Question

Find an equation of the plane. The plane contains the \(y\) -axis and makes an angle of \(\pi / 6\) with the positive \(x\) -axis

Short answer

The equation of the plane which contains the y-axis and makes an angle of \(\pi / 6\) with the positive x-axis is \(\sqrt{3}x + z = 0\).

Step-by-step solution

Calculate the normal vector to the plane

From the problem statement, we know that the plane contains y-axis, which means the normal to the plane is in the xz-plane. So, we can assume the normal direction vector \(\mathbf{n} = [n_x, 0, n_z]\). We also know from the problem that the angle between the normal to the plane and the positive x-axis is \(\pi / 6\). Using the cosine formula for the dot product, we have \(\cos(\pi / 6) = \mathbf{n} \cdot \mathbf{i} / |\mathbf{n}| = n_x / \sqrt{n_x^2 + n_z^2}\) where \(\mathbf{i} = [1, 0, 0]\) is the unit vector along x-axis. Therefore, we get \(n_x = \sqrt{3} n_z\).

Determine the Direction Ratio of the Normal Vector

As \(\mathbf{n}\) is a direction vector and not a unit vector, we can choose the simplest case where \(n_z = 1\). Therefore, \(n_x = \sqrt{3}\) and our normal vector is \(\mathbf{n}=[\sqrt{3}, 0, 1]\) .

Formulate the Equation of the Plane

To form the equation of a plane using a point and a normal vector, we use the formula for the direction ratios of a plane: \(a(x-x_0) + b(y-y_0) + c(z-z_0) = 0\), where \((x_0, y_0, z_0)\) is a point on the plane and \([a, b, c]\) is the normal vector. Since this plane contains the y-axis, we can choose a point on the y-axis (0, y, 0) to simplify the equation, resulting in: \(\sqrt{3}x + z = 0\) as the equation of the plane.