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Question

If b and c are non-zero and non-collinear vectors, 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×c)+(ab)b=(42xsiny)b+(x21)c

(ac)b(ab)c+(ab)b
=(42xsiny)b+(x21)c

Now, (cc)a=c. Therefore,
(cc)(ac)=(cc) or ac=1

1+ab=42xsiny and x21=(ab)
1=42xsiny+x21
siny=x22x+2=(x1)2+1
But siny1x=1, siny=1
y=(4n+1)π2, nI

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