Understanding Direction Cosines of the Z-axis

This question asks for the direction cosines of the z-axis. Let's first understand what direction cosines are in the context of 3D geometry.

What are Direction Cosines?

Direction cosines of a line or a vector in three-dimensional space are the cosines of the angles that the line or vector makes with the positive x, y, and z axes. If a line makes angles \(\alpha\), \(\beta\), and \(\gamma\) with the positive x, y, and z axes respectively, then \(\cos \alpha\), \(\cos \beta\), and \(\cos \gamma\) are its direction cosines. These are usually denoted by \(l, m, n\), respectively.

The direction cosines \(l, m, n\) of any line or vector satisfy the property:

Finding Direction Cosines for the Z-axis

Consider the positive z-axis. This is a line passing through the origin and extending infinitely along the positive z-direction. We need to find the angles it makes with the positive x, y, and z axes.

Angle with the positive x-axis: The positive z-axis is perpendicular to the positive x-axis. The angle between them is \(90^\circ\) or \(\frac{\pi}{2}\) radians. So, \(\alpha = 90^\circ\).

Angle with the positive y-axis: Similarly, the positive z-axis is perpendicular to the positive y-axis. The angle between them is \(90^\circ\) or \(\frac{\pi}{2}\) radians. So, \(\beta = 90^\circ\).

Angle with the positive z-axis: The positive z-axis is parallel to itself in the positive direction. The angle it makes with itself is \(0^\circ\) or \(0\) radians. So, \(\gamma = 0^\circ\).

Now, let's calculate the cosines of these angles to find the direction cosines \(l, m, n\).

\(l = \cos \alpha = \cos(90^\circ) = 0\)

\(m = \cos \beta = \cos(90^\circ) = 0\)

\(n = \cos \gamma = \cos(0^\circ) = 1\)

Thus, the direction cosines of the positive z-axis are \(l=0, m=0, n=1\). These are represented as the triplet <0, 0, 1>.

Comparing with Options

Let's look at the given options:

Option 1: < 1, 1, 1 >

Option 2: < 1, 0, 0 >

Option 3: < 0, 1, 0 >

Option 4: < 0, 0, 1 >

Our calculated direction cosines for the z-axis are <0, 0, 1>, which matches Option 4.

Let's quickly check the sum of squares for the correct option: \(0^2 + 0^2 + 1^2 = 0 + 0 + 1 = 1\), which satisfies the property \(l^2 + m^2 + n^2 = 1\).

Conclusion

The direction cosines of the z-axis are <0, 0, 1>.

Revision Table: Direction Cosines of Axes

Additional Information: Direction Cosines and Ratios

Understanding direction cosines is fundamental in 3D geometry. Here's some related information:

Direction Ratios: Any three numbers \(a, b, c\) that are proportional to the direction cosines \(l, m, n\) of a line are called its direction ratios. That is, \(a = k l, b = k m, c = k n\) for some non-zero constant \(k\).

Relationship: If \(a, b, c\) are direction ratios of a line, its direction cosines can be found using the formulas: \[l = \frac{a}{\pm\sqrt{a^2 + b^2 + c^2}}, \quad m = \frac{b}{\pm\sqrt{a^2 + b^2 + c^2}}, \quad n = \frac{c}{\pm\sqrt{a^2 + b^2 + c^2}}\] The sign (\(\pm\)) depends on the direction along the line you are considering. For the coordinate axes, we usually consider the positive direction.

Direction Vector: A vector parallel to a line has direction ratios as its components. For the z-axis, a direction vector could be \(\vec{k} = <0, 0, 1>\). The components of this vector are \(a=0, b=0, c=1\). The magnitude is \(\sqrt{0^2 + 0^2 + 1^2} = 1\). The direction cosines are \(l = \frac{0}{1} = 0\), \(m = \frac{0}{1} = 0\), \(n = \frac{1}{1} = 1\), which match our result <0, 0, 1>.