The d.C.s of the line 3x-1=2y-3 = 4z-1 are

Options

A. `-2/sqrt(61), 3/sqrt(61), 4/sqrt(61)`
B. `1/sqrt(61), 4/sqrt(61), 3/sqrt(61)`
C. `4/sqrt(61), 6/sqrt(61), 3/sqrt(61)`
D. `1/sqrt(61), -2/sqrt(61), 3/sqrt(61)`

Answer

To find the direction cosines of the line given by the equation \(3x - 1 = 2y - 3 = 4z - 1\), we will follow these steps:

### Step 1: Rewrite the equation in terms of a parameter
We start with the equation:
\[ 3x - 1 = 2y - 3 = 4z - 1 \]
Let us denote this common value as \(k\). Therefore, we can express \(x\), \(y\), and \(z\) in terms of \(k\):
\[ 3x - 1 = k \implies x = \frac{k + 1}{3} \]
\[ 2y - 3 = k \implies y = \frac{k + 3}{2} \]
\[ 4z - 1 = k \implies z = \frac{k + 1}{4} \]

### Step 2: Identify the direction ratios
From the equations for \(x\), \(y\), and \(z\), we can express them in terms of \(k\):
- For \(x\): \(x = \frac{1}{3}k + \frac{1}{3}\)
- For \(y\): \(y = \frac{1}{2}k + \frac{3}{2}\)
- For \(z\): \(z = \frac{1}{4}k + \frac{1}{4}\)

The coefficients of \(k\) give us the direction ratios of the line:
\[ \text{Direction Ratios} = (3, 2, 4) \]

### Step 3: Normalize the direction ratios to find direction cosines
To find the direction cosines, we need to normalize the direction ratios. The direction cosines are given by:
\[ l = \frac{a}{\sqrt{a^2 + b^2 + c^2}}, \quad m = \frac{b}{\sqrt{a^2 + b^2 + c^2}}, \quad n = \frac{c}{\sqrt{a^2 + b^2 + c^2}} \]
where \(a\), \(b\), and \(c\) are the direction ratios.

Calculating the magnitude:
\[ \sqrt{3^2 + 2^2 + 4^2} = \sqrt{9 + 4 + 16} = \sqrt{29} \]

Now, substituting the values:
\[ l = \frac{3}{\sqrt{29}}, \quad m = \frac{2}{\sqrt{29}}, \quad n = \frac{4}{\sqrt{29}} \]

### Final Result
Thus, the direction cosines of the line are:
\[ \left( \frac{3}{\sqrt{29}}, \frac{2}{\sqrt{29}}, \frac{4}{\sqrt{29}} \right) \]