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

Options

A. 14
B. `sqrt7`
C. `sqrt14`
D. 2

Answer

C

Explanation

Projection of `vecv` along `vecu=(vecv.vecu)/(|vecu|)=(vecv.vecu)/2`
projection of `vecw` along `vecu=(vecw.vecu)/(|vecu|)=(vecw.vecu)/2`
Given `(vecv.vecu)/2=(vecw.vecu)/2`…(i)
Also, `vecv.vecw=0` …(ii)
Now `|vecu-vecv+vecw|^2`
`=|vecu|^2+|vecv|^2 +|vecw|^2 -2vecu.vecv-2vecv.vecw+2vecu.vecw`
=1+4+9+0 [From (1) and (2) ] =14
`therefore |vecu-vecv+vecw|=sqrt14`