Let `a=i-2j+k" and "b=i-j+lambdak`, (where `lambda in Z`) be two vectors. If c is a vector such that `a times b=c timesb, c*a=0" and "2b*c+1=0," then "lambda=`________

Answer

To solve the problem, we need to find the value of \( \lambda \) given the vectors \( \mathbf{a} = \mathbf{i} - 2\mathbf{j} + \mathbf{k} \) and \( \mathbf{b} = \mathbf{i} - \mathbf{j} + \lambda \mathbf{k} \), along with the conditions provided.

### Step 1: Understand the conditions
We have the following conditions:
1. \( \mathbf{a} \times \mathbf{b} = \mathbf{c} \times \mathbf{b} \)
2. \( \mathbf{c} \cdot \mathbf{a} = 0 \)
3. \( 2 \mathbf{b} \cdot \mathbf{c} + 1 = 0 \)

### Step 2: Express \( \mathbf{c} \)
From the first condition, we can rearrange it to find \( \mathbf{c} \):
\[ \mathbf{a} \times \mathbf{b} = \mathbf{c} \times \mathbf{b} \implies \mathbf{a} \times \mathbf{b} - \mathbf{c} \times \mathbf{b} = \mathbf{0} \]
This implies that \( \mathbf{a} - \mathbf{c} \) is parallel to \( \mathbf{b} \). Therefore, we can express \( \mathbf{c} \) as:
\[ \mathbf{c} = \mathbf{a} - \alpha \mathbf{b} \]
for some scalar \( \alpha \).

### Step 3: Use the second condition
Substituting \( \mathbf{c} \) into the second condition \( \mathbf{c} \cdot \mathbf{a} = 0 \):
\[ (\mathbf{a} - \alpha \mathbf{b}) \cdot \mathbf{a} = 0 \]
Expanding this, we get:
\[ \mathbf{a} \cdot \mathbf{a} - \alpha (\mathbf{b} \cdot \mathbf{a}) = 0 \]
Let’s calculate \( \mathbf{a} \cdot \mathbf{a} \):
\[ \mathbf{a} \cdot \mathbf{a} = (1)^2 + (-2)^2 + (1)^2 = 1 + 4 + 1 = 6 \]
Now, we need to calculate \( \mathbf{b} \cdot \mathbf{a} \):
\[ \mathbf{b} \cdot \mathbf{a} = (1)(1) + (-1)(-2) + (\lambda)(1) = 1 + 2 + \lambda = 3 + \lambda \]
Substituting these into the equation:
\[ 6 - \alpha(3 + \lambda) = 0 \implies \alpha(3 + \lambda) = 6 \implies \alpha = \frac{6}{3 + \lambda} \]

### Step 4: Use the third condition
Now, we substitute \( \mathbf{c} \) into the third condition \( 2 \mathbf{b} \cdot \mathbf{c} + 1 = 0 \):
\[ 2 \mathbf{b} \cdot (\mathbf{a} - \alpha \mathbf{b}) + 1 = 0 \]
Expanding this gives:
\[ 2(\mathbf{b} \cdot \mathbf{a}) - 2\alpha(\mathbf{b} \cdot \mathbf{b}) + 1 = 0 \]
We already have \( \mathbf{b} \cdot \mathbf{a} = 3 + \lambda \). Now, let’s calculate \( \mathbf{b} \cdot \mathbf{b} \):
\[ \mathbf{b} \cdot \mathbf{b} = (1)^2 + (-1)^2 + (\lambda)^2 = 1 + 1 + \lambda^2 = 2 + \lambda^2 \]
Substituting these values gives:
\[ 2(3 + \lambda) - 2\left(\frac{6}{3 + \lambda}\right)(2 + \lambda^2) + 1 = 0 \]
This simplifies to:
\[ 6 + 2\lambda - \frac{12 + 6\lambda^2}{3 + \lambda} + 1 = 0 \]
Combining terms leads to:
\[ 7 + 2\lambda - \frac{12 + 6\lambda^2}{3 + \lambda} = 0 \]
Multiplying through by \( 3 + \lambda \) to eliminate the fraction:
\[ (7 + 2\lambda)(3 + \lambda) - (12 + 6\lambda^2) = 0 \]
Expanding this:
\[ 21 + 7\lambda + 6\lambda + 2\lambda^2 - 12 - 6\lambda^2 = 0 \]
This simplifies to:
\[ -4\lambda^2 + 13\lambda + 9 = 0 \]
Rearranging gives:
\[ 10\lambda^2 - 13\lambda + 3 = 0 \]

### Step 5: Solve the quadratic equation
Using the quadratic formula:
\[ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{13 \pm \sqrt{(-13)^2 - 4 \cdot 10 \cdot 3}}{2 \cdot 10} \]
Calculating the discriminant:
\[ \sqrt{169 - 120} = \sqrt{49} = 7 \]
Thus:
\[ \lambda = \frac{13 \pm 7}{20} \]
This gives two possible solutions:
\[ \lambda_1 = \frac{20}{20} = 1, \quad \lambda_2 = \frac{6}{20} = \frac{3}{10} \]

Since \( \lambda \in \mathbb{Z} \), the only valid solution is:
\[ \lambda = 1 \]

### Final Answer
\[ \lambda = 1 \]