A) 2
B) \[\sqrt{7}\]
C) \[\sqrt{14}\]
D) 14
Correct Answer: C
Solution :
Given,\[|u|=1,|v|=2,|w|=3\] The projection of v along\[u=\frac{v.u}{|u|}\] and the projection of w along\[u=\frac{w.u}{|u|}\] According to the given condition, \[\frac{v.u}{u}=\frac{w.u}{u}\] \[\Rightarrow \] \[v.u=w.u\] ...(i) and v, w are perpendicular to each other. \[\therefore \] \[v.w=0\] Now, \[|u-v+w{{|}^{2}}=|u{{|}^{2}}+|v{{|}^{2}}+|w{{|}^{2}}\] \[-2u.v+2u.w-2v.w\] \[\Rightarrow \]\[|u-v+w{{|}^{2}}=1+4+9-2u.v+2v.u\] [from Eq. (i)] \[\Rightarrow \] \[|u-v+w{{|}^{2}}=1+4+9\] \[\Rightarrow \] \[|u-v+w|=\sqrt{14}\]
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