Homework Help 3D vectors problem solving

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Homework Help!! 3D vectors problem solving!!

Homework Statement


the vector v=(1,√2,c) makes an angle of 60° with the positive z-axis. Determine the angles that v makes with the positive x-axis and the positive y-axis. Explain how many answers there are.


Homework Equations


dot product, geometry, cosine,sine,tan


The Attempt at a Solution


I started by drawing the 3D vector with the x,y,z axis's and place a point on the z axis for c. I then tried using geometry with the 60 degrees and the drawn vector to find other angles and also used cosine, sine and tan with the values 1 and root 2. None of the methods seemed to work and am very confused as to what I should do next.

Homework Statement





Homework Equations





The Attempt at a Solution


Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
tiny-tim
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welcome to pf!

hi 16leafs! welcome to pf! :smile:
the vector v=(1,√2,c) makes an angle of 60° with the positive z-axis.

the angle between (1,√2,c) and (0,0,1) is 60°

so use the formula for the angle between two vectors! :wink:
 
  • #3
12,811
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start with the dot product of z dot v = |z| |v| cos (60) and recall that z is a unit vector and that cos(60) = 1/2 then solve for c

Now can you see how many solutions for c there from the |z| |v| cos(60) equation?
 
  • #4
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thanks for the replies, I really appreciate the help. I'm still confused as to why you gave c a value of unit vector one though?
 
  • #5
Dick
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thanks for the replies, I really appreciate the help. I'm still confused as to why you gave c a value of unit vector one though?

You can choose z to be any vector along the positive z axis. It doesn't really matter, but picking z=(0,0,1) would be the simplest choice, yes?
 
  • #6
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1


The value for c would actually be different if you chose the vector [0,0,2] as for the positive z-axis. However, I believe the angles would all be the same.
 
  • #7
Dick
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The value for c would actually be different if you chose the vector [0,0,2] as for the positive z-axis. However, I believe the angles would all be the same.

Nah, if you put z=[0,0,2] in then you get 2c in the dot product. But |z|=2. They'll just cancel out.
 
  • #8
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ok (0,0,1) would seem the simplest, but how would i use the dot product method to solve for c because when i tried using (0,0,1), and then (0,0,2) i got different answers?
 
  • #9
133
1


Nah, if you put z=[0,0,2] in then you get 2c in the dot product. But |z|=2. They'll just cancel out.

Yup, you're right. I was doing sqrt(2) for the magnitude for some reason.

@16leafs, you solve for c by doing

##[1, \sqrt{2},c] \cdot [0,0,1] = ||[1, \sqrt{2}, c] || \cdot ||[0,0,1]|| \cos{60}##
 
  • #10
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When i simplify the equation, the c is cancelled out and im left with no variable?
 
  • #11
Dick
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When i simplify the equation, the c is cancelled out and im left with no variable?

That's wrong. Show your work.
 
  • #12
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[1,2√,c]⋅[0,0,1]= (1)(0)+(2√)(0)+(c)(1)= c

lzl=√1^2= 1

lvl= √(1^2+√2^2+c^2)= (√3)c

c=(√3)c*1*cos60

Im using equations we have used in class but it doesnt seem to work with c variable on both sides
 
  • #13
Dick
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[1,2√,c]⋅[0,0,1]= (1)(0)+(2√)(0)+(c)(1)= c

lzl=√1^2= 1

lvl= √(1^2+√2^2+c^2)= (√3)c

c=(√3)c*1*cos60

Im using equations we have used in class but it doesnt seem to work with c variable on both sides

sqrt(1+2+c^2)=sqrt(3+c^2). That isn't equal to sqrt(3)*c.
 
  • #14
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so its √3+c? giving me c=(√3+c)*1*cos60 which c's will still cancel?
 
  • #15
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sorry they dont cancel. so c=√3?
 
  • #16
Dick
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so its √3+c? giving me c=(√3+c)*1*cos60 which c's will still cancel?

No offense, but your algebra is atrocious. You get c=sqrt(3+c^2)*1*cos(60). Figure out what cos(60) is and try to solve for c.
 
  • #17
133
1


so its √3+c? giving me c=(√3+c)*1*cos60 which c's will still cancel?

No, √(3 + c^2) ≠ √3 + c. Just leave it as √(3 + c^2) since you cannot simplify it any more. From here, trying squaring both sides.
 
  • #18
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0


c^2=3/2+0.5c^2

0.5c^2=3/2

c=√3?
 
  • #19
133
1


No, that's not right. When I said square both sides, I didn't say to ignore the 1/2 that was from the cos(60). You still need to square that. Come on!
 

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