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If veca , vecb and vecc are three non-ze...

If `veca , vecb and vecc` are three non-zero, non- coplanar vectors and `vecb_(1)=vecb-(vecb.veca)/(|veca|^(2))veca, \ vecb_(2)=vecb+(vecb.veca)/(|veca|^(2))veca, \ vecc_(1)=vecc-(vecc.veca)/(|veca|^(2))veca+ (vecb.vecc)/(|vecc|^(2))vecb_(1),` `vecc_(2)=vecc-(vecc.veca)/(|veca|^(2)) veca-(vecbvecc)/(|vecb_(1)|^(2))vecb_(1), \ vecc_(3)=vecc- (vecc.veca)/(|vecc|^(2))veca + (vecb.vecc)/(|vecc|^(2))vecb_(1),` ` vecc_(4)=vecc - (vecc.veca)/(|vecc|^(2))veca= (vecb.vecc)/(|vecb|^(2))vecb_(1)`, then the set of mutually orthogonal vectors is

A

(a) `(veca,vecb_(1),vecc_(3))`

B

(b) `(veca,vecb_(1),vecc_(2))`

C

(c) `(veca, vecb_(1),vecc_(1))`

D

(d) `(veca,vecb_(2),vecc_(2))`

Text Solution

Verified by Experts

The correct Answer is:
c

we observe that
`veca.vecb_(1)=veca.vecv- ((vecb.veca)/(|veca|^(2)))veca. veca=veca.vecb- veca.vecb=0`
`veca.vecc_(2)=veca(vecc-(vecc.veca)/(|veca|^(2))veca- (vecc.vecb_(1))/(|vecb_(1)|^(2))vecb_(1))`
`veca.vecc- (veca.vecc)/(|veca|^(2))|veca|^(2)-(vecc.vecb_(1))/(|vecb_(1)|^(2))(veca.vecb_(1))`
`veca.vecc-veca.vecc-0 " " (therefore veca.vecb_(1)=0)`
`and vecb_(1).vecc_(2)=vecb_(1)(vecc- (vecc.veca)/(|veca|^(2))veca- (vecc.vecb_(1))/(|vecb_(1)|^(2))vecb_(1))`
`vecb_(1).vecc-((vecc.veca)(vecb_(1).veca))/(|veca|^(2))- (vecc.vecb_(1))/(|vecb_(1)|^(2))vecb_(1).vecb_(1)`
`vecb_(1).vecc-0-vecb_(1).vecc` (using `vecb_(1).veca =0)`
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