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Practice Questions for Exam 1 Math 231 1. Given the position function of an object � � 3 26 sinr t t t t � �i j k with 0 t Sd d find the

velocity and acceleration vectors. 2. Let 3, 2, 1 and 5, 3, 5u v � � � �

a. Find a unit vector in the direction of u

b. Find u v� .

c. Find u vu .

d. Find the projection of v onto u .

e. Find the angle between u and v .

f. Find a vector parallel to v with length 5 units.

g. Compute the lengths of the diagonals of the parallelogram determined by u and v .

F = < t3, ti Sint >

TH )=<3t2,

2t . Cost ?

ATT) = 4 bt,

2,

- Sint >

Kull = Et =F4 ¥3 ,-2

,-17 = site ,÷r , ,r÷ >

Tt - I ÷ 3 ( 5 ) t t2X - 3) t L - 1k - 5) = 15+6+5 = 26

Text .. |¥gIg[§f- Tho - 3) - jl - ist 5) tkt 9+101=7 it Njth

¥lf Pro j¥(Y¥u) tea,

= 2,4<3 ,

-2¥It 3,

-2,

-17

cos of,oj÷m=r!6#⇒ a- cos 'E¥a)

FFH,-3

, -57= seer , ?¥ , -2¥ >

http = <3 , -2 . -1 >t < 5. - 3,

-57 = ( 8,

-5,

-6 >

FETE .÷f*.to - 8 = <3

,-2 .

- 1 > + < - 5.

3,

57 = ( -2,

1, 4 >

2

h. Find a vector orthogonal to both u and v .

i. Compute the area of the parallelogram determined by u and v .

2. Find a vector equation of the line that passes through the points � �2,6, 1� and � �6,4,0� . 3. Find the parametric equations of the line that passes through the point � �0,1,1 and parallel

to the vector 2, 1,3� . 4. Find a vector equation of the line that passes through � �2,1,0 and perpendicular to both

�i j and �j k . 5. Find the equation of the plane through � �1,6, 5A � � and parallel to the plane

2 0x y z� � � .

Text .. |¥s§g[§|=tln - 3) - jH5t5)tkt9tio)=7itNjtk

115×711=49+100+1 =f5o

7=(8,

2,

- 1 > Plt )=( 2,6 , -17++28,2 , -17 tell

21=2-6, Y=l . t z=1t3t TER

/ }ftp./=i(ho1.jll.o)+ILi.o )MH=h21o7tt< I ,

. hotter

=< 1

, -1,17

T=< 1,1 ,1 > < 1.1 ,

1 >.< Xtl , Y -6,3+57=0

→ Xtlty -6+3+5=0⇒ x+y+z=o

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6. Find an equation of the plane through � �2,3,1P � , � �1,1,0Q , and � �1,0,1R � . 7. Find the equation of the sphere with center � �2, 3,4� and tangent to the xz-plane. 8. Find the equation of the sphere with the line segment � �1,5, 2� and � �3,9, 1� is the

diameter. 9. Find the distance between the points � �1,2, 3P � and � �4,7, 3Q � . 10. Find an equation of the line of intersection of the two plane 5 3y z� and 6 7 5x y� .

11. Find an equation of the line tangent to the curve � � sin2 ,cos2 ,r t t t t at 4

t S

POT = <3,

-2, -17 Plane : - 1<3

, 1,77 . < xt2, y -3

, z - 17=0

FR= < 1,

-3, 0 > ⇒ 31×+2 ) +1g - 3) +713 . 1 )=O

T = FQXPTZ =L - 3, -1

,-7 > ⇒

3x+y+7z=4|Xz - plane is yso Sphere :

radius is how for from pt 12.

-3,4 ) &-25 Hyt3Pt&-412=9to plane y=o which is 3 units .

Center is midpoint of the diameter radius is £ distance of diameter .

x=lEt=2 , g- 5+21=7 z=-¥=z Efta.se#tFEJiE( 2,7

,-312 )

-

Sphere :( x -32 Hy - 75+(2+312) ? 214

P#s3 , 5,0 > llpglk -9+25=534

DFCO ,1

,-57 NIA , -7,0 >

X= 's , y=3 , 3=0 ← pt ofintersection

T=M×M=/bt.tt/z.3s ,

. so , -6 > Line t.TL's6 -7 0 rlt )=ltE , 3,07++4-35

,-30 , -67 te*R

F 'H=s2cos2t ,- 2sm2t ,

1 > FLIY )=< 1

, 0,447P 'M4)=l2cos7z .

- 2sintk,

1 >

=L 0 , -2 , 17Line

Ltt ( 1,0, '7y7tt( 0, -2,17

planes

6×-79=5& y - 52=3

Find pt of intersection !64-74+03=5+

0×+7y -353=21

64 - 353=26 ← let z=o

6X=26⇒X=133

& if 3=0 y - 5z=3⇒y=3

Check 6( 113 ) -713 )=5 -

pt of intersection 113/3,310 )

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12. Determine the distance from the point � �6,3,0P � to the line given by

� � 3 4 , 2 6 ,r t t t t � � � . 13. Show that the lines with equation

1 3 22 4 5x y z� � �

�

and 4 13 2 3x y z� �

are skew, fine the equations of the parallel planes containing the lines, and find the distance between the lines.

14. Let ( ) 2 , ( 1)/ , ln( 1)tt t e t t � � �r .

a. Find the domain of r. b. Find

0lim ( )tr t

o

AT =L - 9,5107 RI×T=< 5, +9

,-747

T =L 4,6 ,I )

p*ua*o' e ¥ ,=sm0⇒l=" RP " " " now " ?[YT¥mw⇒m•

1°'

so

l=HFPxTH=T5t8E7÷N3'

' %),<4.6.17

" T " 16M¥

VT=< 2 , -4,57 Since Ntafkvs the lines are not parallel .

Tz=L 3 , 2,37

× : 24=35+4 } soeuefrtes HE§st,€ -41355+4+3=25-1

y! -4++3=251 - 6g - 10+3=254

z : 5++2=35- 85=-6

t.se#tI 5- 618=314

check 51¥ )+2=3l±, ) =±s+2F=2§'¥+2 FF ; . The 21in do not intersect & aresklw

=

see next page forthe rest of the solution

A) Ft ← ts2

ettanyredexaptto }DMdnCfoYutto, ,]

,

lnlttl ) th

Dltimortthtltsmrtibimoetftikinenuttsurnio >

T,

= < 2 , -4,5 > NTT , XJz=< - 22,9 ,167

Tz=L 3 , 2,37

X : 2+-1=35+4 RI -1,3 , 2) BC 4,

-1,

0 )

y! -4++3=2+1

z : 5++2=35

Plhmslgglyt, )+q1y . 31+1612-3=0

-221 X - 4) +91 YH) +162=0

Distance Between 2 planes

ftp. - < £, -4 ,-÷

T.PH/=.//o+t36)t=Oh.

tea

= 178

* ,

5

15. Given the points A(1, 0 ,1), B(2, 3, 0) and C(-1, 1, 4), find the area of the triangle with vertices A, B, and C.

16. Find the length of the curve 3/2r( ) 2 , cos2 , sin2t t t t for 0 1td d . 17. Find the distance from the point � �2,0, 3� to the plane � � 3 4 5 1x y z . 18. A particle moves in space with parametric equations 2 3/24

3, ,x t y t z t . Find the curvature of its trajectory, the unit tangent vector, and the unit normal vector when t = 1.

ATB --< 1,3 , -1 > Area of triangleAT=f2 , 1,37

ATBXST - < 10 , -1,7 > £ IIAAXFC 11=12 Aso

Pkt )=<3t 'M - 2sin2t ,2cos2t7

HF 'HH=ft+4nztT4os¥9tt4tL=f' rated .at . :( 9++4134.1=3+[13%-8]

Defi Tt . T = HKHHPIICOSO o

plzoidpD=h3i -4157so l= IN.RPI =

15+0+451In # -R - l - HRTHWSO 9+16+25

RLBPR)

EP=s%,

0,3 >=

'¥o=sF2=rF:BFta ,2t ,

st't > FLH=+÷⇒< 1,242T

' 's >

HF 'll - Fett= 1+2t

FU)= fll , 2,27=(5,2-3,25)

Ill )=< ta , ÷ ,

- ta >

d£¥ -111+2+534<1 ,2t ,H "27t -

µ÷+< 0,2 ,E' ' 2> Etkin

h÷tITi" " ' ttso . } , >

= ' T ,

sits>

K(D=1lT 'll )H= ' '±a' to 't >+a5Y§Y=s } , } , } ,

FiF¥=ta

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19. Find the binormal vector and osculating plane for 2 3/243, ,x t y t z t at 1t .

20. Find the distance from the point � �6,3,0� to the line given by � � 1 5 , 6 , 4r t t t t � � � � 21. A particle starts at the origin with initial velocity 3� �i j k . Its acceleration is

2( ) 6 12 6t t t t � �a i j k . Find its position function. 22. Find the tangential and normal components of acceleration for a particle moving along a

conical helix defined by � � cos , sin ,r t t t t t t .

Pll ,1,413)

D= txn

A=<-÷,

. t , ; >

Fists ' } '±z >

ox ,

plane

P=G÷i5i's >

-

%←, ).Bly.

t+%lz-4131=0 .

atuiineurp" "r• " ftp.#l / :{tgteyf;47 &=48#⇐± = ( -36 ,

-28,

-52g.gg,dAIh5'"

Fp=< -7,907 25+1+16= -4<9,7 ,

137

-LEHJAEHH Flt )=fTHa, .

=(3E+C , ,4t3tCz ,-3EtC

,> =< ettttx,,t4tHX,

. EHTTB

VIO )=< C , ,cz ,↳y=< i, -1,3 >

PH 't

VltI=< 3+2+1,4+3.1 ,-34+3 >

Flt )=Ct3tt,t4t ,- ETH

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23. The helix 1( ) cos sint t t t � �r i j k intersects the curve 2 32( ) (1 )t t t t � � �r i j k at

point (1, 0, 0). Find the angle of intersection of these curves.

For additional problems, check out the review problems for Chapters 10 and 11. Note the questions above are simply a sample of questions possible for the exam; it is possible that other types of questions may appear on your exam.

HE fsint ,costa > Pilot Calin coso=Fi'rI

Ehl ,2t

,3+27 Filo ) =L 1,0107 1151111% 'll

rTo#= O Thus the angle of intersection is Is,

Cos O=E÷EHHTH