Finding the Acute Angle Between Two Possible Straight Lines

This problem involves finding the angle between two straight lines in three-dimensional space. We are given information about the direction cosines of the lines, which helps us determine their orientation.

Understanding Direction Cosines

A straight line in 3D space can be uniquely defined by its direction cosines. The direction cosines are the cosines of the angles that the line makes with the positive x-axis, y-axis, and z-axis. Let these angles be \(\alpha\), \(\beta\), and \(\gamma\), respectively. The direction cosines are \(\cos \alpha\), \(\cos \beta\), and \(\cos \gamma\). These values satisfy the fundamental relationship:

\(\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1\)

Any set of three numbers proportional to the direction cosines are called direction ratios of the line. The direction cosines themselves can be thought of as the components of a unit vector parallel to the line.

Using the Given Information

We are given the following facts about the straight line:

1. The line passes through the point (1, 1, 1). Note that the angle between two lines depends only on their directions, not the point they pass through. So, this information is not directly needed to find the direction cosines.
2. The line makes an angle of 60° with the positive direction of the z-axis. So, \(\gamma = 60^\circ\).
3. The cosine of the angle made with the positive y-axis and the cosine of the angle made with the positive x-axis are in the ratio \(\sqrt{3} : 1\). This means \(\frac{\cos \beta}{\cos \alpha} = \frac{\sqrt{3}}{1}\).

Calculating the Direction Cosines

From the given information, we have:

• \(\cos \gamma = \cos 60^\circ = \frac{1}{2}\).
• \(\cos \beta = \sqrt{3} \cos \alpha\).

Now, we use the fundamental relationship of direction cosines:

\(\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1\)

Substitute the values we know:

\(\cos^2 \alpha + (\sqrt{3} \cos \alpha)^2 + \left(\frac{1}{2}\right)^2 = 1\)

\(\cos^2 \alpha + 3 \cos^2 \alpha + \frac{1}{4} = 1\)

\(4 \cos^2 \alpha = 1 - \frac{1}{4}\)

\(4 \cos^2 \alpha = \frac{3}{4}\)

\(\cos^2 \alpha = \frac{3}{16}\)

Taking the square root of both sides, we get two possible values for \(\cos \alpha\):

\(\cos \alpha = \pm \sqrt{\frac{3}{16}} = \pm \frac{\sqrt{3}}{4}\)

Identifying the Two Possible Directions

Based on the two possible values for \(\cos \alpha\), we find the corresponding values for \(\cos \beta\) and \(\cos \gamma\) for each case.

Case 1: Let \(\cos \alpha_1 = \frac{\sqrt{3}}{4}\).

• \(\cos \beta_1 = \sqrt{3} \cos \alpha_1 = \sqrt{3} \left(\frac{\sqrt{3}}{4}\right) = \frac{3}{4}\).
• \(\cos \gamma_1 = \frac{1}{2}\).

The direction cosines for the first possible line are \(\left(\frac{\sqrt{3}}{4}, \frac{3}{4}, \frac{1}{2}\right)\). Let this direction vector be \(\mathbf{d}_1 = \left(\frac{\sqrt{3}}{4}, \frac{3}{4}, \frac{1}{2}\right)\).

Case 2: Let \(\cos \alpha_2 = -\frac{\sqrt{3}}{4}\).

• \(\cos \beta_2 = \sqrt{3} \cos \alpha_2 = \sqrt{3} \left(-\frac{\sqrt{3}}{4}\right) = -\frac{3}{4}\).
• \(\cos \gamma_2 = \frac{1}{2}\).

The direction cosines for the second possible line are \(\left(-\frac{\sqrt{3}}{4}, -\frac{3}{4}, \frac{1}{2}\right)\). Let this direction vector be \(\mathbf{d}_2 = \left(-\frac{\sqrt{3}}{4}, -\frac{3}{4}, \frac{1}{2}\right)\).

These are the two possible positions (directions) of the straight line that satisfy the given conditions.

Finding the Angle Between the Two Lines

The angle \(\theta\) between two lines with direction vectors \(\mathbf{d}_1 = (\cos \alpha_1, \cos \beta_1, \cos \gamma_1)\) and \(\mathbf{d}_2 = (\cos \alpha_2, \cos \beta_2, \cos \gamma_2)\) is given by the dot product formula:

\(\cos \theta = \frac{\mathbf{d}_1 \cdot \mathbf{d}_2}{||\mathbf{d}_1|| \cdot ||\mathbf{d}_2||}\)

Since \(\mathbf{d}_1\) and \(\mathbf{d}_2\) are vectors of direction cosines, they are unit vectors. Their magnitudes are 1.

\(||\mathbf{d}_1|| = \sqrt{\left(\frac{\sqrt{3}}{4}\right)^2 + \left(\frac{3}{4}\right)^2 + \left(\frac{1}{2}\right)^2} = \sqrt{\frac{3}{16} + \frac{9}{16} + \frac{4}{16}} = \sqrt{\frac{16}{16}} = 1\)

\(||\mathbf{d}_2|| = \sqrt{\left(-\frac{\sqrt{3}}{4}\right)^2 + \left(-\frac{3}{4}\right)^2 + \left(\frac{1}{2}\right)^2} = \sqrt{\frac{3}{16} + \frac{9}{16} + \frac{4}{16}} = \sqrt{\frac{16}{16}} = 1\)

Now, calculate the dot product \(\mathbf{d}_1 \cdot \mathbf{d}_2\):

\(\mathbf{d}_1 \cdot \mathbf{d}_2 = \left(\frac{\sqrt{3}}{4}\right)\left(-\frac{\sqrt{3}}{4}\right) + \left(\frac{3}{4}\right)\left(-\frac{3}{4}\right) + \left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\)

\(\mathbf{d}_1 \cdot \mathbf{d}_2 = -\frac{3}{16} - \frac{9}{16} + \frac{1}{4}\)

To add these fractions, find a common denominator (16):

\(\mathbf{d}_1 \cdot \mathbf{d}_2 = -\frac{3}{16} - \frac{9}{16} + \frac{4}{16}\)

\(\mathbf{d}_1 \cdot \mathbf{d}_2 = \frac{-3 - 9 + 4}{16} = \frac{-12 + 4}{16} = \frac{-8}{16} = -\frac{1}{2}\)

Now, substitute the dot product and magnitudes into the angle formula:

\(\cos \theta = \frac{-\frac{1}{2}}{1 \cdot 1} = -\frac{1}{2}\)

The angle whose cosine is \(-\frac{1}{2}\) is \(120^\circ\). So, one angle between the lines is \(\theta = 120^\circ\).

Determining the Acute Angle

When considering the angle between two lines, we usually refer to the acute angle. The angles between two lines can be \(\theta\) or \(180^\circ - \theta\). The acute angle is the smaller of these two values.

If \(\theta = 120^\circ\), the other angle is \(180^\circ - 120^\circ = 60^\circ\).

The acute angle between the two possible positions of the line is \(60^\circ\).

Conclusion

Based on the calculations using direction cosines and the relationship \(\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1\), we found two possible sets of direction cosines for the line. The angle between the lines represented by these two directions is \(120^\circ\). The acute angle is \(180^\circ - 120^\circ = 60^\circ\).

Revision Table: Straight Line Direction Cosines

Additional Information: 3D Line Geometry

Understanding straight lines in 3D space is a key part of vector algebra and 3D geometry. A line can be represented in various forms:

• Vector Form: A line passing through a point \(\mathbf{a}\) and parallel to a vector \(\mathbf{d}\) can be represented as \(\mathbf{r} = \mathbf{a} + t\mathbf{d}\), where \(\mathbf{r}\) is the position vector of any point on the line and \(t\) is a scalar parameter. The vector \(\mathbf{d}\) gives the direction ratios (or direction cosines if it's a unit vector) of the line.
• Cartesian Form: If the line passes through \((x_1, y_1, z_1)\) and has direction ratios \((a, b, c)\), its Cartesian equation is \(\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}\). If \(a, b,\) or \(c\) is zero, the numerator must be zero, indicating the line is parallel to a coordinate plane or axis. For example, if \(a=0\), the equation becomes \(x - x_1 = 0\) and \(\frac{y - y_1}{b} = \frac{z - z_1}{c}\).

Direction cosines \((\cos \alpha, \cos \beta, \cos \gamma)\) are essentially the components of the unit direction vector of the line. If the direction ratios are \((a, b, c)\), the direction cosines are given by:

\(\cos \alpha = \frac{a}{\sqrt{a^2+b^2+c^2}}, \quad \cos \beta = \frac{b}{\sqrt{a^2+b^2+c^2}}, \quad \cos \gamma = \frac{c}{\sqrt{a^2+b^2+c^2}}\)

The angle between two lines in 3D is found using the dot product of their direction vectors. If the direction vectors are \(\mathbf{d}_1\) and \(\mathbf{d}_2\), the cosine of the angle \(\theta\) between them is \(\frac{\mathbf{d}_1 \cdot \mathbf{d}_2}{||\mathbf{d}_1|| ||\mathbf{d}_2||}\). To find the acute angle, we take the absolute value of the cosine: \(\cos \theta_{acute} = \left|\frac{\mathbf{d}_1 \cdot \mathbf{d}_2}{||\mathbf{d}_1|| ||\mathbf{d}_2||}\right|\).

In this problem, the given conditions lead to two distinct sets of direction cosines, representing two different orientations for the line, even though they pass through the same point (1, 1, 1). We then calculated the angle between these two orientation vectors to find the required angle.