A parallelogram is a slanted rectangle: opposite sides are equal and parallel. Draw both diagonals — each runs from one corner to the opposite corner. The question is: how do we describe these diagonals using the two side vectors that start from the same corner?
The Setup
Take a parallelogram with vertices A, B, C, D in order, with A at the origin. From A, two vectors emerge:
a goes from A to B (one side)
b goes from A to D (the other side)
Because opposite sides are equal, B to C is also b and D to C is also a, so the fourth vertex C sits at a+b. The two diagonals run from A to C and from B to D.
The Diagonal from the Common Vertex
From A to C you go a then b, ending at the opposite corner:
d1=a+b
That's the vector sum of the two sides — walk along one side then the other and you land on the opposite corner.
The Other Diagonal
B is at a and D is at b. To go from B to D, you travel from a to b:
d2=b−a
The reverse, from D to B, is a−b. Both are correct; they just differ in direction.
Note
The two diagonals are not the same length in general. They are equal only in a rectangle. The sum and difference of the side vectors give the two diagonals.
The Precise Statement
For a parallelogram with adjacent side vectors a and b from a common vertex:
The diagonal from that common vertex to the opposite vertex is a+b.
The other diagonal (connecting the other two vertices) is b−a (or a−b, depending on direction).
Why This Matters
Vector addition — the diagonal from the common vertex is the sum of the sides. This is the parallelogram law of vector addition.
Finding midpoints — the diagonals bisect each other; both midpoints are the same point, 2a+b.
Physics — the resultant of two forces acting at a point is the diagonal of the parallelogram formed by the force vectors.
A Quick Check
Take a=(3,0) (horizontal) and b=(1,2) (slanted). Then:
Diagonal from the common vertex: (3,0)+(1,2)=(4,2) …
Mistake 1: Reporting the obtuse angle instead of the acute one
Why it's wrong: the diagonals make two supplementary angles (45∘ and 135∘); the question asks for the acute one. Correct approach: if cosθ<0, take the supplement (or use ∣cosθ∣) to report the acute value.
Mistake 2: Using the sides (or a×b) instead of the diagonals
Why it's wrong: the angle is between a+b and a−b, not between a and b. Correct approach: form the diagonals first, then apply the dot-product formula. …
Q.Two sides of a parallelogram are along the lines x+y=5 and x−y=−5. If the diagonals of the parallelogram intersects at (3,6) then one of its vertices, is at
(A) (6,5)
(B) (7,6)
(C) (7,5)
(D) (6,7)
(E) (5,7)
›Reveal solutionSolution
Adjacent sides intersect at a vertex (0,5); the diagonals' intersection (3,6) is the midpoint of the diagonal, giving the opposite vertex (6,7).
Find one vertex. Solve x+y=5 and x−y=−5: adding gives 2x=0⇒x=0,y=5, so a vertex is (0,5). …