Finding the Angle Between Given Lines

We are asked to find the angle between two lines given by the equations:

1. Line 1: \(x\cos \alpha + y\sin \alpha = a\)
2. Line 2: \(x\sin \beta - y\cos \beta = a\)

There are several ways to find the angle between two lines. We can use their slopes or the angles their normal vectors make with the x-axis.

Method 1: Using Slopes

The general form of a linear equation is \(Ax + By + C = 0\). The slope \(m\) of this line is given by \(m = -\frac{A}{B}\).

For Line 1: \(x\cos \alpha + y\sin \alpha - a = 0\)

Here, \(A_1 = \cos \alpha\) and \(B_1 = \sin \alpha\). The slope \(m_1\) is:

\(m_1 = -\frac{\cos \alpha}{\sin \alpha} = -\cot \alpha\)

For Line 2: \(x\sin \beta - y\cos \beta - a = 0\)

Here, \(A_2 = \sin \beta\) and \(B_2 = -\cos \beta\). The slope \(m_2\) is:

\(m_2 = -\frac{\sin \beta}{-\cos \beta} = \frac{\sin \beta}{\cos \beta} = \tan \beta\)

The angle \(\theta\) between two lines with slopes \(m_1\) and \(m_2\) is given by:

\(\tan \theta = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right|\)

Substitute the slopes \(m_1 = -\cot \alpha\) and \(m_2 = \tan \beta\):

\(\tan \theta = \left| \frac{\tan \beta - (-\cot \alpha)}{1 + (\tan \beta)(-\cot \alpha)} \right|\)

\(\tan \theta = \left| \frac{\tan \beta + \cot \alpha}{1 - \tan \beta \cot \alpha} \right|\)

To simplify, we can convert \(\tan \beta\) and \(\cot \alpha\) into sine and cosine terms:

\(\tan \theta = \left| \frac{\frac{\sin \beta}{\cos \beta} + \frac{\cos \alpha}{\sin \alpha}}{1 - \frac{\sin \beta}{\cos \beta} \frac{\cos \alpha}{\sin \alpha}} \right|\)

Find a common denominator for the numerator and denominator:

Numerator: \(\frac{\sin \beta \sin \alpha + \cos \alpha \cos \beta}{\cos \beta \sin \alpha}\)

Denominator: \(\frac{\cos \beta \sin \alpha - \sin \beta \cos \alpha}{\cos \beta \sin \alpha}\)

So, the expression for \(\tan \theta\) becomes:

\(\tan \theta = \left| \frac{\frac{\sin \beta \sin \alpha + \cos \alpha \cos \beta}{\cos \beta \sin \alpha}}{\frac{\cos \beta \sin \alpha - \sin \beta \cos \alpha}{\cos \beta \sin \alpha}} \right|\)

\(\tan \theta = \left| \frac{\cos \alpha \cos \beta + \sin \alpha \sin \beta}{\sin \alpha \cos \beta - \cos \alpha \sin \beta} \right|\)

Using trigonometric identities, \(\cos(A-B) = \cos A \cos B + \sin A \sin B\) and \(\sin(A-B) = \sin A \cos B - \cos A \sin B\):

\(\tan \theta = \left| \frac{\cos(\alpha - \beta)}{\sin(\alpha - \beta)} \right|\)

\(\tan \theta = \left| \cot(\alpha - \beta) \right|\)

We know that \(\cot x = \tan(\pi/2 - x)\) or \(\cot x = \tan(\pi/2 + x)\). Let's use \(\cot x = \tan(x + \pi/2)\).

\(\cot(\alpha - \beta) = \tan((\alpha - \beta) + \pi/2) = \tan(\alpha - \beta + \pi/2)\)

So, \(\tan \theta = \left| \tan(\alpha - \beta + \pi/2) \right|\). The angle \(\theta\) between the lines can be \(\alpha - \beta + \pi/2\) (modulo \(\pi\)).

The angle between the lines is one of the values \(k\pi + (\alpha - \beta + \pi/2)\) or \(k\pi - (\alpha - \beta + \pi/2)\) for integer \(k\). Let's check the options.

Option 4 is \(\frac{\pi - 2\beta + 2\alpha}{2} = \frac{\pi}{2} - \beta + \alpha\). This is exactly \(\alpha - \beta + \pi/2\).

Method 2: Using Normal Vectors

The equation of a line \(Ax + By = C\) has a normal vector \((A, B)\). The angle the normal vector makes with the positive x-axis is \(\phi\) such that \(\cos \phi = A/ \sqrt{A^2+B^2}\) and \(\sin \phi = B / \sqrt{A^2+B^2}\).

For Line 1: \(x\cos \alpha + y\sin \alpha = a\). The normal vector is \((\cos \alpha, \sin \alpha)\). The magnitude is \(\sqrt{\cos^2 \alpha + \sin^2 \alpha} = 1\). The angle the normal makes with the x-axis, let's call it \(\phi_1\), satisfies \(\cos \phi_1 = \cos \alpha\) and \(\sin \phi_1 = \sin \alpha\). So, \(\phi_1 = \alpha\).

For Line 2: \(x\sin \beta - y\cos \beta = a\). The normal vector is \((\sin \beta, -\cos \beta)\). The magnitude is \(\sqrt{\sin^2 \beta + (-\cos \beta)^2} = 1\). The angle the normal makes with the x-axis, let's call it \(\phi_2\), satisfies \(\cos \phi_2 = \sin \beta\) and \(\sin \phi_2 = -\cos \beta\).

We can rewrite \(\sin \beta\) and \(-\cos \beta\) using angle subtraction formulas: \(\sin \beta = \cos(\pi/2 - \beta)\) and \(-\cos \beta = \sin(\pi/2 - \beta)\) is incorrect due to signs. Let's try angle addition/subtraction to \(\beta\).

\(\cos \phi_2 = \sin \beta = \cos(\pi/2 - \beta)\)

\(\sin \phi_2 = -\cos \beta = \sin(\beta - \pi/2)\)

So we need an angle \(\phi_2\) such that \(\cos \phi_2 = \cos(\pi/2 - \beta)\) and \(\sin \phi_2 = \sin(\beta - \pi/2)\). We know \(\sin(\beta - \pi/2) = - \sin(\pi/2 - \beta) = - \cos \beta\). Also \(\cos(\pi/2 - \beta) = \sin \beta\). So, \(\phi_2 = \pi/2 - \beta\). This satisfies \(\cos(\pi/2 - \beta) = \sin \beta\) and \(\sin(\pi/2 - \beta) = \cos \beta\), but we need \(\sin \phi_2 = -\cos \beta\). Let's try \(\phi_2 = \beta - \pi/2\). \(\cos(\beta - \pi/2) = \cos(\pi/2 - \beta) = \sin \beta\). This matches. \(\sin(\beta - \pi/2) = -\cos \beta\). This matches. So, the angle the second normal vector makes with the x-axis is \(\phi_2 = \beta - \pi/2\).

The angle between the two lines is equal to the angle between their normal vectors. The angle between two vectors making angles \(\phi_1\) and \(\phi_2\) with the x-axis is \(\left| \phi_1 - \phi_2 \right|\) or \(\pi - \left| \phi_1 - \phi_2 \right|\).

The difference between the angles of the normals is \(\phi_1 - \phi_2 = \alpha - (\beta - \pi/2) = \alpha - \beta + \pi/2\).

The angle between the lines is \(\left| \alpha - \beta + \pi/2 \right|\) or its supplement. The options suggest a general angle, not necessarily acute.

Comparing \(\alpha - \beta + \pi/2\) with the options:

Option 4: \(\frac{\left( \pi -2\beta +2\alpha \right)}{2} = \frac{\pi}{2} - \beta + \alpha\)

This matches \(\alpha - \beta + \pi/2\).

Both methods yield the same result, which matches option 4.

Conclusion

The angle between the given lines \(x\cos \alpha + y\sin \alpha = a\) and \(x\sin \beta - y\cos \beta = a\) is \(\frac{\left( \pi -2\beta +2\alpha \right)}{2}\).

Revision Table: Key Concepts for Angle Between Lines

Additional Information: Normal Form of a Line

The equation of a line can be written in the normal form as \(x \cos \phi + y \sin \phi = p\), where \(p\) is the length of the perpendicular from the origin to the line, and \(\phi\) is the angle that the perpendicular (normal) makes with the positive x-axis.

Comparing Line 1: \(x\cos \alpha + y\sin \alpha = a\) with the normal form, we see that the perpendicular distance from the origin is \(a\) (assuming \(a \ge 0\)), and the angle the normal makes with the x-axis is \(\phi_1 = \alpha\).

Comparing Line 2: \(x\sin \beta - y\cos \beta = a\) with the normal form requires transformation. We need the coefficients of x and y to be \(\cos \phi\) and \(\sin \phi\) respectively, and their sum of squares must be 1 (which \(\sin^2 \beta + (-\cos \beta)^2 = 1\) satisfies). We want \(x \cos \phi_2 + y \sin \phi_2 = a\).

We need \(\cos \phi_2 = \sin \beta\) and \(\sin \phi_2 = -\cos \beta\). As derived earlier, the angle \(\phi_2 = \beta - \pi/2\) satisfies these conditions.

The angle between two lines in normal form \(x \cos \phi_1 + y \sin \phi_1 = p_1\) and \(x \cos \phi_2 + y \sin \phi_2 = p_2\) is the angle between their normals, which is \(\left| \phi_1 - \phi_2 \right|\) or \(\pi - \left| \phi_1 - \phi_2 \right|\).

Using this, the angle between the given lines is the angle between normals making angles \(\alpha\) and \(\beta - \pi/2\) with the x-axis. The difference is \(\alpha - (\beta - \pi/2) = \alpha - \beta + \pi/2\).

This confirms the angle calculation obtained using slopes.