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Find two vectors satisfying given conditions

Let

    \[ A = ( 1,2,3,4,5), \qquad B = \left( 1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5} \right). \]

Find two vectors C, D \in \mathbb{R}^5 such that

    \[ B = C + 2D, \qquad D \cdot A = 0 \]

and C is parallel to A.


Let C = (c_1, c_2, c_3, c_4, c_5) and D = (d_1, d_2, d_3, d_4, d_5). Then we have

    \begin{align*}   B = C + 2D && \implies && c_1 + 2d_1 &= 1 \\  &&&& c_2 + 2d_2 &= 2 \\  &&&& c_3 + 2d_3 &= 3 \\  &&&& c_4 + 2d_4 &= 4 \\  &&&& c_5 + 2d_5 &= 5. \end{align*}

We also have

    \[ D \cdot A = 0 \quad \implies \quad d_1 + 2d_2 + 3d_3 + 4d_4 + 5d_5 = 0. \]

And, since C is parallel to A there is a nonzero x such that

    \[ C = xA \quad \implies \quad c_1 = x, c_2 = 2x, c_3= 3x, c_4 = 4x, c_5 = 5x. \]

So, putting these together we get expressions for the d_i,

    \begin{align*}  d_1 &= \frac{1}{2}(1-x) \\  d_2 &= \frac{1}{2}\left( \frac{1}{2} - 2x \right) \\  d_3 &= \frac{1}{2}\left( \frac{1}{3} - 3x \right) \\  d_4 &= \frac{1}{2}\left( \frac{1}{4} - 4x \right) \\  d_5 &= \frac{1}{2}\left( \frac{1}{5} - 5x \right). \end{align*}

Therefore,

    \begin{align*}  &&\frac{1}{2}(1-x) + \left( \frac{1}{2} - 2x \right) + \frac{3}{2} \left( \frac{1}{3} - 3x \right) + 2 \left( \frac{1}{4} - 4x \right) + \frac{5}{2} \left( \frac{1}{5} - 5x \right) &= 0 \\  \implies && 1-x + 2 \left( \frac{1}{2} - 2x \right) + 3 \left( \frac{1}{3} - 3x \right) + 4 \left( \frac{1}{4} - 4x \right) + 5\left( \frac{1}{5} - 5x \right) &= 0 \\  \implies && 55x &= 5 \\  \implies && x &= \frac{1}{11}. \end{align*}

Therefore, we have

    \[ C = \frac{1}{11} (1,2,3,4,5), \qquad D = \left( \frac{5}{11}, \frac{7}{44}, \frac{1}{33}, \frac{-5}{88}, \frac{-7}{55} \right). \]

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