If `vecr and vecs` are non-zero constant vectors and the scalar b is chosen such that `|vecr+bvecs|` is minimum, then the value of `|bvecs|^(2)+|vecr+bvecs|^(2)` is equal to

Options

A. `2|vecr|^(2)`
B. `|vecr|^(2)//2`
C. `3|vecr|^(2)`
D. `|vecr|^(2)`

Explanation

To solve the problem, we need to find the value of \( |b \vec{s}|^2 + |\vec{r} + b \vec{s}|^2 \) given that \( |\vec{r} + b \vec{s}| \) is minimized.

### Step-by-step Solution:

1. **Understanding the Condition for Minimum**:
We are given that \( |\vec{r} + b \vec{s}| \) is minimized. The minimum value of a vector's magnitude is zero, which occurs when the vector itself is zero. Therefore, we set:
\[ |\vec{r} + b \vec{s}| = 0 \]
This implies:
\[ \vec{r} + b \vec{s} = \vec{0} \quad \Rightarrow \quad \vec{r} = -b \vec{s} \]

2. **Substituting into the Expression**:
We need to evaluate:
\[ |b \vec{s}|^2 + |\vec{r} + b \vec{s}|^2 \]
Since we have established that \( \vec{r} + b \vec{s} = \vec{0} \), we can substitute this into the expression:
\[ |\vec{r} + b \vec{s}|^2 = |\vec{0}|^2 = 0 \]
Thus, we can simplify our expression to:
\[ |b \vec{s}|^2 + 0 = |b \vec{s}|^2 \]

3. **Calculating \( |b \vec{s}|^2 \)**:
The magnitude of \( b \vec{s} \) can be expressed as:
\[ |b \vec{s}|^2 = b^2 |\vec{s}|^2 \]

4. **Finding \( |\vec{r}|^2 \)**:
From the earlier step, we know that \( \vec{r} = -b \vec{s} \). Therefore, the magnitude of \( \vec{r} \) is:
\[ |\vec{r}|^2 = |-b \vec{s}|^2 = b^2 |\vec{s}|^2 \]

5. **Final Result**:
Since we have \( |b \vec{s}|^2 = b^2 |\vec{s}|^2 \), we can conclude that:
\[ |b \vec{s}|^2 + |\vec{r} + b \vec{s}|^2 = b^2 |\vec{s}|^2 + 0 = b^2 |\vec{s}|^2 \]
Therefore, the final value is:
\[ |b \vec{s}|^2 + |\vec{r} + b \vec{s}|^2 = |\vec{r}|^2 \]

### Conclusion:
The value of \( |b \vec{s}|^2 + |\vec{r} + b \vec{s}|^2 \) is equal to \( |\vec{r}|^2 \).