If `vec(a)` is a vector and x is a non-zero scalar, then

Options

A. x `vec(a)` is a vector in the direction of `vec(a)`
B. x `vec(a)` is a vector collinear to `vec(a)`
C. x `vec(a)` and `vec(a)` have independent directions
D. None of these

Answer

To solve the question, we need to analyze the properties of a vector when it is multiplied by a non-zero scalar. Let's denote the vector as \(\vec{A}\) and the scalar as \(x\).

### Step-by-Step Solution:

1. **Understanding the Vector and Scalar**:
- A vector \(\vec{A}\) has both magnitude and direction.
- A scalar \(x\) is a real number that can stretch or compress the vector but does not change its direction.

2. **Statement 1: \(x\vec{A}\) is in the direction of \(\vec{A}\)**:
- When we multiply the vector \(\vec{A}\) by the scalar \(x\), the direction of the resulting vector \(x\vec{A}\) remains the same as that of \(\vec{A}\) if \(x > 0\).
- If \(x < 0\), the direction of \(x\vec{A}\) will be opposite to that of \(\vec{A}\).
- Therefore, statement 1 is **true**.

3. **Statement 2: \(x\vec{A}\) and \(\vec{A}\) are collinear**:
- Two vectors are collinear if they lie along the same line, which means they have the same or opposite direction.
- Since multiplying \(\vec{A}\) by a non-zero scalar \(x\) (whether positive or negative) does not change the line along which the vector points, \(x\vec{A}\) and \(\vec{A}\) are indeed collinear.
- Thus, statement 2 is **true**.

4. **Statement 3: \(x\vec{A}\) and \(\vec{A}\) have independent direction**:
- This statement suggests that the direction of \(x\vec{A}\) is not related to the direction of \(\vec{A}\).
- However, as established, multiplying a vector by a scalar does not change its direction; it only changes its magnitude.
- Therefore, statement 3 is **false**.

### Conclusion:
- The correct options are:
- Statement 1 is true.
- Statement 2 is true.
- Statement 3 is false.

### Final Answer:
The correct statements are 1 and 2.

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