CBSE Class 12 Math 2011 Solved Paper - Question 20 | ExamSiri

CBSE Class 12 Math 2011 Solved Paper

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Question : 20 of 29

Marks: +1, -0

Find a unit vector perpendicular to each of the vector

a↖{→}+b↖{→}

and

a↖{→}-b↖{→}

, where

a↖{→}

=

3i↖{\^}+2j↖{\^}+2k↖{\^}

and

b↖{→}

=

i↖{\^}+2j↖{\^}-2k↖{\^}

Solution:

a↖{→}

=

3i↖{\^}+2j↖{\^}+2k↖{\^}

,

b↖{→}

=

i↖{\^}+2j↖{\^}-2k↖{\^}

∴

a↖{→}+b↖{→}

=

4i↖{\^}+4j↖{\^}

and

a↖{→}-b↖{→}

=

2i↖{\^}+4k↖{\^}

(a↖{→}+b↖{→}) × (a↖{→}-b↖{→})

=

|\table i↖{\^},j↖{\^},k↖{\^};4,4,0;2,0,4|

=

i↖{\^}

(16) -

j↖{\^}

(16) +

k↖{\^}

(-8) =

16i↖{\^}-16j↖{\^}-8k↖{\^}

∴

|(a↖{→}+b↖{→}) × (a↖{→}-b↖{→})|

=

√{16^2+(-16)^2+(-8)^2}

=

√{256+256+64}

=

√{576}

= 24
So the unit vector, perpendicular to each of the vectors

a↖{→}+b↖{→}

and

a↖{→}-b↖{→}

is given by ±

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

= ±

{16i↖{\^}-16j↖{\^}-8k↖{\^}}/{24}

= ±

{2i↖{\^}-2j↖{\^}-k↖{\^}}/3

= ±

2/3i↖{\^}∓2/3j↖{\^}∓1/3k↖{\^}

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