11.1 Three-Dimensional Coordinate System In three dimensions, a point has three coordinates: (x, y, z). The normal orientation of the x, y, and z-axes is shown below. The three axes divide the region into 8 octants.

z = 0 is the equation of the xy-plane. y = 0 is the equation of the xz-plane. x = 0 is the equation of the yz-plane. What is the projection of the point (2, 6, −5) onto the xy-plane? the yz-plane? the xz-plane?

What do the following represent in R3?

y = 7 2x+z = 8

1 x2 + y2 = 9

z ≥ y2

q 2 2 2 The distance between two points (x1, y1, z1) and (x2, y2, z2) is (x2 − x1) + (y2 − y1) + (z2 − z1) . x + x y + y z + z  The midpoint of the points (x , y , z ) and (x , y , z ) is 1 2 , 1 2 , 1 2 . 1 1 1 2 2 2 2 2 2 The equation of a sphere with center (h, k, l) and radius r is:

(x − h)2 + (y − k)2 + (z − l)2 = r2

Find an equation of the sphere that has center (3, −2, 7) and touches the xz-plane.

2 Find the equation of the sphere that has diameter passing through the points (1, 4, −10) and (−3, 6, −2).

What is the intersection of this sphere with the yz-plane?

Find the center and radius of the sphere x2 + y2 + z2 + 6x − 8z = 11

Describe mathematically the top half of a solid sphere of radius 4 centered at the origin.

3 11.2 Vectors and the Dot Product

In 3 dimensions a vector has 3 components: a =< a1, a2, a3 >. where a1 is the x-component, a2 is the y-component, and a3 is the z-component. An equivalent way of writing a vector is by using the standard unit basis vectors: i =< 1, 0, 0 >, j =< 0, 1, 0 >, and k =< 0, 0, 1 >.

The vector a =< a1, a2, a3 > can be written as a = a1i + a2j + a3k.

q 2 2 2 The magnitude (or length) of the vector a =< a1, a2, a3 > is |a| = a1 + a2 + a3. −−→ Given the points A(x1, y1, z1) and B(x2, y2, z2), the vector from A to B is AB =< x2 − x1, y2 − y1, z2 − z1 >. a To find a unit vector (vector of length 1) in the direction of a vector a, compute . |a| Let a be the vector from the point P (2, −4, −7) to the point Q(1, 3, −5) and let b = −4i + 2j − 6k. Fid a unit vector in the same direction as a.

Find a vector of length 4 in the same direction as the vector a + 2b.

4 Two vectors are parallel if one vector is a scalar multiple of the other.

D 4 E For example, a = h4, −3, 6i is parallel to b = − 3 , 1, −2 since a = −3b.

Given two vectors a =< a1, a2, a3 > and b =< b1, b2, b3 >, the dot product (or scalar product) of a and b, denoted a · b, can be found in either of the following ways:

a · b = |a||b| cos θ where θ is the angle between a and b

a · b = a1b1 + a2b2 + a3b3

A dot product can only be performed on two vectors and the result is a scalar. Note: In the context of this section, a · b would make sense, but |a| · b would not since |a| is not a vector. The · here does not mean multiplication. It means dot product. The first formula above rearranged gives us a formula for finding the cosine of the angle between two nonzero vectors.

a · b cos θ = |a||b|

Two vectors a and b are orthogonal (or perpendicular), if a · b = 0. For what values of x are the vectors < x, 3x, 4 > and < x, 4, 5 > orthogonal?

A triangle has vertices A(0, 3, 9),B(1, −2, 1), and C(3, 1, 2). Find 6 ABC.

5 Given two vectors a =< a1, a2, a3 > and b =< b1, b2, b3 >, The scalar projection of b onto a is given by a · b compab = |a| The vector projection of b onto a is given by a · b a a · b projab = = a |a| |a| |a|2

Given the vectors a = 10i − 2k and b =< 3, −4, 1 >, find the scalar and vector projections of b onto a.

6 11.3 The Cross Product A determinant for a 2 × 2 array of numbers (matrix):

a b = ad − bc c d

2 −3 Example: = 5 −9 A determinant for a 3 × 3 array of numbers (matrix):

a a a 1 2 3 b2 b3 b1 b3 b1 b2 b1 b2 b3 = a1 − a2 + a3 c2 c3 c1 c3 c1 c2 c1 c2 c3

2 3 1

Example: 0 2 1 =

5 3 0

The cross product of two vectors a =< a1, a2, a3 > and b =< b1, b2, b3 > is:

i j k

a × b = a1 a2 a3

b1 b2 b3

Example: Find a × b if a =< 3, 2, −1 > and b =< 4, 1, 1 >.

The cross product of two vectors is a VECTOR!

7 Very Important Fact: The vector a × b is orthogonal to both a and b. Find a vector that is perpendicular to the plane containing the points A(1, 2, 3),B(−2, 1, −1), and C(1, −1, 1).

The direction in which the cross product points can be determined by the right-hand rule.

The right-hand rule helps us to see that a × b 6= b × a. What is true is a × b = −b × a. If θ is the angle between two vectors a and b, then

|a × b| = |a||b| sin θ

This above fact tells us the following: (1) Two nonzero vectors a and b are parallel if and only if |a × b| = 0. (2) The area of the parallelogram formed by the vectors a and b is |a × b|.

Find the area of the triangle from the previous example with vertices A(1, 2, 3),B(−2, 1, −1), and C(1, −1, 1).

8 The scalar triple product of the vectors a, b, and c is a · (b × c). The volume of the parallelipiped determined by the vectors a, b, and c is the absolute value of the scalar triple product:

V = |a · (b × c)|

When finding the scalar triple product, you can either first find b × c and then dot with a, or you can find it all in one step by computing the determinant below where a =< a1, a2, a3 >, b =< b1, b2, b3 >, and c =< c1, c2, c3 >.

a a a 1 2 3

b1 b2 b3

c1 c2 c3 Find the volume of the parallelipiped formed by the vectors a =< 1, 3, 1 >, b =< 4, −1, 2 > and c =< 2, 2, 0 >.

What does it mean, then, if the scalar triple product of three vectors is 0?

Do the points P (3, 0, 1), Q(−1, 2, 5), R(5, 1, −1) and S(0, 4, 2) all lie in the same plane? i.e. Are they coplanar?

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