Author Topic: Vector Proof (Read 1198 times) Tweet Share

Flaming_Arrow · « **on:** January 12, 2009, 08:20:05 pm »

can you check if this vector proof is sufficient?

Prove that the sum of the squares of the lengths of the diagonals of any parallelogram is
equal to the sum of the squares of the lengths of the sides.

$\mbox {Let} \vec{OA}=a \mbox {and} \vec{OC}=c$

$\vec{OB} = \vec{OA} + \vec{OC} = a+c$

$\vec{CA} = \vec{CO} + \vec{OA} = a-c$

$\vec{|OB|} = a^2 +c^2 \vec{|CA|} = a^2 +c^2$

$\vec{|OB|}^2 + \vec{|CA|}^2 = 2(a^2+c^2)$

$\vec{|OA|} = \vec{|CB|} \vec{|OC|} = \vec{|AB|}$

$\vec{|OA|}^2 = \vec{|CB|}^2 = a^2 \vec{|OC|}^2 = \vec{|AB|}^2 = c^2$

$\vec{|OA|}^2 +\vec{|CB|}^2 + \vec{|OC|}^2 + \vec{|AB|}^2 = 2(a^2+c^2)$

$\vec{|OA|}^2 +\vec{|CB|}^2 + \vec{|OC|}^2 + \vec{|AB|}^2 = \vec{|OB|}^2 + \vec{|CA|}^2$

$\therefore \mbox{ sum of the squares of the lengths of the diagonals of any parallelogram is<br />equal to the sum of the squares of the lengths of the sides}$

Mao · « **Reply #1 on:** January 12, 2009, 08:41:16 pm »

I don't think yours is right. Your expression for both |OB| and |AC| are wrong.

Here's how I would go about it:

OB.OB = |OB|² = a²+c²+2a.c
AC.AC = |AC|² = a²+c²-2a.c

|OB|²+|AC|²=2a²+2c²=|OA|²+|AB|²+|BC|²+|CD|²

QED

/0 · « **Reply #2 on:** January 12, 2009, 08:42:33 pm »

The magnitude formula only works for vectors at right angles I think, since it's derived using pythagoras.

Flaming_Arrow · « **Reply #3 on:** January 12, 2009, 08:50:42 pm »

oops i forgot to square them

Mao · « **Reply #4 on:** January 12, 2009, 09:06:13 pm »

sorry mate, still not right. You can't work out magnitude like that. That only works when the components are in right angles, i.e. i and j

when they are not (such as in this case), you can only use dot product.

Flaming_Arrow · « **Reply #5 on:** January 12, 2009, 09:08:22 pm »

Quote from: Mao on January 12, 2009, 09:06:13 pm

sorry mate, still not right. You can't work out magnitude like that. That only works when the components are in right angles, i.e. i and j

when they are not (such as in this case), you can only use dot product.

oh thanks

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Author Topic: Vector Proof (Read 1198 times) Tweet Share

Flaming_Arrow

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