Hi, I'm struggling to do this proof. I've tried using pythagoras as my first step to find an expression for the magnitude of b (v2), but it just ends up cancelling out to zero on the left hand side of the equation. Have I chosen the wrong first step? I'm not sure what else to do... thanks!

Original post by blizzardliz

Hi, I'm struggling to do this proof. I've tried using pythagoras as my first step to find an expression for the magnitude of b (v2), but it just ends up cancelling out to zero on the left hand side of the equation. Have I chosen the wrong first step? I'm not sure what else to do... thanks!

Probably help to upload what you tried. It sounds similar to the proof of the cos rule.

Note the altitude is v2 sin(theta), not v1 sin(theta)

Original post by blizzardliz

Hi, I'm struggling to do this proof. I've tried using pythagoras as my first step to find an expression for the magnitude of b (v2), but it just ends up cancelling out to zero on the left hand side of the equation. Have I chosen the wrong first step? I'm not sure what else to do... thanks!

As mqb says, upload what you've tried. But I think you'd do better to make your first step finding the length of v.

Edit: and to be explicit, mqb is saying the diagram is slightly wrong - where it says "v1 sin theta" it should say "V2 sin theta".

(edited 9 months ago)

Original post by mqb2766

Probably help to upload what you tried. It sounds similar to the proof of the cos rule.

Note the altitude is v2 sin(theta), not v1 sin(theta)

Note the altitude is v2 sin(theta), not v1 sin(theta)

I didn't notice the altitude was wrong, thank you! I'll change it and have another go and if I get stuck again I'll upload what I've done

Original post by mqb2766

Probably help to upload what you tried. It sounds similar to the proof of the cos rule.

Note the altitude is v2 sin(theta), not v1 sin(theta)

Note the altitude is v2 sin(theta), not v1 sin(theta)

Here's my working, I've never really done proof with vectors before, only with scalars, so I'm not too sure if there are any rules I'm going wrong with?

Original post by blizzardliz

Here's my working, I've never really done proof with vectors before, only with scalars, so I'm not too sure if there are any rules I'm going wrong with?

I think they want you to prove it using scalars, so its basically how the cos rule maps to vector magnitudes. So pythagoras with

a = base = |v1| + |v2|cos(theta)

b = perp height = |v2|sin(theta)

c = hypotenuse = |v|

Should just be a couple of lines.

For the usual cos rule, theta would be the supplementary value 180-theta, and the sign is flipped accordingly so -2abcos(theta)

(edited 9 months ago)

Original post by blizzardliz

Here's my working, I've never really done proof with vectors before, only with scalars, so I'm not too sure if there are any rules I'm going wrong with?

I don't see how this answers the question (I don't think it even moves you in the right direction).

Look at the diagram - can you see a right angled triangle with hypotenuse v that you could use to find the length of v? You basically just need to apply pythagorus to that and rearrange slightly and you're done.

Original post by DFranklin

I don't see how this answers the question (I don't think it even moves you in the right direction).

Look at the diagram - can you see a right angled triangle with hypotenuse v that you could use to find the length of v? You basically just need to apply pythagorus to that and rearrange slightly and you're done.

Look at the diagram - can you see a right angled triangle with hypotenuse v that you could use to find the length of v? You basically just need to apply pythagorus to that and rearrange slightly and you're done.

Oh, I didn't see the big right angled triange I just saw them as two seperate triangles, you can tell I haven't done maths for a while..... thanks for pointing it out!!)

Original post by mqb2766

I think they want you to prove it using scalars, so its basically how the cos rule maps to vector magnitudes. So pythagoras with

a = base = |v1| + |v2|cos(theta)

b = perp height = |v2|sin(theta)

c = hypotenuse = |v|

Should just be a couple of lines.

For the usual cos rule, theta would be the supplementary value 180-theta, and the sign is flipped accordingly so -2abcos(theta)

a = base = |v1| + |v2|cos(theta)

b = perp height = |v2|sin(theta)

c = hypotenuse = |v|

Should just be a couple of lines.

For the usual cos rule, theta would be the supplementary value 180-theta, and the sign is flipped accordingly so -2abcos(theta)

Ohhh I was using the wrong triangle.. sorry silly mistake. Thanks so much for the help!

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