NIMCET 2015 Question Paper with solutions

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Question : 45 of 120

Marks: +1, -0

If

{AC}↖{→}

=

2i↖{\^}+j↖{\^}+k↖{\^}

and

{BD}↖{→}

=

-i↖{\^}+3j↖{\^}+2k↖{\^}

then thearea of the quadrilateral ABCD is;

$5/2 √3$
$5√3$
${15}/2 √3$
$10√3$

Solution:

We have,

AC=2i↖{⋀}+j↖{⋀}+k↖{⋀}

BD=i↖{⋀}+3j↖{⋀}+2k↖{⋀}

Area of quadrilateral ABCD

=|AB×CD|

AC×BD=|\table i↖{⋀},j↖{⋀},k↖{⋀};2,1,1;-1,3,2|

=(2-3)i↖{⋀}-(4+1)j↖{⋀}+(6+1)k↖{⋀}

=i↖{⋀}-5j↖{⋀}+7k↖{⋀}

|AC×BD|=√{1+25+49}

=√75=5√3

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