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Let vecu, vecv, vecw be such that |vecu|...

Let `vecu, vecv, vecw` be such that `|vecu|=1, |vecv|=2 and |vecw|=3`. If the projection of `vecv` along `vecu` is equal to that of `vecw` along `vecv and vecw` are perpendicular to each other then `|vecu-vecv+vecw|` equals

A

2

B

`sqrt(7)`

C

`sqrt(14)`

D

14

Text Solution

Verified by Experts

The correct Answer is:
C
Projection of `vecv` along `vecu and vecw` along `vecu is (vecv. vecu)/(|vecu|) and (vecw.vecu)/(|vecu|)` respectively.
According to the equation `(vecv.vecu)/(|vecu|)=(vecw.vecu)/(|vecu|) rArr vecv.vecu=vecw. vecu and vecv. vecw=0`
`|vecu-vecv+vecw|^(2)=|vecu|^(2)+|vecv|^(2)+|vecw^(2)|-2vecu.vecv-2vecv.vecw+2vecu//vecw=14 rArr |vecu-vecv+vecw|= sqrt(14)`
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