Vector Calculus please help?

Question

Using Vectors, prove that the diagonals of a parallelogram are perpendicular if and only if the parallelogram is a rhombus. (Note: a rhombus is a parallelogram whose four sides all have the same lenth)

I've got no idea how to do this problem and my TA isn't very clear in his explanations. Could anyone give me a hand.

Answer 1

Suppose the parallelogram has vertices O(0,0), A(a,b), B(a+c,b), and C(c,0).  The diagonals are OB(a+c,b) and CA(c-a,-b).  The diagonals are perpendicular iff their dot product is 0, or iff (c+a)(c-a) - b^2 = 0, or c^2 - a^2 - b^2 = 0.  This says iff c^2 = a^2 + b^2.  Now notice that side OA has length sqrt(a^2 + b^2) and side OC has length sqrt(c^2); since c^2 = a^2 + b^2, these sides are of equal length, so this parallelogram is a rhombus.

Answer 2

(1) Denote the two diagonal vectors as DV1 and DV2.

(2) Also denote two adjacent sides of the parallelogram as

vectors SV1 and SV2, whose lengths are S1 and S2,

respectively.

(3) Using vector arithmetic, if DV1 = SV1 + SV2, then

DV2 = SV2 - SV1

(4) Taking the dot product (*) of DV1 and DV2, yields the

following scalar product P:

P = DV1 * DV2 = (SV1 + SV2) * (SV2 - SV1)

= SV1 * SV2 + SV2 * SV2 - SV1 * SV1 - SV2 * SV1

(5) Since SV1 * SV2 = SV2 * SV1, we see that

P = SV2^2 - SV1^2, i.e., P = S2^2 - S1^2

(6) Clearly, if S2 = S1, the product P is zero. This, in turn,

means that the two diagonal vectors are perpendicular to

each other, since cos(90 deg) = 0; i.e., rhombus diagonals

are orthogonal to each other.

(7) On the other hand, if S2 and S1 are unequal, P cannot be

zero, so the two diagonal vectors cannot intersect at 90 deg;

thus non-rhombus parallogram diagonals are not

orthogonal to each other.

The above completes the "if and only" parts of the theorem

proof.