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Find a unit vector perpendicular to a given vector and parallel to a given plane

Find a unit length vector A which is parallel to the plane with Cartesian equation x - y + 5z = 1 and perpendicular to the vector \mathbf{i} + 2 \mathbf{j} - 3 \mathbf{k}.


The plane with Cartesian equation x - y + 5z = 1 has intercepts (1,0,0), \ (0,-1,0), \ \left( 0,0, \frac{1}{5} \right). Therefore, the plane is the set of points

    \[ M = \left\{ (1,0,0) + s(1,1,0 + t\left( 1, 0, -\frac{1}{5} \right) \right\}. \]

So, the vector A being parallel to M implies A is in the span of \left\{ (1,0,0), \ \left(1,0,-\frac{1}{5} \right) \right\}. Therefore,

    \[ A = \left( s + t, s, -\frac{1}{5}t \right) \]

for some s,t. Then, A perpendicular to \mathbf{i} + 2 \mathbf{j} - 3 \mathbf{k} implies

    \[ A \cdot (1,2,-3) = 0 \quad \implies \quad s+t + 2s + \frac{3}{5} t = 0 \quad \implies \quad s = -\frac{8}{15} t. \]

Finally, since A has unit length we have

    \begin{align*}  \lVert A \rVert = 1 && \implies && \sqrt{ (s+t)^2 + s^2 + \frac{t^2}{25} } &= 1 \\  && \implies && s^2 + 2st + t^2 + s^2 + \frac{t^2}{25} &= 1 \\  && \implies && 2s^2 + \left( \frac{26}{25} \right)t^2 + 2st &= 1 \\  && \implies && t^2 \left( \frac{128}{225} + \frac{26}{25} - \frac{16}{15} \right) &= 1 \\  && \implies && t &= \frac{15}{\sqrt{122}}. \end{align*}

This then gives us

    \[ s = -\frac{8}{\sqrt{122}}. \]

Hence,

    \[ A = \frac{1}{\sqrt{122}} (7,-8,-3). \]

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