HWQ Find the xy trace:

HWQ Find the standard form of the equation of a sphere with endpoints of a diameter (2,-2,2) and (-1,4,6)

Vectors and the Geometry of Space 2015 Section 10.2 Vectors in Space

Vectors in Three-Dimensional Space Now that we have an understanding of the three-dimensional system, we are ready to discuss vectors in the three-dimensional system. All the information you learned about vectors in Chapter 6 will apply, only now we will add in the third component. Vectors in component form in three dimensions are written as ordered triples, in other words, now a vector in component form is . In three dimensions the zero vector is O = < 0, 0, 0> and the standard unit vectors are . z Each of the unit vectors represents one unit of change in the direction of each of their respective positive axes. y x

Component form of a vector Given the initial point, and the terminal point, , the component form of the vector can be found the same way it was on the Cartesian Plane. Component form of a vector Be sure to subtract the initial point’s coordinates from the terminal point’s coordinates. The same vector can be written as a combination of the unit vectors. Standard Unit Vector Notation We will look at examples using both forms.

More on Vectors in Three-Dimensions Let and let c be a scalar. Vector Equality: Magnitude or Length of a Vector: Vector Addition: Scalar Multiplication: Unit Vector in the Direction of : Note: This is simply the vector multiplied by the reciprocal of its magnitude.

Let’s look at some example problems involving vectors. Sketch the vector with initial point P(2, 1, 0) and terminal point Q(3, 5, 4). Then find the component form of the vector, the standard unit vector form and a unit vector in the same direction. Solution: First draw a 3D system and plot P and Q. The vector connects P to Q. Q P

Example 1 Continued: Second, find the component form of the vector. Do this by subtracting the initial point’s coordinates P(2, 1, 0) from the terminal point’s coordinates Q(3, 5, 4). Component form Standard Unit Vector Form Last, find a unit vector in the same direction. Do this by multiplying the vector by the reciprocal of the magnitude. Note: You can verify it’s a unit vector by finding its magnitude.

Example 2: Given the vectors find the following: a. b. c. Solution: a. b. c.

Parallel Vectors You may recall that a nonzero scalar multiple of a vector has the same direction as the vector (positive scalar) or the opposite direction as the vector (negative scalar). Since this is the case, any nonzero scalar multiple of a vector is considered a parallel vector. In other words, if two vectors, , are parallel, then there exists some scalar, c such that . The zero vector does not have direction so it cannot be parallel. To get the idea, look at these parallel vectors on the Cartesian Plane. y x

Example 3: Determine if the vector with initial point, P(3,2,-2) and terminal point, Q(7,5,-3) is parallel to the vector . Solution: First find the component form of the vector from P to Q. Second, if the two vectors are parallel, then there exists some scalar, c, such that Then –12 = 4c c = -3 And -9 = 3c c = -3 And 3 = -c c = -3 For the two vectors to be parallel, c would have to be the same for each coordinate. Since it is, the two vectors are parallel.

Example 4: Determine whether the points A(2,3,-1), B(0,1,3) and C(-3,-2,8) are collinear. Solution: We need to find two vectors between the three points and determine if they are parallel. If the two vectors are parallel and pass through a common point then the three points must be in the same line. The vector from A to B is Now we need to find the vector from A to C or B to C. The vector from A to C is To be parallel: -2 = -5c c = 2/5 -2 = -5c c = 2/5 4 = 9c c = 4/9 Since c is not the same in each case, the vectors are not parallel and the points are not collinear.

Example 4b You Try: Determine whether the points P(2, -1,4), Q(5,4,6) and R(-4,-11,0) are collinear.

Example 5: Find a vector parallel to the vector with magnitude 5. Solution: Be careful. We might quickly assume that all we need to do is to multiply the vector by 5. This would be fine if we were dealing with a unit vector. Since we are not, we need to multiply by the reciprocal of the magnitude first to get a unit vector and then multiply by 5.

Solution to Example 5 Continued: You can verify the new vector is parallel if you look at the form: Obviously the scalar multiple is . You can verify the magnitude is 5 by finding the magnitude of the form:

More on Vectors in Three-Dimensions Let and let c be a scalar. Dot Product of u and v: Angle between 2 vectors u and v : Orthogonal vectors have a dot product of 0.

Example 6: Find the dot product of Example 7: Find the angle between

Example 6: The weight of an 80lb. chandelier hanging 2.5 feet from the ceiling is distributed over 3 chains. If the chains are located as shown below, represent the force exerted on each chain with a vector. (-1,-1,0) 1 ft 1 ft 1 ft (0,1,0) (1,-1,0) 1 ft (0,0,-2.5)

Solution to Example 6: First find the vectors from the chandelier to the three points on the ceiling. Each force is a multiple of the vector since we can find the direction, but we don’t know the magnitude. (-1,-1,0) 1 ft 1 ft 1 ft (0,1,0) (1,-1,0) 1 ft (0,0,-2.5)

The sum of the three forces must negate the downward force of the chandelier from its weight. (Add vectors by adding individual components.) So, This gives us a system of three equations in three unknowns, a,b and c. Solving the system, you get a = 16, b = 8 and c = 8. Thus the three forces are

Homework: 10.1-10.2 Worksheet (evens) For extra practice do: 10.2 Pg.719 1-45 odd