A unit vector parallel to vector `vecV=(2,-3,6)` is vector

Options

A. `(-2,3,-6)`
B. `(6,-3,2)`
C. `(-0.29,0.43,-0.86)`
D. `(0.29,0.43,-0.86)`

Explanation

To find a unit vector parallel to the vector \(\vec{V} = (2, -3, 6)\), we will follow these steps:

### Step 1: Calculate the Magnitude of the Vector
The magnitude of a vector \(\vec{V} = (a, b, c)\) is given by the formula:
\[ |\vec{V}| = \sqrt{a^2 + b^2 + c^2} \]
For our vector \(\vec{V} = (2, -3, 6)\):
\[ |\vec{V}| = \sqrt{2^2 + (-3)^2 + 6^2} \]
Calculating each term:
\[ = \sqrt{4 + 9 + 36} \]
\[ = \sqrt{49} \]
\[ = 7 \]

### Step 2: Find the Unit Vector
A unit vector in the direction of \(\vec{V}\) is given by:
\[ \hat{V} = \frac{\vec{V}}{|\vec{V}|} \]
Substituting the values we have:
\[ \hat{V} = \frac{1}{7} (2, -3, 6) \]
This gives us:
\[ \hat{V} = \left(\frac{2}{7}, \frac{-3}{7}, \frac{6}{7}\right) \]

### Step 3: Calculate the Components
Calculating the components:
\[ \frac{2}{7} \approx 0.29, \quad \frac{-3}{7} \approx -0.43, \quad \frac{6}{7} \approx 0.86 \]
Thus, the unit vector is:
\[ \hat{V} \approx (0.29, -0.43, 0.86) \]

### Step 4: Consider Both Directions
Since a unit vector can point in either direction, we also consider the negative of the unit vector:
\[ -\hat{V} = \left(-\frac{2}{7}, \frac{3}{7}, -\frac{6}{7}\right) \approx (-0.29, 0.43, -0.86) \]

### Conclusion
The two unit vectors parallel to \(\vec{V}\) are:
1. \( (0.29, -0.43, 0.86) \)
2. \( (-0.29, 0.43, -0.86) \)

Among the options given, option 3, \((-0.29, 0.43, -0.86)\), is the correct answer.