The area of the parallelogram whose diagonals are a^{\to} = 3i^{\to}+j^{\to}-2k^{\to} and b^{\to} = i^{\to}-3j^{\to}+4k^{\to} is :

NIMCET 2013 Question Paper with solutions

Question : 32 of 120

Marks: +1, -0

The area of the parallelogram whose diagonals are

a↖{→}

3i↖{→}+j↖{→}-2k↖{→}

and

b↖{→}

i↖{→}-3j↖{→}+4k↖{→}

is :

Solution:

The given diagonals of the parallelogram are

{ a }↖{→}=3 { i }↖{∧}+{ j }↖{∧}-2 { k }↖{∧}

and

{ b }↖{→}={ i }↖{∧}-3 { j }↖{∧}+4 { k }↖{∧}

.
Using the formula for the area of a parallelogram whose diagonals

{a}↖{→}

and

{b}↖{→}

are given, we get:
Area

=\;{1}/{2}|{ a }↖{→} × { b }↖{→}|=\;{1}/{2}|( a _{2} b _{3}- a _{3} b _{2}) { i }↖{∧}+( a _{3} b _{1}- a _{1} b _{3}) { j }↖{∧}+( a _{1} b _{2}- a _{2} b _{1}) { k }↖{∧}|

=\;{1}/{2} √{( a _{2} b _{3}- a _{3} b _{2})^{2}+( a _{3} b _{1}- a _{1} b _{3})^{2}+( a _{1} b _{2}- a _{2} b _{1}})^{2}

=\;{1}/{2} √{[(1)(4)-(-2)(-3)]^{2}+[(-2)(1)-(3)(4)]^{2}+[(3)(-3)-(1)(1)]^{2}}

=\;{1}/{2} √{(4-6)^{2}+(-2-12)^{2}+(-9-1)^{2}}

=\;{1}/{2} √{4+196+100}

=\;{1}/{2} √{300}

=5 √{3}

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