NIMCET 2017 Question Paper with solutions

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Question : 103 of 120

Marks: +1, -0

If

a↖{→}

=

(i↖{\^}+2j↖{\^}-3k↖{\^})

and

b↖{→}

=

(3i↖{\^}-j↖{\^}+2k↖{\^})

then the angle between (a + b) and (a - b) is

$π/3$
$π/4$
$π/2$
${2π}/3$

Solution:

a↖{→}+b↖{→}

=

(i+2j↖{\^}-3k↖{\^})

+

(3i↖{\^}-j↖{\^}+2k↖{\^})

=

4i↖{\^}+j↖{\^}-k↖{\^}

and

a↖{→}-b↖{→}

=

(i↖{\^}+2j↖{\^}-3k↖{\^})

-

(3i↖{\^}-j↖{\^}+2k↖{\^})

=

-2i↖{\^}+3j↖{\^}-5k↖{\^}

Let θ be the angle between

(a↖{→}+b↖{→})

and

(a↖{→}-b↖{→})

∴ cos θ =

{(a↖{→}+b↖{→}).(a↖{→}.b↖{→})}/{|a↖{→}+b↖{→}||a↖{→}-b↖{→}|}

=

{(4i↖{\^}+j↖{\^}-k↖{\^}).(-2i↖{\^}+3j↖{\^}-5k↖{\^})}/{|4i↖{\^}+j↖{\^}-k↖{\^}| |-2i↖{\^}+3j↖{\^}-5k↖{\^}|}

=

{-8+3+5}/√{4^2+1^2+(-1)^2√{(-2)^2+3^2+(-5)^2}}

= 0
∴ θ = 90°

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