Statement 1: Any vector in space can be uniquely written as the linear combination of three non-coplanar vectors.
Stetement 2: If `veca, vecb, vecc` are three non-coplanar vectors and `vecr` is any vector in space then
`[(veca,vecb, vecc)]vecc+[(vecb, vecc, vecr)]veca+[(vecc, veca, vecr)]vecb=[(veca, vecb, vecc)]vecr`

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To solve the given problem, we need to analyze both statements and provide a step-by-step explanation.

### Step 1: Analyze Statement 1
**Statement 1:** Any vector in space can be uniquely written as the linear combination of three non-coplanar vectors.

**Explanation:**
- Let \(\vec{r}\) be any vector in space.
- If \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) are three non-coplanar vectors, then any vector \(\vec{r}\) can be expressed as:
\[ \vec{r} = x \vec{a} + y \vec{b} + z \vec{c} \]
where \(x\), \(y\), and \(z\) are scalars.
- Since \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) are non-coplanar, they span a three-dimensional space, allowing for a unique representation of \(\vec{r}\).

**Conclusion:**
- Therefore, Statement 1 is **True**.

### Step 2: Analyze Statement 2
**Statement 2:** If \(\vec{a}, \vec{b}, \vec{c}\) are three non-coplanar vectors and \(\vec{r}\) is any vector in space, then:
\[ [(\vec{a}, \vec{b}, \vec{c})]\vec{c} + [(\vec{b}, \vec{c}, \vec{r})]\vec{a} + [(\vec{c}, \vec{a}, \vec{r})]\vec{b} = [(\vec{a}, \vec{b}, \vec{c})]\vec{r} \]

**Explanation:**
1. **Understanding the Notation:**
- The notation \([\vec{a}, \vec{b}, \vec{c}]\) represents the scalar triple product, which gives the volume of the parallelepiped formed by the vectors.

2. **Finding the Components:**
- We can express \(\vec{r}\) in terms of the basis formed by \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\):
\[ \vec{r} = x \vec{a} + y \vec{b} + z \vec{c} \]
- We will find the coefficients \(x\), \(y\), and \(z\) using the scalar triple product.

3. **Calculating Each Term:**
- Calculate \((\vec{b} \times \vec{c}) \cdot \vec{r}\) to find \(x\):
\[ x = \frac{(\vec{b} \times \vec{c}) \cdot \vec{r}}{(\vec{a} \times \vec{b}) \cdot \vec{c}} \]
- Calculate \((\vec{c} \times \vec{a}) \cdot \vec{r}\) to find \(y\):
\[ y = \frac{(\vec{c} \times \vec{a}) \cdot \vec{r}}{(\vec{a} \times \vec{b}) \cdot \vec{c}} \]
- Calculate \((\vec{a} \times \vec{b}) \cdot \vec{r}\) to find \(z\):
\[ z = \frac{(\vec{a} \times \vec{b}) \cdot \vec{r}}{(\vec{a} \times \vec{b}) \cdot \vec{c}} \]

4. **Substituting Back:**
- Substitute \(x\), \(y\), and \(z\) back into the equation to verify the equality:
\[ \vec{r} = \frac{(\vec{b} \times \vec{c}) \cdot \vec{r}}{(\vec{a} \times \vec{b}) \cdot \vec{c}} \vec{a} + \frac{(\vec{c} \times \vec{a}) \cdot \vec{r}}{(\vec{a} \times \vec{b}) \cdot \vec{c}} \vec{b} + \frac{(\vec{a} \times \vec{b}) \cdot \vec{r}}{(\vec{a} \times \vec{b}) \cdot \vec{c}} \vec{c} \]

**Conclusion:**
- Thus, Statement 2 is also **True**.

### Final Conclusion
Both Statement 1 and Statement 2 are true, and Statement 1 serves as the explanation for Statement 2.

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