The distance from the origin to (8,4,4) is sqrt(96).
The distance from (8,4,4) to the z-axis is sqrt(80).

Problem 3

The components of the vector from (1,1,1) to (2,3,4) is [1,2,3]. The distance between the points is

Problem 4

Problem 5

(a)

(b)

Problem 6, 7, and 8

6

Do this in your head! 0.5 * |(3)(1)-(6)(7)| = 0.5 * |3-42| = 0.5 * 39 = 19.5

7

Just one example, [-7,-1] and [-4,5]

8

The longest altitude goes to the shortest side. Therefore,

Problem 9

It could be a Rhombus, or an Isosceles Trapezoid, or just an ordinary quadrilateral with perpendicular diagonals.

I will wait patiently for your proofs.

Problem 10

(a)

[-5,8]

(b)

[7,6,13]

Problem 11

Problem 12

This configuration is a sphere.

Problem 13

Parallelogram.

*created by raphael and maggie p

Problem 14

(a)

Rectangle

(b)

Rhombus

Page 30

Problem 1

Use the dot product. The vector from B to A is [5,-3,6]. The vector from B to C is [3,7,1]. The dot product of these two vectors is (3)(5)+(7)(-3)+(6)(1)=15-21+6=0. Hence, the two vectors are perpendicular, so the angle ABC is a right angle.

Problem 2

The length of [2,3,6] is

To make a vector of length 21, you multiply the vector by the scalar 3, producing the vector [6,9,18].

Problem 3

(a)

The vector from K to L is [4,7,4] and has a length 9. Therefore, the point on the segment 5 units from K would be

(b)

Notice that the vector from K to L has the same components, so we simply add.

Problem 4

In the applet below, type u+v+w into the input line, then press enter. Drag point A to the origin. Drag the other points around. This is what vector addition is all about.

Problem 5

The figure must be a parallelogram, and as a result, the vectors AD and BC are equal.

Problem 6 and 7

In the input line below, enter each vector sum listed in Problem 6. Be certain to drag the vectors around. You should experiment with other vector sums. Then use this same applet for Problem 7.

Problem 8, 9, and 10

8

The equation of the perpendicular bisector is

This line intersects x=3 at

9

You just need to do this problem. However, since the perpendicular bisector will never be perpendicular to the x-axis, because the segment FN will never have a slope of 0, there will always be an intersection point.

10

Do you agree that the distance from F to Q is greater than the distance from F to P? Do you agree that the distance from F to P is equal to the distance from P to N? Do you agree that the distance from Q to F is equal to the distance from Q to N? Why must the distance from Q to the x-axis be less than the distance from Q to N? Why must the distance from Q to the x-axis be less than the distance from Q to F?

This calls for the dot product, which equals 0, because we essentially DID this problem at the top of this page.

Problem 13

In the applet below, click on the arrows to play through the construction one step at a time. Once you have figured out if this construction makes sense, type Reflect[P,a] in the input line.

## Table of Contents

## Page 29

## Problem 1

(a) The coordinates for the other six vertices are:

B(6,0,0), C(6,3,0), D(0,3,0), E(0,0,2), F(6,0,2), and H(0,3,2)

(b) The lengths are:

## Problem 2

brought to you by Livescribe

The distance from the origin to (8,4,4) is sqrt(96).

The distance from (8,4,4) to the z-axis is sqrt(80).

## Problem 3

The components of the vector from (1,1,1) to (2,3,4) is [1,2,3]. The distance between the points is

## Problem 4

## Problem 5

## (a)

## (b)

## Problem 6, 7, and 8

## 6

Do this in your head! 0.5 * |(3)(1)-(6)(7)| = 0.5 * |3-42| = 0.5 * 39 = 19.5## 7

Just one example, [-7,-1] and [-4,5]## 8

The longest altitude goes to the shortest side. Therefore,## Problem 9

It could be a Rhombus, or an Isosceles Trapezoid, or just an ordinary quadrilateral with perpendicular diagonals.

I will wait patiently for your proofs.

## Problem 10

## (a)

[-5,8]## (b)

[7,6,13]## Problem 11

## Problem 12

This configuration is a sphere.

## Problem 13

Parallelogram.

*created by raphael and maggie p

## Problem 14

## (a)

Rectangle## (b)

Rhombus## Page 30

## Problem 1

Use the dot product. The vector from B to A is [5,-3,6]. The vector from B to C is [3,7,1]. The dot product of these two vectors is (3)(5)+(7)(-3)+(6)(1)=15-21+6=0. Hence, the two vectors are perpendicular, so the angle ABC is a right angle.

## Problem 2

The length of [2,3,6] is

To make a vector of length 21, you multiply the vector by the scalar 3, producing the vector [6,9,18].

## Problem 3

## (a)

The vector from K to L is [4,7,4] and has a length 9. Therefore, the point on the segment 5 units from K would be## (b)

Notice that the vector from K to L has the same components, so we simply add.## Problem 4

In the applet below, type u+v+w into the input line, then press enter. Drag point A to the origin. Drag the other points around. This is what vector addition is all about.

## Problem 5

The figure must be a parallelogram, and as a result, the vectors AD and BC are equal.

## Problem 6 and 7

In the input line below, enter each vector sum listed in Problem 6. Be certain to drag the vectors around. You should experiment with other vector sums. Then use this same applet for Problem 7.

## Problem 8, 9, and 10

## 8

The equation of the perpendicular bisector isThis line intersects

x=3 at## 9

You just need to do this problem. However, since the perpendicular bisector will never be perpendicular to the x-axis, because the segment

FNwill never have a slope of 0, there will always be an intersection point.## 10

Do you agree that the distance from F to Q is greater than the distance from F to P? Do you agree that the distance from F to P is equal to the distance from P to N? Do you agree that the distance from Q to F is equal to the distance from Q to N? Why must the distance from Q to the x-axis be less than the distance from Q to N? Why must the distance from Q to the x-axis be less than the distance from Q to F?

## Problem 11

you just need to draw this.

brought to you by Livescribe

## Problem 12

This calls for the dot product, which equals 0, because we essentially DID this problem at the top of this page.

## Problem 13

In the applet below, click on the arrows to play through the construction one step at a time. Once you have figured out if this construction makes sense, type

Reflect[P,a]in the input line.