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Find a Cartesian equation for a plane through a point with normal vector making given angles

Let M be the plane whose normal vector N makes angles \frac{\pi}{3}, \frac{\pi}{4}, \frac{\pi}{3} with the unit coordinate vectors and which contains the point (1,1,1). Find a Cartesian equation for the plane.


Since the normal vector to the plane makes angles \frac{\pi}{3}, \frac{\pi}{4}, \frac{\pi}{3} with the unit coordinate vectors, we have

    \begin{align*}  N \cdot \mathbf{i} &= \cos \left( \frac{\pi}{3} \right) = \frac{1}{2} & \implies && x_1 &= \frac{1}{2} \\  N \cdot \mathbf{j} &= \cos \left( \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2} & \implies && x_2 &= \frac{\sqrt{2}}{2} \\  N \cdot \mathbf{k} &= \cos \left( \frac{\pi}{3} \right) = \frac{1}{2} & \implies && x_3 &= \frac{1}{2}. \end{align*}

Hence, N = (1, \sqrt{2}, 1). So, the plane has a Cartesian equation of the form

    \[ x + \sqrt{2} y + z = d. \]

Since it contains (1,1,1) we have d = 1 + \sqrt{2} + 1 = 2 + \sqrt{2}. Therefore, a Cartesian equation of the plane is

    \[ x + \sqrt{2} y + z = 2 + \sqrt{2}. \]

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