Discussion Forum

Suppose  that p,q, and r  three non-coplanar vectors  in R³. Let  the components of a vector  s along p,q and r be 4,3 and 5  , respectively. If the components  of this vector  s along    (-p+q+r)  (p-q+r)  and (-p-q+r)  are x,y and z respectively, then the value of 2x+y+z is 

Answer:

Explanation:

 Here, s = 4p+3q+5r              ........(i)

  and s=   (-p+q+r)x+(p-q+r)y

                                  +(-p-q+r)z        .........(ii)

 $\therefore$    4p+3q+5r = p(-x+y-z)+q(x-y-z)+r(x+y+z)

 On comparing both sides, we get

             -x+y-z=4, x-y-z=3

         and x+y+z=5

  On solving above equations, we get

  $x=4,y=\frac{9}{2},z=\frac{-7}{2}$

 $\therefore$     $2x+y+z=8+\frac{9}{2}-\frac{7}{2}=9$