NIMCET 2021 Question Paper with solutions

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Question : 26 of 120

Marks: +1, -0

The area of the triangle formed by the vertices whose position vector are

3 {i}

+j, 5 i+2 j+k, i-2 j+3 k

is

$\√{21}$
$\√{23}$
$\√{33}$
$\√{29}$

Solution:

Area of

\Δ

formed by adjacent sides is

{1}/{2}|{{AB}↖{→}} \×{{AC}}↖{→}|

∵

{A}↖{→}=3 {i}↖{\^}+{j}↖{\^}

{B↖{→}}=5 {i}↖{\^}+2 {j}↖{\^}+{k}↖{\^}

{C}↖{→}={i}↖{\^}-2 {j}↖{\^}+3 {k}↖{\^}

{{AB}↖{→}}={{OB}↖{→}}-{OA}↖{→}

5 {i}↖{\^}+2 {j}↖{\^}+{k}↖{\^}-3 {i}↖{\^}-{j}↖{\^}

{A B}↖{→}=2 {i}↖{\^}+{j}↖{\^}+{k}↖{\^}

{A C}↖{→}=({i}↖{\^}-2 {j}↖{\^}+3 {k↖{\^}})-(3 {i}↖{\^}+{j}↖{\^})

{A C}↖{→}=-2 {i}↖{\^}-3 {j}↖{\^}+3 {k↖{\^}}

Now,

{A B}↖{→}×{A C}↖{→}=|\table i↖{\^},j↖{\^},k↖{\^};2,1,1;-2,-3,3|

{A B}↖{→}×{A C}↖{→}=6i↖{\^}-8j↖{\^}-4k↖{\^}

\Δ=1/2|{A B}↖{→}×{A C}↖{→}|

\Δ={1}/{2} \√{36+64+16}

\Δ={1}/{2} \√{116}

\Δ={2}/{2} \√{29}

\Δ=\√{29}

Unit

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