If `vec(r_(1))=lambda vec(i)+2hat(j)+hat(k), vec(r_(2))=hat(i)+(2-lambda)hat(j)+2hat(k)` are such that `|vec(r_(1))| gt |vec(r_(2))|`, then `lambda` satisfies which one of the following?

Options

A. `lambda=0 ` only
B. `lambda=1`
C. `lambda lt 1`
D. `lambda lt 1`

Explanation

To solve the problem, we need to find the values of \( \lambda \) such that the magnitude of vector \( \vec{r_1} \) is greater than the magnitude of vector \( \vec{r_2} \).

### Step-by-step Solution:

1. **Write the vectors**:
- Given \( \vec{r_1} = \lambda \hat{i} + 2 \hat{j} + \hat{k} \)
- Given \( \vec{r_2} = \hat{i} + (2 - \lambda) \hat{j} + 2 \hat{k} \)

2. **Calculate the magnitudes**:
- The magnitude of \( \vec{r_1} \) is calculated as:
\[ |\vec{r_1}| = \sqrt{\lambda^2 + 2^2 + 1^2} = \sqrt{\lambda^2 + 4 + 1} = \sqrt{\lambda^2 + 5} \]
- The magnitude of \( \vec{r_2} \) is calculated as:
\[ |\vec{r_2}| = \sqrt{1^2 + (2 - \lambda)^2 + 2^2} = \sqrt{1 + (2 - \lambda)^2 + 4} \]
Expanding \( (2 - \lambda)^2 \):
\[ (2 - \lambda)^2 = 4 - 4\lambda + \lambda^2 \]
Therefore,
\[ |\vec{r_2}| = \sqrt{1 + 4 - 4\lambda + \lambda^2 + 4} = \sqrt{\lambda^2 - 4\lambda + 9} \]

3. **Set up the inequality**:
- We need to solve the inequality:
\[ |\vec{r_1}| > |\vec{r_2}| \]
This translates to:
\[ \sqrt{\lambda^2 + 5} > \sqrt{\lambda^2 - 4\lambda + 9} \]

4. **Square both sides**:
- Squaring both sides (since both sides are positive):
\[ \lambda^2 + 5 > \lambda^2 - 4\lambda + 9 \]

5. **Simplify the inequality**:
- Cancel \( \lambda^2 \) from both sides:
\[ 5 > -4\lambda + 9 \]
- Rearranging gives:
\[ 4\lambda > 9 - 5 \]
\[ 4\lambda > 4 \]
- Dividing by 4:
\[ \lambda > 1 \]

### Conclusion:
Thus, the value of \( \lambda \) satisfies the condition \( \lambda > 1 \).