Geometry question

This is a very simple question but I am not sure if it is me or the solution that is wrong
Here is what I get and the solution:
EBEA3574-BAE3-4758-A563-62164F8FED7A.jpg.jpeg

EBEA3574-BAE3-4758-A563-62164F8FED7A.jpg.jpeg

DD495BA4-D648-4CD3-84CA-24DEC8395CF7.jpg.jpeg

Reply 1

mqb2766

THe magnitude of a+b and a-b should be the same (isosceles), also its half the semicircle so 1/4 rather than 1/2 on the right.

(edited 10 months ago)

Reply 2

grhas98

Original post by mqb2766

THe magnitude of a+b and a-b should be the same (isosceles), also its half the semicircle so 1/4 rather than 1/2 on the right.

I don’t see how the the magnitude of a + b is equal to a-b?

Reply 3

mqb2766

Original post by grhas98

I don’t see how the the magnitude of a + b is equal to a-b?

Isosceles triangle or pythagoras. As Thales said, theyre at right angles.

Reply 4

davros

Original post by grhas98

I don’t see how the the magnitude of a + b is equal to a-b?

Presumably you can see from the diagram that a-b is just a vector along the diameter - essentially the base of a right-angled triangle, so hypotenuse length is |a-b|. Now imagine that vector b is joined to the tip of vector a but points in the opposite direction. Now you can form another right-angled triangle with sides a, b and a+b (all vectors) so the hypotenuse length is |a+b|.

Alternatively, in vector notation |a+b|^2 = (a+b).(a+b) = a^2 + 2a.b + b^2 = a^2 + b^2 = a^2 - 2a.b + b^2 = |a-b|^2 since a.b = 0