Understanding Direction Cosines and Ratios

Direction cosines are the cosines of the angles made by a line with the positive x, y, and z axes. If a line makes angles α, β, and γ with the x, y, and z axes respectively, its direction cosines are \(l = \cos \alpha\), \(m = \cos \beta\), and \(n = \cos \gamma\). A fundamental property of direction cosines is that \(l^2 + m^2 + n^2 = 1\).

Direction ratios are numbers proportional to the direction cosines. If (a, b, c) are direction ratios of a line, then the direction cosines (l, m, n) can be found using the formulas:

• \(l = \dfrac{a}{\sqrt{a^2 + b^2 + c^2}}\)
• \(m = \dfrac{b}{\sqrt{a^2 + b^2 + c^2}}\)
• \(n = \dfrac{c}{\sqrt{a^2 + b^2 + c^2}}\)

The standard symmetric form of a line in 3D space passing through point \((x_0, y_0, z_0)\) with direction ratios (a, b, c) is given by:

\(\dfrac{x - x_0}{a} = \dfrac{y - y_0}{b} = \dfrac{z - z_0}{c}\)

Finding Direction Ratios from the Given Line Equation

The given equation of the line is \(x - 1 = 2(y + 3) = 1 - z\). To find the direction ratios, we need to convert this equation into the standard symmetric form. Let's rewrite the terms:

• The first part is already in the form \((x - x_0)/a\): \(x - 1 = \dfrac{x - 1}{1}\). Here \(x_0 = 1\) and \(a = 1\).
• The second part is \(2(y + 3)\). To get it into the form \((y - y_0)/b\), we can write \(2(y + 3) = \dfrac{y + 3}{1/2}\). Here \(y_0 = -3\) and \(b = 1/2\).
• The third part is \(1 - z\). To get it into the form \((z - z_0)/c\), we need the numerator to be \(z - z_0\). So, \(1 - z = -(z - 1) = \dfrac{z - 1}{-1}\). Here \(z_0 = 1\) and \(c = -1\).

Equating these parts, we get the standard symmetric form:

\(\dfrac{x - 1}{1} = \dfrac{y + 3}{1/2} = \dfrac{z - 1}{-1}\)

From this form, we can identify the direction ratios (a, b, c) as \((1, 1/2, -1)\).

Calculating Direction Cosines (l, m, n)

Now that we have the direction ratios \((a, b, c) = (1, 1/2, -1)\), we can find the direction cosines (l, m, n). First, we calculate the magnitude of the direction ratio vector:

\(\sqrt{a^2 + b^2 + c^2} = \sqrt{1^2 + \left(\dfrac{1}{2}\right)^2 + (-1)^2}\)

\(= \sqrt{1 + \dfrac{1}{4} + 1}\)

\(= \sqrt{2 + \dfrac{1}{4}}\)

\(= \sqrt{\dfrac{8}{4} + \dfrac{1}{4}}\)

\(= \sqrt{\dfrac{9}{4}}\)

\(= \dfrac{3}{2}\)

Now we can find the direction cosines:

• \(l = \dfrac{a}{\sqrt{a^2 + b^2 + c^2}} = \dfrac{1}{3/2} = 1 \times \dfrac{2}{3} = \dfrac{2}{3}\)
• \(m = \dfrac{b}{\sqrt{a^2 + b^2 + c^2}} = \dfrac{1/2}{3/2} = \dfrac{1}{2} \times \dfrac{2}{3} = \dfrac{1}{3}\)
• \(n = \dfrac{c}{\sqrt{a^2 + b^2 + c^2}} = \dfrac{-1}{3/2} = -1 \times \dfrac{2}{3} = -\dfrac{2}{3}\)

So, the direction cosines of the line are \(l = 2/3\), \(m = 1/3\), and \(n = -2/3\).

Calculating \(l^4 + m^4 + n^4\)

We need to find the value of \(l^4 + m^4 + n^4\). Let's calculate each term:

• \(l^4 = \left(\dfrac{2}{3}\right)^4 = \dfrac{2^4}{3^4} = \dfrac{16}{81}\)
• \(m^4 = \left(\dfrac{1}{3}\right)^4 = \dfrac{1^4}{3^4} = \dfrac{1}{81}\)
• \(n^4 = \left(-\dfrac{2}{3}\right)^4 = \dfrac{(-2)^4}{3^4} = \dfrac{16}{81}\)

Now, sum these values:

\(l^4 + m^4 + n^4 = \dfrac{16}{81} + \dfrac{1}{81} + \dfrac{16}{81}\)

\(= \dfrac{16 + 1 + 16}{81}\)

\(= \dfrac{33}{81}\)

We can simplify this fraction by dividing the numerator and denominator by their greatest common divisor, which is 3:

\(\dfrac{33 \div 3}{81 \div 3} = \dfrac{11}{27}\)

Thus, \(l^4 + m^4 + n^4\) is equal to \(\dfrac{11}{27}\).

Revision Table: Key Concepts for Direction Cosines

Additional Information on Lines in 3D

The equation of a line in 3D space can be represented in different forms:

1. Vector Form: \(\vec{r} = \vec{a} + \lambda \vec{v}\), where \(\vec{r}\) is the position vector of any point on the line, \(\vec{a}\) is the position vector of a known point on the line, \(\vec{v}\) is a vector parallel to the line (direction vector), and \(\lambda\) is a scalar parameter.
2. Cartesian Form (Symmetric): \(\dfrac{x - x_0}{a} = \dfrac{y - y_0}{b} = \dfrac{z - z_0}{c}\), where \((x_0, y_0, z_0)\) is a point on the line and \((a, b, c)\) are the direction ratios.
3. Cartesian Form (Non-symmetric): This form comes from the intersection of two planes. The given form \(x - 1 = 2(y + 3) = 1 - z\) is a variation that can be converted to the symmetric form.

Understanding how to convert between these forms is crucial for solving problems involving lines in three-dimensional geometry.