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Question

If a is perpendicular to b and r is a non – zero vector such that pr+(r.b)a=c, then r is equal to

A
cp(b.c)ap2
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B
ap(c.a)bp2
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C
bp(a.c)cp2
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D
cp2(b.c)ap
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Solution

The correct option is A cp(b.c)ap2
We have, pr+(r.b)a=c
p(r.b)+(a.b)(a.b)=c.bp(r.b)=c.b[ab,a.b=0]
r.b=c.bp
Substituting the value of r.b in Eq. (i), we get
pr+(c.bp)a=cr=cpc.bp2a

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