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Vector Algebra

Question
CBSEENMA12034162

Show that the diagonals of a rhombus bisect each other at right angles.

Solution
Let ABCD be the rhombus. Take A as origin.

    Let straight b with rightwards arrow on top comma space straight d with rightwards arrow on top be the position vectors of B and D respectively so that
                           AB with rightwards arrow on top space equals space straight b with rightwards arrow on top comma space space AD with rightwards arrow on top space equals space straight d with rightwards arrow on top
Now,        AC with rightwards arrow on top space equals space AB with rightwards arrow on top space space plus space BC with rightwards arrow on top space equals space AB with rightwards arrow on top space plus space AD with rightwards arrow on top
                                                               [because BC is equal and parallel to AD]
therefore space space space AC with rightwards arrow on top space equals space straight b with rightwards arrow on top space plus space straight d with rightwards arrow on top
Position vector of mid-point of diagonal AC is fraction numerator 0 with rightwards arrow on top plus left parenthesis straight b with rightwards arrow on top plus straight d with rightwards arrow on top right parenthesis over denominator 2 end fraction space space or space space fraction numerator straight b with rightwards arrow on top plus straight d with rightwards arrow on top over denominator 2 end fraction
Also position vector of mid-point of diagonal BD is fraction numerator straight b with rightwards arrow on top plus straight d with rightwards arrow on top over denominator 2 end fraction

∴    position vector of mid-point of diagonal AC is same as position vector of mid-point of diagonals BD.
∴  diagonals AC and BD bisect each other.
Also,  AC with rightwards arrow on top. space BD with rightwards arrow on top space equals space left parenthesis straight b with rightwards arrow on top plus straight d with rightwards arrow on top right parenthesis. space left parenthesis straight d with rightwards arrow on top space minus space straight b with rightwards arrow on top right parenthesis
                                          open square brackets because space space BD with rightwards arrow on top space equals space straight P. straight V. space of space straight D space minus space straight P. straight V. space of space straight B space equals space straight d with rightwards arrow on top minus straight b with rightwards arrow on top close square brackets
equals space straight d with rightwards arrow on top squared space minus space straight b with rightwards arrow on top squared space equals space AD squared space minus space AB squared
equals space space 0
                                           [because AD = AB as all sides of rhombus are equal]
therefore space space space AC with rightwards arrow on top space is space perpendicular space to space BD with rightwards arrow on top.
therefore diagonals AC and BD are perpendicular to each other. 
Hence the result.