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 is \(\sqrt{3}/2x ± 1/2z = 0\).

Step-by-step solution

Determine the normal vector

Since the plane contains the y-axis, the normal vector to the plane will have a y-component equal to 0. For the x-component, we use the trigonometric identity that cos(\(\pi / 6\)) is equal to the x-component of the normal vector over its magnitude, and since the magnitude of the normal vector is 1, the value of cos(\(\pi / 6\)) will be the x-component. As we also know that cos(\(\pi / 6\)) equals \(\sqrt{3}/2\), the normal vector becomes \<\(\sqrt{3}/2\), 0, z\>, where z is the z-component of the normal vector.

Find the z-component of the normal vector

Given the length of the normal vector as 1, we use the equation of the magnitude of a vector to find the z-component. The equation states that 1 (length of the vector) is equal to the square root of (x-component^2 + y-component^2 + z-component^2). Substituting our known values, we get \(1 = \sqrt{(\sqrt{3}/2)^2 + 0 + z^2}\), which simplifies to \(1 = \sqrt{3/4 + z^2}\). Solving for z gives us \(z = ± \sqrt{1/4}\), therefore the z-component is ±1/2. Consequently, we can have two possible normal vectors: \<\(\sqrt{3}/2\), 0, 1/2\> or \<\(\sqrt{3}/2\), 0, -1/2\>.

Form the equation of the plane

The standard equation of a plane is given as \(ax + by + cz + d = 0\), where \ represents the normal vector and d represents the plane's distance from the origin. Substituting for the normal vector and setting d = 0 (since the plane contains the y-axis, hence it intercepts the origin), our equation becomes \(\sqrt{3}/2x ± 1/2z = 0\).