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Find vectors which satisfy given relations

by RoRi
  1. Find all of the vectors a \mathbf{i} + b \mathbf{j} + c \mathbf{k} which satisfy

        \[ (a \mathbf{i} + b \mathbf{j} + c \mathbf{k}) \cdot \mathbf{k} \times (6 \mathbf{i} + 3 \mathbf{j} + 4 \mathbf{k}) = 3. \]

  2. Find the shortest length vector a \mathbf{i} + b \mathbf{j} + c \mathbf{k} which satisfies the relation in part (a).
  1. We can compute,

        \begin{align*} && (a \mathbf{i} + b \mathbf{j} + c \mathbf{k}) \cdot ((0,0,1) \times (6,3,4)) &= 3 \\ \implies && (a,b,c) \cdot (-3,6,0) &=3 \\ \implies && -3a + 6b &= 3 \\ \implies && a &= 2b-1. \end{align*}

    So, b,c can take any value, and then we must have a = 2b-1 for the relation to be satisfied. Therefore, any vector

        \[ (2b-1) \mathbf{i} + b \mathbf{j} + c \mathbf{k} \]

    satisfies the relations.

  2. So, we know the vectors satisfying the relation are of the form (2b-1)\mathbf{i} + b \mathbf{j} + c \mathbf{k}. This means we want to minimize

        \[ (2b-1)^2 + b^2 + c^2 = 5b^2 -4b + 1 + c^2. \]

    This is minimal when c = 0 (since c^2 > 0 for any other value of c). Then we want to find the value of b which minimizes 5b^2 - 4b + 1. Taking the derivative and setting it equal to 0 we have

        \[ 10b - 4 = 0 \qquad \implies \qquad b = \frac{2}{5} \quad \implies \quad a = -\frac{1}{5}. \]

    Hence, the vector of minimal length which satisfies the given relation is

        \[ -\frac{1}{5} \mathbf{i} + \frac{2}{5} \mathbf{j}. \]

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