3. Now we will prove in general that the diagonals in a rhombus are perpendicula

Question

3. Now we will prove in general that the diagonals in a rhombus are perpendicular. Assume the figure at the right is a rhombus spanned by the vectors A and B. Since we are assuming that the figure is a rhombus, what are we assuming about the vectors A and B? The diagonals in the rhombus are illustrated as vectors D and E. Express each of those vectors in terms of vectors A and B Now use the dot product to show that the vectors Dand E are perpendicular. To do this, you will use your expressions for D and E in terms of A and B, you will simplify using the algebraic rules for dot product (see Theorem 12.2 and Exercise 80 in Section 12.3), and you will use the above assumption about A and B that reflects that the quadrilateral is assumed to be a rhombus

Explanation / Answer

Solution:

Consider an arbitrary parallelogram. As given in figure one corner to be the origin and let the vectors to the adjacent corners be A and B. Then one diagonal is A B and the other diagonal is A - B.

(A B) . (A - B)
= A.A + B.A - A.B - B.B
= A.A - B.B (since A.B = B.A)
= ||A||^2 - ||B||^2

But for a rhombus, A and B are the same length so ||A|| = ||B||.
Hence (A B) . (A - B) = 0, and therefore the diagonals are perpendicular.

Conversely, if the diagonals are perpendicular then ||A||^2 = ||B||^2, hence ||A|| = ||B|| and all the sides of the parallelogram have the same length. So it is a rhombus.

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