Vector problems

Tensaiga

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

andrevdh

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.

Israfil

third question: the forces must be arranged in a way that the sum of them is the zero vector.

Israfil

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]

Israfil

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]

Tensaiga

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

Israfil

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...