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Permform some computations with vectors in R3

Given vectors

    \[ A = (1,1,1), \qquad B = (0,1,1), \qquad C = (1,1,0) \]

in \mathbb{R}^3, let D = xA + yB + zC where x,y,z \in \mathbb{R} are scalars.

  1. Compute the components of D.
  2. If D = O, prove that we must have x = y = z = 0.
  3. Find values for x,y,z such that D = (1,2,3).

  1. We compute,

        \[ D = x(1,1,1) + y(0,1,1) + z(1,1,0) = (x + z, x + y + z, x + y). \]

  2. Proof. D = O = (0,0,0) implies that we have

        \begin{align*}  x + z &= 0 & \implies && x &= -z\\  x+y+z &= 0 & \implies && y &= 0\\  x+y &= 0 & \implies && x &= -y & \implies && x = z = 0. \end{align*}

    Hence, we have x = y = z = 0. \qquad \blacksquare

  3. If D = (1,2,3) then we have the equations

        \begin{align*}  x + z &= 1 & \implies && z = 1-x \\  x + y + z &= 2 &\implies && x+y+1-x &=2 & \implies y&= 1\\  x + y &= 3 & \implies && x+1 &= 3 & \implies x &= 2. \end{align*}

    Finally this leaves z = -1. So, we have x = 2, \ y = 1, \ z = -1.

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