if `veca` and `vecb` non-zero ad non-collinear vectors, then

Options

A. `vecaxxvecb=[vecavecbhati]hati+[vecavecbhatj]hatj+[vecavecbhatk]hatk`
B. veca.vecb=(veca.hati)(vecb.hati)+(veca.hatj)(vecb.hatj)+(veca.hatk)(vecb.hatk)`
C. if `vecu=hata-(hata.hatb)hatb` and `vecv=hataxxhatb` then `|vecv|=|vecu|`
D. if `vecc=vecaxx(vecaxxvecb)`, then `vecc.veca=0`

Explanation

To solve the problem regarding the non-zero and non-collinear vectors \(\vec{a}\) and \(\vec{b}\), we will analyze the properties of vector operations, specifically the cross product and dot product.

### Step-by-Step Solution:

1. **Understanding Non-Collinear Vectors**:
- Non-collinear vectors \(\vec{a}\) and \(\vec{b}\) means they do not lie on the same line. This implies that their cross product \(\vec{a} \times \vec{b}\) will not be zero.

2. **Cross Product**:
- The cross product of two vectors \(\vec{a}\) and \(\vec{b}\) is given by:
\[ \vec{a} \times \vec{b} = |\vec{a}||\vec{b}|\sin(\theta) \hat{n} \]
where \(\theta\) is the angle between the two vectors and \(\hat{n}\) is the unit vector perpendicular to the plane containing \(\vec{a}\) and \(\vec{b}\).

3. **Dot Product**:
- The dot product of two vectors \(\vec{a}\) and \(\vec{b}\) is given by:
\[ \vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos(\theta) \]
This product gives a scalar value and is zero if the vectors are orthogonal.

4. **Verifying Options**:
- **Option A**: The expression \(\vec{a} \times \vec{b} = \vec{a}b_1 \hat{i} + \vec{a}b_2 \hat{j} + \vec{a}b_3 \hat{k}\) can be verified by calculating the components of the cross product using the determinant of a matrix formed by the unit vectors and the components of \(\vec{a}\) and \(\vec{b}\).
- **Option B**: The dot product \(\vec{a} \cdot \vec{b}\) can be computed using the components of \(\vec{a}\) and \(\vec{b}\) and confirming that it equals \(a_1b_1 + a_2b_2 + a_3b_3\).
- **Option C**: For the vectors defined as \(u = \hat{a} - \hat{a} \cdot \hat{b} \hat{b}\) and \(v = \hat{a} \times \hat{b}\), we can show that the magnitudes of \(u\) and \(v\) are equal by using the properties of sine and cosine functions in vector projections.
- **Option D**: The expression \(c = \vec{a} \times (\vec{a} \times \vec{b})\) can be simplified using the vector triple product identity, and we can show that \(c \cdot \vec{a} = 0\).

5. **Conclusion**:
- After verifying all options, we conclude that all options (A, B, C, D) are correct.