• Support PF! Buy your school textbooks, materials and every day products Here!

Vector problems

  • Thread starter Tensaiga
  • Start date
  • #1
43
0
hello, i have a few theories questions to ask. (i don't know where to start for question such as these...)

Question: By considering the angles between the vectors, show that vector A + vector B and vector A - vector B are perpendicular when |A| = |B|.

Question: Prove for any vectors A and B, --->
that |A+B|^2 + |A-B|^2 = 2(|A|^2 +|B|^2)

Question: Three forces of 5N , 7N, 8N, are applied to an object. If the object is in a state of equillibrium, show how must the forces be arranged.

also i wonder why is zero vector's direction is undefined? is it because there is no magnitude?

Thank You
 
Last edited:

Answers and Replies

  • #2
andrevdh
Homework Helper
2,128
116
For your last question the three forces need to form a closed triangle. Of cause the triangle can be rotated in any direction, which means that the direction of the forces are not uniquely defined, but the angles that the vectors make with one another are fixed. Also note that by changing the order of adding the three vectors will produce two triangles that are mirror images of each other.
 
  • #3
14
0
third question: the forces must be arranged in a way that the sum of them is the zero vector.
 
  • #4
14
0
hey :)

to your first question:

[tex]
( \vec A + \vec B ) \cdot (\vec A - \vec B) = (a_1+b_1)*(a_1-b_1) + (a_2+b_2)*(a_2-b_2) + (a_3+b_3)*(a_3-b_3)\\

= a_1^2-b_1^2 + a_2^2 - b_2^2 + a_3^2 - b_3^2\\
= a_1^2+a_2^2+a_3^2 - (b_1^2+b_2^2+b_3^2)
[/tex]

if perpendicular, this is supposed to be 0, so

[tex]
a_1^2+a_2^2+a_3^2 - (b_1^2+b_2^2+b_3^2) = 0\\
a_1^2+a_2^2+a_3^2 = b_1^2+b_2^2+b_3^2\\
\sqrt{a_1^2+a_2^2+a_3^2} = \sqrt{b_1^2+b_2^2+b_3^2}\\
\Leftrightarrow |\vec A| = |\vec B|
[/tex]
 
Last edited:
  • #5
14
0
sorry, there's been linebreaks missing, so here again:

hey :)

to your first question:

[tex]
( \vec A + \vec B ) \cdot (\vec A - \vec B) = (a_1+b_1)*(a_1-b_1) + (a_2+b_2)*(a_2-b_2) + (a_3+b_3)*(a_3-b_3)\\

= a_1^2-b_1^2 + a_2^2 - b_2^2 + a_3^2 - b_3^2\\
= a_1^2+a_2^2+a_3^2 - (b_1^2+b_2^2+b_3^2)
[/tex]

if perpendicular, this is supposed to be 0, so

[tex]
a_1^2+a_2^2+a_3^2 - (b_1^2+b_2^2+b_3^2) = 0\\
a_1^2+a_2^2+a_3^2 = b_1^2+b_2^2+b_3^2\\
\sqrt{a_1^2+a_2^2+a_3^2} = \sqrt{b_1^2+b_2^2+b_3^2}\\
\Leftrightarrow |\vec A| = |\vec B|
[/tex]
 
  • #6
43
0
wait, we just factor them out? wow, that's cool thanks.
But why did you mutiply the two vectors? Because they are perpendicular?

For the last question i know that their sum has to be zero, but where would you place them? why are the angles fixed? it doesn't have to fixed, it could have a degree to it, doesn't it? i know that the resultant force of two forces has to be equal to the last vector, but how...?

Thanks
 
  • #7
14
0
I multiplied them out because if I wanna find something out about the angle between them, the scalar product tells you. So basically I rewrote your task to:

Proof: [tex] (\vec A + \vec B)*(\vec A - \vec B) = 0 [/tex] if [tex] |\vec A|=|\vec B| [/tex]

That's how I read your question...
 

Related Threads for: Vector problems

  • Last Post
Replies
4
Views
499
  • Last Post
Replies
14
Views
647
  • Last Post
Replies
3
Views
806
  • Last Post
Replies
8
Views
804
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
1
Views
760
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
1
Views
827
Top