Question

Suppose that p,q and r are three non-coplanar vectors in $R^3$. 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 1

Here, $s=4p+3q+5r\cdots \left(i\right)\text{and}s=(-p+q+r)x+(p-q+r)y+(-p-q+r)z\cdots \left(ii\right)\therefore 4p+3q+5r=p(-x+y-z)+q(x-y-z)+r(x+y+z)$
On comparing both sides, we get
$-x+y-z=4,x-y-z=3andx+y+z=5$
On solving above equations, we get
$x=4,y=\frac{9}{2},z=\frac{-7}{2}\therefore 2x+y+z=8+\frac{9}{2}-\frac{7}{2}=9$