JEE Main & Advanced AIEEE Solved Paper-2004

  • question_answer
    Let\[u,v,w\]be such that\[|u|=1,|v|=2,\]\[|w|=3.\]If the projection v along u is equal to that of w along u and\[v,w\]are perpendicular to each other, then\[|u-v+w|\]equals

    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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