Parallogram law of vector
Answers
Explanation:
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, i.e. (AB) = (CD) and (BC) = (DA), the law can be stated as.A parallelogram. The sides are shown in blue and the diagonals in red.
If the parallelogram is a rectangle, the two diagonals are of equal lengths (AC) = (BD)
where x is the length of the line segment joining the midpoints of the diagonals. It can be seen from the diagram that x = 0 for a parallelogram, and so the general formula simplifies to the parallelogram law.
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 the 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}}