Suppose that p→,q→ and r→ are three non-coplanar vectors in ℝ3. Let the components of a vector 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,−p→−q→+r→ are x,y and z respectively, then the value of 2x+y+z is
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answer is 9.
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Detailed Solution
Given s¯=4p¯+3q¯+5r¯.Let s¯=x-p¯+q¯+r¯+yp¯-q¯+r¯+z-p¯-q¯+r¯ ⇒s¯=-x+y-zp¯+x-y-zq¯+x+y+zr¯ ⇒-x+y-z=4, x-y-z=3, x+y+z=5 Solving the above equations, we get x=4, y=92, z=-72 ∴2x+y+z=9