Ohhh I've been misreading the question this whole time, I read v and w as a single vector but it is asking "when if v and w are both perpendicular to u then show that v + 2w must be perpendicular, if this proposition is false, give a counter example."
This seems harder, I believe I need to show that v and 2w logically must always have the same gradient as either v or w alone would have.
Thank you.
Here is my attempt:
The basic "plan of attack":
https://imgur.com/a/V5Z5OrAThe write up:
As both v and w have gradients meaning they are perpendicular then v and w must be multiples of each other therefore v + 2w is simple a multiple of both either v or w hence it has the same gradient and therefore as the gradient of both v and w multiplied by the recipricol of the gradient of u is -1 and so are perpendicular the product of the gradient of v + 2w and the recipricol of the gradient of u must also be -1 [ok I suck at wording things]
[Is there a way of doing this algebreicly?]
[[e.g. to check
v = <2 , 5> , w = < 4, 10> , u = < 5 , 2 >
v * 1/[gradient of...?]u = -1 and w * 1/u = -1?
(v + 2w) * 1/u must also be -1?
v + 2w = <10 , 25>
Yeah I now see, how does one get the recipricol of u if that is indeed what I need?
]]