1. 3/3 points | Previous Answers WorldwideCMMVCalc1 1.1.510d.XP. My Notes

Consider the following four points.

P = (6, 6, 6),

Q = (−8, 4, −8),

R = (7, 2, 13),

and

S = (12, 3, −7)

Solution or Explanation

The distance from a point to a coordinate axis is the distance between the point and its orthogonal projection onto the coordinate axis. To find which of the points is closest to each coordinate axis, we will find the distances between the points and each of their orthogonal projections onto the coordinate axes. In each case, the shortest distance will identify the point that is closest to each axis.

Recall that the orthogonal projection of a point

(x, y, z)

onto the x-axis is the point

(x, 0, 0).

Similarly, the orthogonal projection of a point

(x, y, z)

onto the y-axis is the point

(0, y, 0),

and the orthogonal projection of

(x, y, z)

onto the z-axis is the point

(0, 0, z).

Also recall that the Euclidean distance between two points

(x₁, y₁, z₁)

and

(x₂, y₂, z₂)

is given by

d =

(x₁ − x₂)² + (y₁ − y₂)² + (z₁ − z₂)²

.

Therefore, the distance between a point

(x, y, z)

and

(x, 0, 0),

its orthogonal projection onto the x-axis, is

d_x =

(x − x)² + (y − 0)² + (z − 0)²

=

y² + z²

.

Similarly, the distance between a point

(x, y, z)

and

(0, y, 0),

its orthogonal projection onto the y-axis, is

d_y =

x² + z²

.

The distance between a point

(x, y, z)

and

(0, 0, z),

its orthogonal projection onto the z-axis, is

d_z =

x² + y²

.

(a) To find the point closest to the x-axis, we will calculate

d_x =

y² + z²

for each point.

The distance from

P = (6, 6, 6)

to the x-axis is

P_x =

6² + 6²

=

72

.

The distance from

Q = (−8, 4, −8)

to the x-axis is

Q_x =

4² + (−8)²

=

80

.

The distance from

R = (7, 2, 13)

to the x-axis is

R_x =

2² + 13²

=

173

.

The distance from

S = (12, 3, −7)

to the x-axis is

S_x =

3² + (−7)²

=

58

.

Since

S_x < P_x < Q_x < R_x,

the point

S = (12, 3, −7)

is closest to the x-axis.

(b) To find the point closest to the y-axis, we will calculate

d_y =

x² + z²

for each point.

The distance from

P = (6, 6, 6)

to the y-axis is

P_y =

6² + 6²

=

72

.

The distance from

Q = (−8, 4, −8)

to the y-axis is

Q_y =

(−8)² + (−8)²

=

128

.

The distance from

R = (7, 2, 13)

to the y-axis is

R_y =

7² + 13²

=

218

.

The distance from

S = (12, 3, −7)

to the y-axis is

S_y =

12² + (−7)²

=

193

.

Since

P_y < Q_y < S_y < R_y,

the point

P = (6, 6, 6)

is closest to the y-axis.

(c) To find the point closest to the z-axis, we will calculate

d_z =

x² + y²

for each point.

The distance from

P = (6, 6, 6)

to the z-axis is

P_z =

6² + 6²

=

72

.

The distance from

Q = (−8, 4, −8)

to the z-axis is

Q_z =

(−8)² + 4²

=

80

.

The distance from

R = (7, 2, 13)

to the z-axis is

R_z =

7² + 2²

=

53

.

The distance from

S = (12, 3, −7)

to the z-axis is

S_z =

12² + 3²

=

153

.

Since

R_z < P_z < Q_z < S_z,

the point

R = (7, 2, 13)

is closest to the z-axis.

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2. 1/1 points | Previous Answers WorldwideCMMVCalc1 1.2.505a.XP. My Notes

What vector has the same magnitude as the vector

u =

9, −4

and points in the opposite direction?

⟨−9,4⟩

$webMathematica generated answer key$

Solution or Explanation

Recall that the magnitude of a vector is also known as the length of a vector. Given a vector

u =

a, b

,

the magnitude of the vector is

|u| =

a² + b²

.

Also, a vector pointing in the opposite direction of the vector

u =

a, b

can be found by multiplying

u =

a, b

by a negative scalar. This will change the sign of each component of

u

so that the resulting vector is pointing in the opposite direction.

To find the vector that points in the opposite direction of

u

and has the same magnitude as

u,

we multiply

u

by the scalar

−1.

This gives

−u =

−a, −b

.

Notice that

|−u| =

(−a)² + (−b)²

=

a² + b²

= |u|.

Thus, these vectors have the same magnitude.

In this case, the vector with the same magnitude as

u =

9, −4

that points in the opposite direction of

u

is

−u =

−9, 4

.

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3. 1/1 points | Previous Answers WorldwideCMMVCalc1 1.2.511b.XP. My Notes

Consider the function

f(x) = cos(x) + 5.

Sketch the displacement vector whose tail is at the point on the curve where

x = π

and whose head is at the point on the curve where

x = −2π.

Solution or Explanation

Recall that a displacement vector is a directed line segment from an initial point, its tail, to a terminal point, its head.

It is given that the tail of the vector is the point on the curve

f(x) = cos(x) + 5

where

x = π,

and the terminal point of the vector is the point on

f(x)

where

x = −2π.

Since

f(π) = cos(π) + 5 = 4

and

f(−2π) = cos(−2π) + 5 = 6,

the initial point of the vector is

(π, 4),

and the terminal point of the vector is

(−2π, 6).

To draw this vector, first label the points

(π, 4)

and

(−2π, 6).

Then, connect the points with a line segment. Lastly, draw an arrowhead with its point at

(−2π, 6),

the terminal point of the vector, to indicate the direction of the vector.

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4. 4/4 points | Previous Answers WorldwideCMMVCalc1 1.2.526a.XP. My Notes

Match each of the graphs below with the appropriate vector sum using the Triangle Rule.

(i)		(ii)
(iii)		(iv)

(a)	u + v	=	(iv)

(b)	u − v	=	(iii)

(c)	u + u	=	(ii)

(d)	v − v	=	(i)

Solution or Explanation

The given question asks us to match the vectors

u + v,

u − v,

u + u,

and v − v to their appropriate graph using the Triangle Rule.

Recall that the Triangle Rule is a method used to find the sum of two vectors. If vectors u and v are oriented such that the initial point of v is at the terminal point of u, then the sum

u + v

is defined by the vector that has initial point at the initial point of u, and terminal point at the terminal point of v.

(a) By first drawing the vector u, then placing the initial point of v at the initial point of u, and lastly drawing the vector with the same initial point as u and the same terminal point as v, we obtain the sum

u + v

in the following graph.

(b) The vector difference u − v can be written as the sum

u + (−v)

in order to use the Triangle Rule. The vector −v is the same length, but the opposite direction, as the vector v. It is oriented so that its initial point is at the terminal point of u. Then the vector

u − v = u + (−v)

is drawn from the initial point of u to the terminal point of −v.

(c) The vector sum

u + u

is found by first drawing u, then drawing a second vector u with initial point at the terminal point of the first u. The vector sum

u + u

is drawn from the initial point of the first u to the terminal point of the second u.

(d) The vector difference v − v can be written as the sum

v + (−v)

in order to use the Triangle Rule. The vector −v is the same length, but the opposite direction as the vector v. The initial point of −v is oriented at the terminal point of v. Therefore the terminal point of −v is at the initial point of v, which means that the sum

v + (−v)

is equal to

0.

Thus, the vectors are matched up with their corresponding graph as follows.
(a) - (iv)

(b) - (iii)

(c) - (ii)

(d) - (i)

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5. –/5 points WorldwideCMMVCalc1 1.3.510a.XP. My Notes

Let

u = i − 3k,

v =

8, 5

,

and

w = 7j + 5k.

Compute the expressions. (If an answer is undefined, enter UNDEFINED.)

(a)

u · v

(b)

u · w

(c)

v · v

(d)

v · w

(e)

(u − w) · 5w

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6. –/1 points WorldwideCMMVCalc1 1.3.513a.XP. My Notes

Determine the acute angle between the lines

y = 3x + 7

and

y = 8 − x.

(Round your answer to one decimal place.)
°

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7. –/1 points WorldwideCMMVCalc1 1.3.518c.XP. My Notes

If

u =

26, −3, −4

and if

w =

26,

1

2

, −

1

2

is the component of u orthogonal to v, find a unit vector parallel to v.

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8. –/1 points WorldwideCMMVCalc1 3.4.506d.XP.Tut. My Notes

Evaluate

f(x, y, z) dV

for the specified function f and box calligraphic B

.

f(x, y, z) =

z

x

; 1 ≤ x ≤ 5, 0 ≤ y ≤ 2, 0 ≤ z ≤ 2

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9. –/1 points WorldwideCMMVCalc1 3.6.504d.XP.Tut. My Notes

Convert from spherical to rectangular coordinates.

8,

π

3

,

π

4

(x, y, z) =

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10. –/1 points WorldwideCMMVCalc1 3.7.501d.XP. My Notes

Find the average value of the function

f(x, y) =

y

x² + y²

over the region in the xy-plane bounded by the x-axis,

y = x,

and

x = 1.

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Massey - Worldwide Multivariable Calculus 1/e (Homework)

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