Let `veca,vecb,vecc` be three unit vectors such that `3veca+4vecb+5vecc=vec0`. Then which of the following statements is true? (A) `veca` is parrallel to vecb` (B) `veca` is perpendicular to vecb` (C) `veca` is neither parralel nor perpendicular to `vecb` (D) `veca,vecb,vecc` are copalanar

Options

A. `veca` is parallel to `vecb`
B. `veca` is perpendicular to `vecb`
C. `veca` is neither parallel nor perpendicular to `vecb`
D. none of these

Explanation

To solve the problem, we start with the equation given:

\[ 3\vec{a} + 4\vec{b} + 5\vec{c} = \vec{0} \]

### Step 1: Rearranging the Equation
We can rearrange the equation to express one vector in terms of the others. Let's isolate \(\vec{c}\):

\[ 5\vec{c} = - (3\vec{a} + 4\vec{b}) \]

Now, dividing both sides by 5, we get:

\[ \vec{c} = -\frac{3}{5}\vec{a} - \frac{4}{5}\vec{b} \]

### Step 2: Analyzing the Vectors
Since \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) are unit vectors, we know that:

\[ |\vec{a}| = 1, \quad |\vec{b}| = 1, \quad |\vec{c}| = 1 \]

### Step 3: Understanding Collinearity
The equation \(3\vec{a} + 4\vec{b} + 5\vec{c} = 0\) implies that the vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) are linearly dependent. This means that they lie in the same plane, which is a property of coplanarity.

### Step 4: Conclusion About the Options
From the analysis, we see that:

- The vectors are not guaranteed to be parallel because they are not scalar multiples of each other.
- The vectors are not guaranteed to be perpendicular because there is no information indicating that the dot product is zero.
- The only conclusion we can draw is that the vectors are coplanar since they are linearly dependent.

Thus, the correct answer is:

**(D) \(\vec{a}, \vec{b}, \vec{c}\) are coplanar.**