A unit vector parallel to the straight line `(x-2)/3 =(3+y)/(-1)=(z-2)/(-4)` is

Options

A. `1/sqrt26(3hati-hatj+4hatk)`
B. `1/sqrt26(hati+3hatj-hatk)`
C. `1/sqrt26(3hati-hatj-4hatk)`
D. `1/sqrt26(3hati+hatj+4hatk)`

Answer

To find a unit vector parallel to the straight line given by the equation \((x-2)/3 = (3+y)/(-1) = (z-2)/(-4)\), we will follow these steps:

### Step 1: Identify the direction ratios
The equation of the line can be expressed in the form:
\[ \frac{x - 2}{3} = \frac{3 + y}{-1} = \frac{z - 2}{-4} \]
From this, we can identify the direction ratios of the line, which are the coefficients in the denominators. The direction ratios are:
- For \(x\): \(3\)
- For \(y\): \(-1\)
- For \(z\): \(-4\)

Thus, the direction ratios of the line are \(3, -1, -4\).

### Step 2: Write the direction vector
The direction vector \( \mathbf{a} \) parallel to the line can be expressed in vector form as:
\[ \mathbf{a} = 3\mathbf{i} - 1\mathbf{j} - 4\mathbf{k} \]

### Step 3: Calculate the magnitude of the direction vector
To find the unit vector, we first need to calculate the magnitude (or modulus) of the vector \( \mathbf{a} \):
\[ |\mathbf{a}| = \sqrt{(3)^2 + (-1)^2 + (-4)^2} \]
Calculating this:
\[ |\mathbf{a}| = \sqrt{9 + 1 + 16} = \sqrt{26} \]

### Step 4: Find the unit vector
The unit vector \( \mathbf{u} \) in the direction of \( \mathbf{a} \) is given by:
\[ \mathbf{u} = \frac{\mathbf{a}}{|\mathbf{a}|} \]
Substituting the values we found:
\[ \mathbf{u} = \frac{3\mathbf{i} - 1\mathbf{j} - 4\mathbf{k}}{\sqrt{26}} \]

### Final Answer
Thus, the unit vector parallel to the given straight line is:
\[ \mathbf{u} = \frac{3}{\sqrt{26}} \mathbf{i} - \frac{1}{\sqrt{26}} \mathbf{j} - \frac{4}{\sqrt{26}} \mathbf{k} \]