mcv4u - practice mastery test #8 -...

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Name: ________________________ Class: ___________________ Date: __________ ID: A 1 MCV4U - Practice Mastery Test #8 Multiple Choice Identify the choice that best completes the statement or answers the question. ____ 1. If x - 3 ( ) 2 - x ( ) 2 2x - 1 ( ) < 0 then a. x > 3 or 1 2 < x < 2 c. 1 2 < x < 2 or 2 < x < 3 b. x < 1 2 or 2 < x < 3 or x > 3 d. x < 1 2 or 1 2 < x < 2 or x > 3 ____ 2. If f = {(2,4), (4,-3), (-3,2)} then f -1 (-3) = a. 4 b. 2 c. -3 d. -2 ____ 3. A function is defined by y = x - 3 . Its domain is a. {x | x 0, x ò} c. {x | x 3, x ò} b. {x | x > 0, x ò} d. {x | x < 3, x ò} ____ 4. Which of the following limits do not exist? i) lim x 3 + 3 - x ii) lim x 3 - 3 - x iii) lim x 2 3 - x a. i) only b. ii) only c. iii) only d. i) and iii) ____ 5. The expression lim h 0 x + h - x h is most likely to be... a. the derivative of a function c. slope of a secant b. the value of a derivative d. none of the above ____ 6. If y = x 3 x - 2 ( ) + 3 then dy dx = a. 4x 3 - 6x 2 b. 4x 3 - 2 c. -6x 2 + 1 d. 3x 2 - 4x + 3 ____ 7. The number lines for y, dy dx , and d 2 y dx 2 are shown below. All zeros are shown. -4 -3 -2 -1 0 1 2 3 4 5 y - - - 0 - - - - - - - - - - - 0 + + + + + dy dx + + + 0 - - - - - - - 0 + + + + + + + + + d 2 y dx 2 - - - 0 - - - 0 + + + + + + + + + 0 - - - The x-coordinates of all points of inflection are a. -3, -1 and 4 b. -1 and 4 c. -3 and 1 d. -3 and 3 e. none

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Page 1: MCV4U - Practice Mastery Test #8 - PBworksmrblakely.pbworks.com/w/file/fetch/54332729/PMT8_MCV4U(2012).pdf · MCV4U - Practice Mastery Test #8 Multiple Choice Identify the choice

Name: ________________________ Class: ___________________ Date: __________ ID: A

1

MCV4U - Practice Mastery Test #8

Multiple ChoiceIdentify the choice that best completes the statement or answers the question.

____ 1. If x − 3( ) 2− x( )2 2x − 1( ) < 0 then

a. x > 3 or12

< x < 2 c.12

< x < 2 or 2 < x < 3

b. x < 12

or 2 < x < 3 or x > 3 d. x < 12

or12

< x < 2 or x > 3

____ 2. If f = {(2,4), (4,-3), (-3,2)} then f-1(-3) = a. 4 b. 2 c. -3 d. -2

____ 3. A function is defined by y = x − 3. Its domain is a. {x | x ≥ 0, x ∈ ò} c. {x | x ≥ 3, x ∈ ò}b. {x | x > 0, x ∈ ò} d. {x | x < 3, x ∈ ò}

____ 4. Which of the following limits do not exist?

i) limx → 3

+3− x ii) lim

x → 3 −3− x iii) lim

x → 23− x

a. i) only b. ii) only c. iii) only d. i) and iii)

____ 5. The expression limh → 0

x + h − xh

is most likely to be...

a. the derivative of a function c. slope of a secantb. the value of a derivative d. none of the above

____ 6. If y = x3 x − 2( ) + 3 then d yd x

=

a. 4x3 − 6x2 b. 4x3 − 2 c. −6x2 + 1 d. 3x2 − 4x + 3

____ 7. The number lines for y, d yd x

, and d2y

d x2 are shown below. All zeros are shown.

-4 -3 -2 -10 1 2 3 4 5

y - - - 0 - - - - - - - - - - - 0 + + + + +

d yd x

+ + + 0- - - - - - -

0 + + + + + + + + +

d2y

d x2- - -

0- - -

0 + + + + + + + + + 0- - -

The x-coordinates of all points of inflection are

a. -3, -1 and 4 b. -1 and 4 c. -3 and 1 d. -3 and 3 e. none

Page 2: MCV4U - Practice Mastery Test #8 - PBworksmrblakely.pbworks.com/w/file/fetch/54332729/PMT8_MCV4U(2012).pdf · MCV4U - Practice Mastery Test #8 Multiple Choice Identify the choice

Name: ________________________ ID: A

2

____ 8. The number lines for y, d yd x

, and d2y

d x2 are shown below. All zeros are shown.

-4 -3 -2 -10 1 2 3 4 5

y + + + + + + + 0 - - - - - - - - - - - - -

d yd x

- - -0

- - - - - - -0 + + + + + + + + +

d2y

d x2

+ + + 0- - -

0 + + + + + + + 0- - - - -

The x-coordinates of the absolute minimum is

a. 1 b. 3 c. -3 d. -1 e. none

____ 9. The number lines for y, d yd x

, and d2y

d x2 are shown below. All zeros are shown. Assume the function is

continuous for all x ∈ ò. “U” means undefined.

-4 -3 -2 -10 1 2 3 4 5

y + + + 0 + + + + + + + 0 - - - - - 0 + + +

d yd x

- - -0 + + + U

- - - - - - -0 + + + + +

d2y

d x2

+ + + 0 + + + U + + + + + + + + + + + + +

The x-coordinates of all zeros are

a. -3,1,3 and 4 b. --3 and 3 c. -3 and 4 d. -3, 1 and 4e. none

____ 10. If y = 3x2 + 1ÊËÁÁÁ

ˆ¯˜̃̃ x + 3( ) , then

d yd x

=

a. 6x( ) x + 3( ) + 1( ) 3x2 + 1ÊËÁÁÁ

ˆ¯˜̃̃ c. 6x( ) 1( ) + x + 3( ) 3x2 + 1Ê

ËÁÁÁ

ˆ¯˜̃̃

b. 6x( )(1) 3x2 + 1ÊËÁÁÁ

ˆ¯˜̃̃ x + 3( ) d. 6x + 1( ) 3x2 + 1Ê

ËÁÁÁ

ˆ¯˜̃̃ + x + 3( )

____ 11. If y = 3x + 1( ) x2 + 3ÊËÁÁÁ

ˆ¯˜̃̃ , then

d yd x

=

a. 3( ) 2x + 3( ) + 3x + 1( ) x2 + 3ÊËÁÁÁ

ˆ¯˜̃̃ c. 3( ) x2 + 3Ê

ËÁÁÁ

ˆ¯˜̃̃ + 2x + 3( ) 3x( )

b. 3( ) x2 + 3ÊËÁÁÁ

ˆ¯˜̃̃ + 2x( ) 3x + 1( ) d. 3( )(2x) x2 + 3Ê

ËÁÁÁ

ˆ¯˜̃̃ 3x + 1( )

Page 3: MCV4U - Practice Mastery Test #8 - PBworksmrblakely.pbworks.com/w/file/fetch/54332729/PMT8_MCV4U(2012).pdf · MCV4U - Practice Mastery Test #8 Multiple Choice Identify the choice

Name: ________________________ ID: A

3

____ 12. Given f(x) = x − 4 and g(x) = x2 .If k(x) = f û g(x), then the domain of y=k(x) is ...

a. x| x ≥ 2 or x ≤ −2, x ∈ òÏÌÓ

ÔÔÔÔ

¸˝˛

ÔÔÔÔ c. x| − 2 ≤ x ≤ 2 ,x ∈ ò

ÏÌÓ

ÔÔÔÔ

¸˝˛

ÔÔÔÔ

b. x| x ≥ 2, x ∈ òÏÌÓ

ÔÔÔÔ

¸˝˛

ÔÔÔÔ d. x| x ≤ 2 , x ∈ ò

ÏÌÓ

ÔÔÔÔ

¸˝˛

ÔÔÔÔ

____ 13. If y = − 2x3 + 1ÊËÁÁÁ

ˆ¯˜̃̃

−2

, then d yd x

=

a. 3 2x3 + 1ÊËÁÁÁ

ˆ¯˜̃̃

−1

6x2 + 1ÊËÁÁÁ

ˆ¯˜̃̃ c. 2 2x3 + 1Ê

ËÁÁÁ

ˆ¯˜̃̃

−1

6x2 + 1ÊËÁÁÁ

ˆ¯˜̃̃

b. 3 2x3 + 1ÊËÁÁÁ

ˆ¯˜̃̃

−1

6x2ÊËÁÁÁ

ˆ¯˜̃̃ d. 2 2x3 + 1Ê

ËÁÁÁ

ˆ¯˜̃̃

−3

6x2ÊËÁÁÁ

ˆ¯˜̃̃

____ 14. sin(7)a. sin(6) + sin(1) c. sin(6) sin(1) + cos(1) cos(6)b. sin(6) cos(1) + sin(1) cos(6) d. sin(6) cos(1) − sin(1) cos(6)

____ 15. If y = sin 3x2 + 2ÊËÁÁÁ

ˆ¯˜̃̃ then

a.d yd x

= 6x cos 3x2 + 2ÊËÁÁÁ

ˆ¯˜̃̃

b.d yd x

= −6x sin 3x2 + 2ÊËÁÁÁ

ˆ¯˜̃̃

c.d yd x

= 3x2 + 2ÊËÁÁÁ

ˆ¯˜̃̃ cos 6x( ) + 6x sin 3x2 + 2Ê

ËÁÁÁ

ˆ¯˜̃̃

d.d yd x

= 3x2 + 2ÊËÁÁÁ

ˆ¯˜̃̃ sin 6x( ) − 6x cos 3x2 + 2Ê

ËÁÁÁ

ˆ¯˜̃̃

____ 16. If y = cos 4x2 + 3ÊËÁÁÁ

ˆ¯˜̃̃ then

a.d yd x

= 8x cos 4x2 + 3ÊËÁÁÁ

ˆ¯˜̃̃

b.d yd x

= −8x sin 4x2 + 3ÊËÁÁÁ

ˆ¯˜̃̃

c.d yd x

= 4x2 + 3ÊËÁÁÁ

ˆ¯˜̃̃ cos 8x( ) + 8x sin 4x2 + 3Ê

ËÁÁÁ

ˆ¯˜̃̃

d.d yd x

= 4x2 + 3ÊËÁÁÁ

ˆ¯˜̃̃ sin 8x( ) − 8x cos 4x2 + 3Ê

ËÁÁÁ

ˆ¯˜̃̃

____ 17. Evaluate −52

a. − 1

25 b. 25 c.1

25 d. -25

____ 18. If y = ln(2x − 3), then d yd x

= (It is not necessary to state restrictions)

a.2

2x − 3b.

12x − 3

c.ln(2x − 3)

2x − 3d.

1ln(2x − 3)

Page 4: MCV4U - Practice Mastery Test #8 - PBworksmrblakely.pbworks.com/w/file/fetch/54332729/PMT8_MCV4U(2012).pdf · MCV4U - Practice Mastery Test #8 Multiple Choice Identify the choice

Name: ________________________ ID: A

4

____ 19. If y = 5e2x + 1, then d yd x

= (It is not necessary to state restrictions)

a. 10e2x + 1 b. 10xe x2

c. 2e2x + 1 d. 10e2x

____ 20. Given vector a→

as shown ... then 23

a→

is which vector below ?

a. b. c. d.

____ 21. Given the following vectors...

then −a→

− b→

is illustrated by which c→

below ?

a. b. c. d.

Page 5: MCV4U - Practice Mastery Test #8 - PBworksmrblakely.pbworks.com/w/file/fetch/54332729/PMT8_MCV4U(2012).pdf · MCV4U - Practice Mastery Test #8 Multiple Choice Identify the choice

Name: ________________________ ID: A

5

____ 22.

In the prism shown, DE→

is equal to...

a. −a→

+ b→

+ c→

c. −a→

− b→

− c→

b. −a→

+ b→

− c→

d. −a→

− b→

+ c→

____ 23.

In the prism shown, CH→

is equal to...

a. −a→

+ b→

+ c→

c. −a→

− b→

− c→

b. −a→

+ b→

− c→

d. −a→

− b→

+ c→

____ 24. Given a→

= (−1,−2,4) then a→|

||||| =

a. 21 c. 7b. 1 d. 21

Page 6: MCV4U - Practice Mastery Test #8 - PBworksmrblakely.pbworks.com/w/file/fetch/54332729/PMT8_MCV4U(2012).pdf · MCV4U - Practice Mastery Test #8 Multiple Choice Identify the choice

Name: ________________________ ID: A

6

____ 25. Given the following vectors where aä| | = 10 and bä|||| = 9 then a

→•b

→is approximately

a. 77.9 b. 45 c. 72.7 d. 39.8

____ 26. Given the following vectors where aä| | = 6 and bä|||| = 8 then a

→× b

→|||

||| is approximately

a. 39.3 b. 27.5 c. 36.9 d. 32.3

____ 27.

In the prism shown, a→|

||||| = b

→|||

||| = c

→|||

||| = 1

c→

× c→

is equal to...

a. 0→

b. BA→

c. AD→

d. CB→

Page 7: MCV4U - Practice Mastery Test #8 - PBworksmrblakely.pbworks.com/w/file/fetch/54332729/PMT8_MCV4U(2012).pdf · MCV4U - Practice Mastery Test #8 Multiple Choice Identify the choice

Name: ________________________ ID: A

7

____ 28.

In the prism shown, a→|

||||| = b

→|||

||| = c

→|||

||| = 1

c→

× b→

is equal to...

a. GH→

b. DG→

c. 0→

d. HE→

____ 29.

In the prism shown, a→|

||||| = b

→|||

||| = c

→|||

||| = 1

b→

× a→

is equal to...

a. GD→

b. DC→

c. 0→

d. EF→

____ 30. Given a→

= (2,1,3) and b→

= (1,2,−1) then a→

× b→

is equal to

a. −7,5,3ÊËÁÁ ˆ

¯˜̃ c. 3,−7,−5Ê

ËÁÁ ˆ

¯˜̃

b. 5,1,5ÊËÁÁ ˆ

¯˜̃ d. 1,−2,0Ê

ËÁÁ ˆ

¯˜̃

Page 8: MCV4U - Practice Mastery Test #8 - PBworksmrblakely.pbworks.com/w/file/fetch/54332729/PMT8_MCV4U(2012).pdf · MCV4U - Practice Mastery Test #8 Multiple Choice Identify the choice

ID: A

1

MCV4U - Practice Mastery Test #8Answer Section

MULTIPLE CHOICE

1. ANS: C

12 2 3

x − 3( ) - - - - - - - - - - - - - - - - 0 + + + +

2− x( )2 + + + + + + + + + + 0 + + + + + + + + + +

2x − 1( ) - - - 0 + + + + + + + + + + + + + + + + +

x − 3( ) 2− x( )2 2x − 1( ) + + + 0 - - - - - - 0 - - - - - 0 + + + +

So... x − 3( ) 2− x( )2 2x − 1( ) < 0 if

12

< x < 2 or 2 < x < 3

PTS: 1 2. ANS: A

The ordered pair (4,-3) shows us that f takes 4 and produces -3. Therefore, f-1 takes -3 and produces 4. I.e., f-1(-3) = 4.

PTS: 1 3. ANS: A

In the real numbers, x is defined only if x ≥ 0, so the domain is all real numbers that are greater or equal to 0.

PTS: 1

Page 9: MCV4U - Practice Mastery Test #8 - PBworksmrblakely.pbworks.com/w/file/fetch/54332729/PMT8_MCV4U(2012).pdf · MCV4U - Practice Mastery Test #8 Multiple Choice Identify the choice

ID: A

2

4. ANS: A

As x → 3+, x is greater than 3, so 3− x will negative, so 3− x will be undefined, so limx → 3

+3− x does not

exist.

As x → 3−, x is less than 3, so 3− x will positive, so 3− x will be defined and will approach 0, so

limx → 3 −

3− x = 0

As x → 2, from either side, x is less than 3, so 3− x will positive, 3− x will be defined, and will approach

1, so limx → 3 −

3− x = 1

PTS: 1 5. ANS: A

If f(x) = x , then the expression is of the form limh → 0

f(x + h) − f(x)h

which is the derivative.

PTS: 1 6. ANS: A

if y = x3 x − 2( ) + 3 then

y = x4 − 2x3 + 3 so d yd x

= 4x3 − 6x2

PTS: 1 7. ANS: B

At a point of inflection, the second derivative will change sign, (it changes from concave up/down to concave down/up), so they are at x=-1 and x=4.

PTS: 1 8. ANS: A

To the left of x=1, the function is decreasing or stationary (the derivative is negative or zero), to the right of x=1, the function is increasing (derivative is positive), so there is an absolute minimum at x=1.

PTS: 1 9. ANS: D

For a zero, the y coordinate is 0, so the zeros are at -3, 1 and 4

PTS: 1

Page 10: MCV4U - Practice Mastery Test #8 - PBworksmrblakely.pbworks.com/w/file/fetch/54332729/PMT8_MCV4U(2012).pdf · MCV4U - Practice Mastery Test #8 Multiple Choice Identify the choice

ID: A

3

10. ANS: A

y = 3x2 + 1ÊËÁÁÁ

ˆ¯˜̃̃ x + 3( ) , so

d yd x

= ddx

3x2 + 1ÊËÁÁÁ

ˆ¯˜̃̃

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃ x + 3( ) + d

dxx + 3( )

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃ 3x2 + 1ÊËÁÁÁ

ˆ¯˜̃̃

= 6x( ) x + 3( ) + 1( ) 3x2 + 1ÊËÁÁÁ

ˆ¯˜̃̃

PTS: 1 11. ANS: B

y = 3x + 1( ) x2 + 3ÊËÁÁÁ

ˆ¯˜̃̃ , so

d yd x

= ddx

3x + 1( )Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃ x2 + 3ÊËÁÁÁ

ˆ¯˜̃̃ + d

dxx2 + 3ÊËÁÁÁ

ˆ¯˜̃̃

Ê

Ë

ÁÁÁÁÁÁ

ˆ

¯

˜̃˜̃˜̃ 3x + 1( )

= 3( ) x2 + 3ÊËÁÁÁ

ˆ¯˜̃̃ + 2x( ) 3x + 1( )

PTS: 1 12. ANS: A

k(x) = f û g(x) , then k(x) = f g(x)ÊËÁÁ ˆ

¯˜̃

so k(x) = f g(x)ÊËÁÁ ˆ

¯˜̃

= f x2ÊËÁÁÁ

ˆ¯˜̃̃

= x2 − 4

For k(x) to be defined, x2 − 4 ≥ 0

x2 ≥ 4

x ≥ 2 or x ≤ −2

The domain of y=k(x) is x| x ≥ 2 or x ≤ −2 , x ∈ òÏÌÓ

ÔÔÔÔ

¸˝˛

ÔÔÔÔ

PTS: 1 13. ANS: D

y = − 2x3 + 1ÊËÁÁÁ

ˆ¯˜̃̃

−2

, so

d yd x

=d − 2x3 + 1Ê

ËÁÁÁ

ˆ¯˜̃̃

−2

d 2x3 + 1ÊËÁÁÁ

ˆ¯˜̃̃

d 2x3 + 1ÊËÁÁÁ

ˆ¯˜̃̃

d x

= 2 2x3 + 1ÊËÁÁÁ

ˆ¯˜̃̃

−3

6x2ÊËÁÁÁ

ˆ¯˜̃̃

PTS: 1 14. ANS: B PTS: 1

Page 11: MCV4U - Practice Mastery Test #8 - PBworksmrblakely.pbworks.com/w/file/fetch/54332729/PMT8_MCV4U(2012).pdf · MCV4U - Practice Mastery Test #8 Multiple Choice Identify the choice

ID: A

4

15. ANS: A

y = sin 3x2 + 2ÊËÁÁÁ

ˆ¯˜̃̃ , so

d yd x

=d sin 3x2 + 2Ê

ËÁÁÁ

ˆ¯˜̃̃

d x

=d sin 3x2 + 2Ê

ËÁÁÁ

ˆ¯˜̃̃

d 3x2 + 2ÊËÁÁÁ

ˆ¯˜̃̃

d 3x2 + 2d x

(chain rule)

= cos 3x2 + 2ÊËÁÁÁ

ˆ¯˜̃̃6x

= 6x cos 3x2 + 2ÊËÁÁÁ

ˆ¯˜̃̃

PTS: 1 16. ANS: B

y = cos 4x2 + 3ÊËÁÁÁ

ˆ¯˜̃̃ , so

d yd x

=d cos 4x2 + 3Ê

ËÁÁÁ

ˆ¯˜̃̃

d x

=d cos 4x2 + 3Ê

ËÁÁÁ

ˆ¯˜̃̃

d 4x2 + 3ÊËÁÁÁ

ˆ¯˜̃̃

d 4x2 + 3d x

(chain rule)

= −sin 4x2 + 3ÊËÁÁÁ

ˆ¯˜̃̃8x

= −8x sin 4x2 + 3ÊËÁÁÁ

ˆ¯˜̃̃

PTS: 1 17. ANS: D

− 52

= −25

Note that order of operations requires you to apply the exponent to the 5 BEFORE the negative is

applied.

PTS: 1 18. ANS: A

d yd x

=d ln(2x − 3)ÊËÁÁ ˆ

¯˜̃

d (2x − 3)d 2x − 3( )

d x

= 12x − 3

2

= 22x − 3

PTS: 1

Page 12: MCV4U - Practice Mastery Test #8 - PBworksmrblakely.pbworks.com/w/file/fetch/54332729/PMT8_MCV4U(2012).pdf · MCV4U - Practice Mastery Test #8 Multiple Choice Identify the choice

ID: A

5

19. ANS: A

d yd x

= 5d e2x + 1ÊËÁÁÁ

ˆ¯˜̃̃

d (2x + 1)d 2x + 1( )

d x

= 5e2x + 12

= 10e2x + 1

PTS: 1 20. ANS: A

The given vector a→

is to be multiplied by 23

, so it will be in the same direction as a→

(because 23

is positive)

and it will be smaller than a→

because 23

is smaller than 1.

PTS: 1 21. ANS: C

To draw −a→

− b→

, draw vector −a→

and then draw vector −b→

with its tail attached to the head of −a→

. Draw

the vector from the tail of −a→

to the head of −b→

.

This is −a→

− b→

.

PTS: 1

Page 13: MCV4U - Practice Mastery Test #8 - PBworksmrblakely.pbworks.com/w/file/fetch/54332729/PMT8_MCV4U(2012).pdf · MCV4U - Practice Mastery Test #8 Multiple Choice Identify the choice

ID: A

6

22. ANS: AIn the prism shown, to get from D to E, follow a path along the edges. Go from D to A, then A to B, and then B to E, so...

DE→

= DA→

+ AB→

+ BE→

= −a→

+ b→

+ c→

because DA→

is in the opposite direction to a→

PTS: 1 23. ANS: D

In the prism shown, to get from C to H, follow a path along the edges. Go from C to B, from B to A and then A to H, so...

CH→

= CB→

+ BA→

+ AH→

= −a→

− b→

+ c→

because

CB→

is in the opposite direction to a→

, and

BA→

is in the opposite direction to b→

.

PTS: 1

Page 14: MCV4U - Practice Mastery Test #8 - PBworksmrblakely.pbworks.com/w/file/fetch/54332729/PMT8_MCV4U(2012).pdf · MCV4U - Practice Mastery Test #8 Multiple Choice Identify the choice

ID: A

7

24. ANS: AStarting at the origin, the 1st component of the vector comes out towards us (if positive) or goes back into the page (if negative). In this case the –1 is shown in green.The 2nd component of the vector moves left or right from there. In this case the –2 is shown in blue.

These two vectors form a right triangle with hypotenuse of length −1( )2 + −2( )

2

When we move up or down using the 3rd component, the former hypotenuse forms a right triangle with the red vector.

The new hypotenuse has length −1( )2 + −2( )

2 + 4( )2

= 21

So the length of the vector in simplest form is 21

PTS: 1 25. ANS: A

Because we know the angle between the vectors, to find a→

•b→

, use the formula ...

a→

•b→

= a→|

||||| b

→|||

||| cos(θ)

= (10)(9) cos(30°)≈ 77.9

PTS: 1

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ID: A

8

26. ANS: B

Because we know the angle between the vectors, to find a→

× b→|

||||| use the formula ...

a→

× b→|

|||||

= a→|

||||| b

→|||

|||sin(θ)

= (6)(8) sin(35°)≈ 27.5

PTS: 1

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ID: A

9

27. ANS: AIf you use the “right hand rule”,

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ID: A

10

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ID: A

11

you will see that

a→

× b→

= c→

b→

× c→

= a→

c→

× a→

= b→

If you try the following, your finger is pointing in the opposite direction of a, b, or c.

b→

× a→

= − c→

c→

× a→

= − a→

a→

× c→

= − b→

When you take the cross product of a vector with itself, the angle between the vectors is 0, so if you use the

formula for magnitude (length) you get... a→

× a→|

||||| = a

→|||

||| a

→|||

||| sin(0°)

= a→|

||||| a

→|||

|||(0)

= 0and if the length is 0, then the result is the 0 vector.

So...

a→

× a→

= 0→

b→

× b→

= 0→

c→

× c→

= 0→

In this case, c→

× c→

= 0→

which is equal to 0→

(same direction & length)

PTS: 1

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ID: A

12

28. ANS: AIf you use the “right hand rule”,

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ID: A

13

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ID: A

14

you will see that

a→

× b→

= c→

b→

× c→

= a→

c→

× a→

= b→

If you try the following, your finger is pointing in the opposite direction of a, b, or c.

b→

× a→

= − c→

c→

× a→

= − a→

a→

× c→

= − b→

When you take the cross product of a vector with itself, the angle between the vectors is 0, so if you use the

formula for magnitude (length) you get... a→

× a→|

||||| = a

→|||

||| a

→|||

||| sin(0°)

= a→|

||||| a

→|||

|||(0)

= 0and if the length is 0, then the result is the 0 vector.

So...

a→

× a→

= 0→

b→

× b→

= 0→

c→

× c→

= 0→

In this case, c→

× b→

= 0→

which is equal to GH→

(same direction & length)

PTS: 1

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ID: A

15

29. ANS: AIf you use the “right hand rule”,

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ID: A

16

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ID: A

17

you will see that

a→

× b→

= c→

b→

× c→

= a→

c→

× a→

= b→

If you try the following, your finger is pointing in the opposite direction of a, b, or c.

b→

× a→

= − c→

c→

× a→

= − a→

a→

× c→

= − b→

When you take the cross product of a vector with itself, the angle between the vectors is 0, so if you use the

formula for magnitude (length) you get... a→

× a→|

||||| = a

→|||

||| a

→|||

||| sin(0°)

= a→|

||||| a

→|||

|||(0)

= 0and if the length is 0, then the result is the 0 vector.

So...

a→

× a→

= 0→

b→

× b→

= 0→

c→

× c→

= 0→

In this case, b→

× a→

= 0→

which is equal to GD→

(same direction & length)

PTS: 1

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ID: A

18

30. ANS: ATo find the cross product, write the components of the vector as shown - the 1st vector in the top row (yes - this matters!) repeating the 1st component on the right. To determine the 1st component of the answer, cover over the first (and last) columns. In the remaining square of four numbers “multiply down to the right and subtract the ‘other’ product”, to get (1)(−1) - (2)(3) =–7To determine the 2nd component of the answer, cover over the second column. In the remaining square of four numbers “ multiply down to the right and subtract the ‘other’ product”, to get (3)(1)− (−1)(2)=5To determine the 3rd component of the answer, cover over the third column. In the remaining square of four numbers “multiply down to the right and subtract the ‘other’ product”, to get (2)(2)− (1)(1)=3Now ‘collect’ the components in the answer:

a→

× b→

=(–7, 5, 3)

It is a good idea to check to make sure your answer is perpendicular to the given vectors by taking the dot

product of (–7, 5, 3) with each of a→

= (2,1,3) and b→

= (1,2,−1) . (The dot product should be 0)

PTS: 1