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Question

The vectors b and c are non-colinear, if a×(b×c)+(ab)b=(42xsiny)b+(x21)c and (cc)a=c then

A
x=1
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B
x=1
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C
y=(4n+1)π2, nI
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D
y=(2n+1)π2, nI
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Solution

The correct option is C y=(4n+1)π2, nI
a×(b×b)+(ab)b =(42xsiny)b+(x21)c(ac)b(ab)c+(ab)b =(42xsiny)b+(x21)c(i)

We know that,
(cc)a=c(cc)ac=ccac=1

1+ab=42xsiny and x21=(ab)

1+1x2=42xsinysiny=x22x+2siny=(x1)2+1

We know that,
1siny1 and
(x1)2+11
So the solution will be,
siny=1y=(4n+1)π2, nI
and x=1

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