If `vec(r )_(1)=lamda hat(i) + 2hat(j) + hat(k). vec(r )_(2)= hat(i) + (2 - lamda) hat(j) +2hat(k)` are such that `|vec(r )_(1)| gt |vec(r )_(2)|` then `lamda` satisfies which one of the following?

Options

A. `lamda = 0`
B. `lamda =1`
C. `lamda lt 1`
D. `lamda gt 1`

Answer

To solve the problem, we need to determine 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. **Define the vectors**:
\[ \vec{r}_1 = \lambda \hat{i} + 2 \hat{j} + \hat{k} \]
\[ \vec{r}_2 = \hat{i} + (2 - \lambda) \hat{j} + 2 \hat{k} \]

2. **Calculate the magnitudes of the vectors**:
The magnitude of \( \vec{r}_1 \) is given by:
\[ |\vec{r}_1| = \sqrt{\lambda^2 + 2^2 + 1^2} = \sqrt{\lambda^2 + 4 + 1} = \sqrt{\lambda^2 + 5} \]

The magnitude of \( \vec{r}_2 \) is given by:
\[ |\vec{r}_2| = \sqrt{1^2 + (2 - \lambda)^2 + 2^2} = \sqrt{1 + (2 - \lambda)^2 + 4} \]
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** (since both sides are positive):
\[ \lambda^2 + 5 > \lambda^2 - 4\lambda + 9 \]

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

### Conclusion
The value of \( \lambda \) must satisfy:
\[ \lambda < 1 \]