Let `|A|=1, |vecb|=4` and `veca xxvecr +vecb=vecr`. If the projection of `vecr` along `veca` is 2, then the projection of `vecr` along `vecb` is

Answer

To solve the problem step by step, we start with the given information:

1. **Given Information**:
- \(|\vec{A}| = 1\)
- \(|\vec{B}| = 4\)
- \(\vec{A} \times \vec{R} + \vec{B} = \vec{R}\)
- The projection of \(\vec{R}\) along \(\vec{A}\) is 2.

2. **Understanding Projections**:
The projection of vector \(\vec{R}\) along vector \(\vec{A}\) is given by:
\[ \text{Projection of } \vec{R} \text{ along } \vec{A} = \frac{\vec{R} \cdot \vec{A}}{|\vec{A}|} \]
Since \(|\vec{A}| = 1\), this simplifies to:
\[ \text{Projection of } \vec{R} \text{ along } \vec{A} = \vec{R} \cdot \vec{A} \]
Given that this projection equals 2, we have:
\[ \vec{R} \cdot \vec{A} = 2 \]

3. **Using the Given Relation**:
We have the relation:
\[ \vec{A} \times \vec{R} + \vec{B} = \vec{R} \]
Rearranging gives:
\[ \vec{A} \times \vec{R} = \vec{R} - \vec{B} \]

4. **Dotting with \(\vec{A}\)**:
Now, we dot both sides of the equation \(\vec{A} \times \vec{R} = \vec{R} - \vec{B}\) with \(\vec{A}\):
\[ \vec{A} \cdot (\vec{A} \times \vec{R}) = \vec{A} \cdot \vec{R} - \vec{A} \cdot \vec{B} \]
The left-hand side is zero because the dot product of a vector with a vector perpendicular to it (cross product) is zero. Thus:
\[ 0 = \vec{A} \cdot \vec{R} - \vec{A} \cdot \vec{B} \]
Rearranging gives:
\[ \vec{A} \cdot \vec{B} = \vec{A} \cdot \vec{R} \]
Substituting \(\vec{A} \cdot \vec{R} = 2\):
\[ \vec{A} \cdot \vec{B} = 2 \]

5. **Finding the Projection of \(\vec{R}\) along \(\vec{B}\)**:
We want to find the projection of \(\vec{R}\) along \(\vec{B}\):
\[ \text{Projection of } \vec{R} \text{ along } \vec{B} = \frac{\vec{R} \cdot \vec{B}}{|\vec{B}|} \]
We need to find \(\vec{R} \cdot \vec{B}\). We can use the relation:
\[ \vec{A} \times \vec{R} = \vec{R} - \vec{B} \]
Dotting with \(\vec{B}\):
\[ \vec{A} \times \vec{R} \cdot \vec{B} = \vec{R} \cdot \vec{B} - \vec{B} \cdot \vec{B} \]
The left-hand side is zero because \(\vec{A} \times \vec{R}\) is perpendicular to both \(\vec{A}\) and \(\vec{B}\). Thus:
\[ 0 = \vec{R} \cdot \vec{B} - |\vec{B}|^2 \]
Since \(|\vec{B}| = 4\), we have:
\[ 0 = \vec{R} \cdot \vec{B} - 16 \]
Therefore:
\[ \vec{R} \cdot \vec{B} = 16 \]

6. **Calculating the Projection**:
Now we can calculate the projection of \(\vec{R}\) along \(\vec{B}\):
\[ \text{Projection of } \vec{R} \text{ along } \vec{B} = \frac{\vec{R} \cdot \vec{B}}{|\vec{B}|} = \frac{16}{4} = 4 \]

### Final Answer:
The projection of \(\vec{R}\) along \(\vec{B}\) is \(4\).