.40 (30 = 6 Nm Thus the magnitude of the torque is 6 Nm.
MATH 100 WORKSHEET 1.5 Lines and Planes in Space The equation of a line requires a point on the line P (x 0, y 0, z 0 and a vector in the direction of the line v = a, b, c. The three forms of equations for a line are the vector equation: x x 0, y y 0, z z 0 = t v, the parametric equations: x = x 0 + at, y = y 0 + bt, z = z 0 + ct, and the symmetric equations: x x 0 = y y 0 = z z 0. a b c The equation of a plane requires a point on the plane P (x 0, y 0, z 0 and a vector normal (orthogonal to the plane n = a, b, c. The equation of a plane is a(x x 0 + b(y y 0 + c(z z 0 = 0. Find the equation of the line that passes through the point (,3,4 and is perpendicular to the plane given by 3x + y z = 6. Solution. First we must determine the vector in the direction of the line based on the information given. Since we want a line that is perpendicular (orthogonal to the given plane, the line is in the same direction as the normal vector to the plane which is n = 3,, 1. Thus we have that the point on the line is P (, 3, 4 and the vector in the direction of the line is 3,, 1. The equation of the line can be written in any of the forms below Vector : x, y, z =, 3, 4 + t 3,, 1 Parametric : x = + 3t y = 3 + t z = 4 t Symmetric : x = y 3 = z 4 3 1 Find the equation of the plane that passes through the points P (1,, 3, Q(, 3, 1 and R(0,, 1. Solution. To determine the normal vector to the plane containing the three points P, Q and R, we must create two vectors using the three points and then take the cross product of those two vectors. We will create the vectors PQ and PR. PQ = 1, 3, 1 ( 3 = 1, 1, 4 PR = 0 1,, 1 ( 3 = 1, 4, PQ PR = î ĵ ˆk 1 1 4 1 4 = 18, 6, 3 = ( ( 16 î ( ( 4 ĵ + ( 4 ( 1 ˆk We will write the equation of the plane using the point P (1,, 3 and the normal vector n = 18, 6, 3. Thus we have 18(x 1 6(y 3(z + 3 = 0 or 18x 6y 3z 15 = 0.
Determine whether the planes are parallel, orthogonal, or neither. If they are neither, find the cosine of the angle of intersection. 3x + y z = 7 x 4y + z = 0 Solution. To determine if the planes are parallel, orthogonal, or neither, we simply compare the normal vectors of the two planes. We have that n 1 = 3,, 1 and n = 1, 4,. Since it is easy to observe that n 1 k n we know that the two vectors, and therefore the two planes, are not parallel. Given that the dot product of the two vectors n 1 n = 3(1 + ( 4 + ( 1 = 3 8 = 7 0 we know that the two planes are not orthogonal. Thus we determine the cosine of the angle of intersection between the planes using the formula from 1.3 as follow cos θ = = n 1 n 1 n1 n 7 9 + 4 + 1 1 + 16 + 4 = 7 14 1 which after simplification gives us the resulting answer cos θ = 1 6. Find the point of intersection, if any, between the plane and the line given below. x + 3y = 10 x 1 3 = y + 1 = z 3 Solution. To find the point or points of intersection, we first write the equation of the line in parametric form. x = 1 3t y = 1 t z = 3 + t We can then substitute the parametric equations for x, y and z into our equation for the plane. We then determine what, if any, t-values satisfy the equation. x + 3y = 10 (1 + 3t + 3( 1 t = 10 + 6t 3 6t = 10 1 = 10 Since in our equation all the t s drop out and the resulting equation is never true, we have that the line and the plane never intersect.
MATH 100 WORKSHEET 1.6 Cylinders and Quadric Surfaces There are three types of surfaces in space we will deal with, cylinders, quadric surfaces and surfaces of revolution. For cylinders, the key is that only variables are present in the equation. The trace in those two variables is the generating curve. For quadric surfaces, there are 6 different types that we will be exploring: ellipsoid, hyperboloid of one sheet, hyperboloid of two sheets, elliptic cone, elliptic paraboloid, and hyperbolic paraboloid. Surfaces of revolution involve a generating curve in the variable of the axis of rotation. Describe and sketch the surface y + z = 4.. Solution. Since one variable is missing in the equation, we have the equation of a cylinder. The equation in the remaining two variables is a parabola, thus our surface is a parabolic cylinder. Identify the quadric surface z x y /4 = 1. Solution. To identify the quadric surface we first look at the traces in the coordinate planes. In the xy-plane z = k k 1 = x + y 4 In the xz-plane y = k if k > 1 ellipse z x = 1 + k 4 In the yz-plane z = k all k hyperbola z y 4 = 1 + k all k hyperbola With the traces being an ellipse and two hyperbolas and the RHS = 1, we look at the number of negative signs on the LHS of the equation to identify the surface as a TWO SHEET HYPERBOLOID.
Given the equation of the surface 4y + z x 16y 4z + 0 = 0, write the equation of the surface in standard form, identify the traces in (and parallel to each coordinate plane, i.e. x = k, y = k, and z = k, and identify the surface itself. Solution. To write the equation in standard form we need to complete the square for the y and z terms. 4y 16y + z 4z = x 0 4 ( y 4y + + ( z 4z + = x 0 + 4 ( + ( 4 ( 4 4y + 4 + ( z 4z + 4 = x 0 + 4 ( 4 + ( 4 4 (y + (z = x (y + (z 4 The traces in the coordinate planes are = x 4 STANDARD FORM z = k In the xy-plane: y = k In the xz-plane: x = k In the yz-plane: (y + k 4 = x k (z + = x (y (z + = k 4 4 4 4 4 all k parabola all k parabola k > 0 ellipse Thus our quadric surface is an ELLIPTICAL PARABOLOID. Given the equation of the surface 4x + 4y 8y + z = 0, write the equation of the surface in standard form, identify the traces in (and parallel to each coordinate plane, i.e. x = k, y = k, and z = k, and identify the surface itself. Solution. To write the equation in standard form we need to complete the square for the y term. 4x + 4y 8y + z = 0 4x 4 ( y y + + z = 0 + 4 ( 4x + 4 ( 4 y + 1 + z = 0 + 4 ( 1 4x + 4 (y 1 + z = 4 x + (y 1 + z 4 The traces in the coordinate planes are = 1 STANDARD FORM In the xy-plane z = k In the xz-plane y = k In the yz-plane x = k x + (y 1 = 1 k x + z 4 4 = 1 (k 1 (y 1 + z 4 = 1 k k < 4 circle (k 1 < 1 ellipse k < 1 ellipse Thus our quadric surface is an ELLIPSOID or CIRCULAR ELLIPSOID.