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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 vectors C and D such that C is parallel to A, D is orthogonal to A, and B = C+D.


Let C = (c_1, c_2, c_3, c_4, c_5) and D = (d_1, d_2, d_3, d_4, d_5). The condition C parallel to A tells us that there is some x \neq 0 such that

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

The condition that B = C+D gives us

    \[ c_1 + d_1 = 1, \quad c_2 + d_2 = \frac{1}{2}, \quad c_3 + d_3 = \frac{1}{3}, \quad c_4 + d_4 = \frac{1}{4}, \quad c_5 + d_5 = \frac{1}{5}. \]

Putting these two sets of equations together we have

    \[ d_1 = 1 - x, \quad d_2 = \frac{1}{2} - 2x, \quad d_3 = \frac{1}{3} - 3x, \quad d_4 = \frac{1}{4} - 4x, \quad d_5 = \frac{1}{5} - 5x. \]

Then the condition that D is orthogonal to A gives us

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

Substituting our expressions for the d_i we then have

    \begin{align*}  &&(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,

    \[ C = \frac{1}{11} (1,2,3,4,5), \qquad D = \frac{1}{11} \left( 10, \frac{7}{2}, \frac{2}{3}, -\frac{5}{4}, -\frac{14}{5} \right). \]

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