Question #db33c

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Answer 1 · 2014-11-14T17:33:41

Let #vec{w}=(a,b,c)# be the vector we are looking for.

Since #vec{w}# is a unit vector,

#|w|=sqrt{a^2+c^2+c^2}=1 => a^2+b^2+c^2=1# ... (Eq.1)

Since #vec{w}# is perpendicular to #vec{u}#,

#vec{w}cdot vec{v}=a+2b+3c=0# ... (Eq.2)

Since #vec{w}# is perpendicular to #vec{v}#,

#vec{w}cdot vec{v}=2a+b+4c=0# ... (Eq.3)

By (Eq.2)#-2#(Eq.3),

#-3a-5c=0 => c=-3/5a#

By solving (Eq.3) for #b# and plugging in #c=-3/5a#,

#b=-2a-4c=-2a+12/5a=2/5a#

By plugging #b=2/5a# and #c=-3/5a# into (Eq.1),

#a^2+4/25a^2+9/25a^2=1 => 38/25a^2=1 => a=pm5/sqrt{38}#

Hence,

#vec{w}=(a,2/5a,-3/5a)=pm1/sqrt{38}(5,2,-3)#

I hope that this was helpful.