91Ó°ÊÓ

Trigonometry is a branch of mathematics focused on the relationships between angles and sides of triangles. In the context of finding a line in a plane, trigonometry helps us understand how the direction of a line relates to angles formed with coordinate axes.

In the exercise, we were given specific angles between the line and the standard basis vectors along the x-axis and y-axis, denoted as i and j. These angles are \(\pi/6\) and \(\pi/3\). The cosine of an angle in trigonometry can give us a relationship between these angles and the components of our direction vector.

• \(\cos(\pi/6) = \sqrt{3}/2\)
• \(\cos(\pi/3) = 1/2\)

Thus, trigonometric values help us set up equations to determine the components of the direction vector. This knowledge is essential for translating geometrical angles into algebraic equations.

The direction vector of a line is crucial because it shows the line's orientation in space. For a line on the plane defined by \(z=3\), its direction vector only needs to account for x and y components, because z is constant at 3.

In the problem, we determined the direction vector \(\mathbf{d} = (a, b, 0)\) by using trigonometric principles from the angles given: \(\theta_i = \pi/6\) and \(\theta_j = \pi/3\).

• Using trigonometry, we found that \(a/\|\mathbf{d}\| = \sqrt{3}/2\)
• Similarly, \(b/\|\mathbf{d}\| = 1/2\)

A unit vector has a norm of 1, which means that \(\|\mathbf{d}\| = 1\). Thus, we calculated that the direction vector is \((\sqrt{3}/2, 1/2, 0)\). This vector describes the line's path parallel to the plane where z doesn't change.

A plane equation represents the entire set of points that make up a particular flat surface in three-dimensional space. For this problem, the plane is defined by \(z = 3\), a constant value.

Any line lying within this plane can be expressed in parametric form based on a point and direction vector. Here, the parametric equations use the direction vector \((\sqrt{3}/2, 1/2, 0)\) derived from our trigonometric analysis.

• The parametric equation for \(x\) is \(x = \sqrt{3}/2 \cdot t\)
• The equation for \(y\) is \(y = 1/2 \cdot t\)
• Finally, \(z = 3\) remains unchanged

By choosing a convenient point on the line, like \((0, 0, 3)\), we simplify our equations, making it easier to visualize the line remaining in the plane as it extends in the direction dictated by our vector.