Let ABCD be a parallelogram whose diagonals intersect at P and let O be the origin. What is \(\overrightarrow {{\rm{OA}}} + \overrightarrow {{\rm{OB}}} + \overrightarrow {{\rm{OC}}} + \overrightarrow {{\rm{OD}}}\) equal to?
Understanding the Problem: Parallelogram and Vectors
The question asks for the value of the vector sum of the position vectors of the vertices of a parallelogram ABCD, relative to an origin O. The diagonals of the parallelogram intersect at point P.
We are given the expression: \(\overrightarrow {{\rm{OA}}} + \overrightarrow {{\rm{OB}}} + \overrightarrow {{\rm{OC}}} + \overrightarrow {{\rm{OD}}}\)
Here, O is the origin, and A, B, C, D are the vertices of the parallelogram. \(\overrightarrow {{\rm{OA}}}\), \(\overrightarrow {{\rm{OB}}}\), \(\overrightarrow {{\rm{OC}}}\), and \(\overrightarrow {{\rm{OD}}}\) are the position vectors of points A, B, C, and D respectively, with respect to the origin O.
Key Property of Parallelogram Diagonals
A fundamental property of any parallelogram is that its diagonals bisect each other. This means the point where the diagonals intersect (point P) is the midpoint of both diagonal AC and diagonal BD.
Using Position Vectors and Midpoint Formula
Let the position vectors of points A, B, C, D, and P with respect to the origin O be \(\vec{a}\), \(\vec{b}\), \(\vec{c}\), \(\vec{d}\), and \(\vec{p}\) respectively. Thus:
- \(\overrightarrow {{\rm{OA}}} = \vec{a}\)
- \(\overrightarrow {{\rm{OB}}} = \vec{b}\)
- \(\overrightarrow {{\rm{OC}}} = \vec{c}\)
- \(\overrightarrow {{\rm{OD}}} = \vec{d}\)
- \(\overrightarrow {{\rm{OP}}} = \vec{p}\)
Since P is the midpoint of the diagonal AC, we can use the midpoint formula for vectors:
\(\vec{p} = \frac{\vec{a} + \vec{c}}{2}\)
Multiplying both sides by 2 gives us:
\(\vec{a} + \vec{c} = 2\vec{p}\) (Equation 1)
Similarly, since P is also the midpoint of the diagonal BD, we have:
\(\vec{p} = \frac{\vec{b} + \vec{d}}{2}\)
Multiplying both sides by 2 gives us:
\(\vec{b} + \vec{d} = 2\vec{p}\) (Equation 2)
Calculating the Vector Sum
Now, let's consider the vector sum we need to find:
\(\overrightarrow {{\rm{OA}}} + \overrightarrow {{\rm{OB}}} + \overrightarrow {{\rm{OC}}} + \overrightarrow {{\rm{OD}}} = \vec{a} + \vec{b} + \vec{c} + \vec{d}\)
We can rearrange the terms and group them as follows:
\((\vec{a} + \vec{c}) + (\vec{b} + \vec{d})\)
Using Equation 1 and Equation 2, we can substitute the values:
\((2\vec{p}) + (2\vec{p})\)
Adding the terms:
\(4\vec{p}\)
Since \(\vec{p} = \overrightarrow {{\rm{OP}}}\), the vector sum is \(4\overrightarrow {{\rm{OP}}}\).
Conclusion
The sum of the position vectors of the vertices of a parallelogram ABCD with respect to an origin O is equal to four times the position vector of the point of intersection of its diagonals P, with respect to the same origin O.
Therefore, \(\overrightarrow {{\rm{OA}}} + \overrightarrow {{\rm{OB}}} + \overrightarrow {{\rm{OC}}} + \overrightarrow {{\rm{OD}}} = 4\overrightarrow {{\rm{OP}}}\).
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