state and prove parallelogram law of vector addition
Answers
In mathematics, the simplest form of the parallelogram law (also called the parallelogram identity) belongs to elementary geometry. It states that the sum of the squares of the lengths of the four sides of a parallelogram equals the sum of the squares of the lengths of the two diagonals. Using the notation in the diagram on the right, the sides are (AB), (BC), (CD), (DA). But since in Euclidean geometry a parallelogram necessarily has opposite sides equal, or (AB) = (CD) and (BC) = (DA), the law can be stated as,
{\displaystyle 2(AB)^{2}+2(BC)^{2}=(AC)^{2}+(BD)^{2}\,} 2(AB)^2+2(BC)^2=(AC)^2+(BD)^2\,
If the parallelogram is a rectangle, the two diagonals are of equal lengths (AC) = (BD) so,
{\displaystyle 2(AB)^{2}+2(BC)^{2}=2(AC)^{2}\,} 2(AB)^2+2(BC)^2=2(AC)^2\,
and the statement reduces to the Pythagorean theorem. For the general quadrilateral with four sides not necessarily equal,
{\displaystyle (AB)^{2}+(BC)^{2}+(CD)^{2}+(DA)^{2}=(AC)^{2}+(BD)^{2}+4x^{2},\,} {\displaystyle (AB)^{2}+(BC)^{2}+(CD)^{2}+(DA)^{2}=(AC)^{2}+(BD)^{2}+4x^{2},\,}
where x is the length of the line segment joining the midpoints of the diagonals. It can be seen from the diagram that, for a parallelogram, x = 0, and the general formula simplifies to the parallelogram law.
• Prove :-
In the parallelogram on the left, let AD=BC=a, AB=DC=b, ∠BAD = α. By using the law of cosines in triangle ΔBAD, we get:
{\displaystyle a^{2}+b^{2}-2ab\cos(\alpha )=BD^{2}} {\displaystyle a^{2}+b^{2}-2ab\cos(\alpha )=BD^{2}}
In a parallelogram, adjacent angles are supplementary, therefore ∠ADC = 180°-α. By using the law of cosines in triangle ΔADC, we get:
{\displaystyle a^{2}+b^{2}-2ab\cos(180^{\circ }-\alpha )=AC^{2}} {\displaystyle a^{2}+b^{2}-2ab\cos(180^{\circ }-\alpha )=AC^{2}}
By applying the trigonometric identity {\displaystyle \cos(180^{\circ }-x)=-\cos x} {\displaystyle \cos(180^{\circ }-x)=-\cos x} to our former result, we get:
{\displaystyle a^{2}+b^{2}+2ab\cos(\alpha )=AC^{2}} {\displaystyle a^{2}+b^{2}+2ab\cos(\alpha )=AC^{2}}
Now the sum of squares {\displaystyle BD^{2}+AC^{2}} {\displaystyle BD^{2}+AC^{2}} can be expressed as:
{\displaystyle BD^{2}+AC^{2}=a^{2}+b^{2}-2ab\cos(\alpha )+a^{2}+b^{2}+2ab\cos(\alpha )} {\displaystyle BD^{2}+AC^{2}=a^{2}+b^{2}-2ab\cos(\alpha )+a^{2}+b^{2}+2ab\cos(\alpha )}
After simplifying this expression, we get:
{\displaystyle BD^{2}+AC^{2}=2a^{2}+2b^{2}} {\displaystyle BD^{2}+AC^{2}=2a^{2}+2b^{2}}
Proof for parallelogram law of vector addition. ... If two vectors are considered to be the adjacent sides of a parallelogram, then the resultant of two vectors is given by the vector that is a diagonal passing through the point of contact of two vectors.
