If the vectors `lambda bar(i)-3bar(j)+5bar(k), and 2lambdabar(i)-lambda(j)-bar(k)`, are perpendicular to each other find `lambda`.

Answer

To find the value of \( \lambda \) for which the vectors \( \lambda \mathbf{i} - 3 \mathbf{j} + 5 \mathbf{k} \) and \( 2\lambda \mathbf{i} - \lambda \mathbf{j} - \mathbf{k} \) are perpendicular, we can follow these steps:

### Step 1: Understand the condition for perpendicular vectors
Two vectors are perpendicular if their dot product is zero.

### Step 2: Write down the vectors
Let:
\[ \mathbf{A} = \lambda \mathbf{i} - 3 \mathbf{j} + 5 \mathbf{k} \]
\[ \mathbf{B} = 2\lambda \mathbf{i} - \lambda \mathbf{j} - \mathbf{k} \]

### Step 3: Set up the dot product
The dot product \( \mathbf{A} \cdot \mathbf{B} \) is given by:
\[ \mathbf{A} \cdot \mathbf{B} = (\lambda)(2\lambda) + (-3)(-\lambda) + (5)(-1) \]

### Step 4: Calculate the dot product
Calculating each term:
- The \( \mathbf{i} \) components: \( \lambda \cdot 2\lambda = 2\lambda^2 \)
- The \( \mathbf{j} \) components: \( -3 \cdot -\lambda = 3\lambda \)
- The \( \mathbf{k} \) components: \( 5 \cdot -1 = -5 \)

Putting it all together:
\[ \mathbf{A} \cdot \mathbf{B} = 2\lambda^2 + 3\lambda - 5 \]

### Step 5: Set the dot product to zero
Since the vectors are perpendicular:
\[ 2\lambda^2 + 3\lambda - 5 = 0 \]

### Step 6: Solve the quadratic equation
To solve the quadratic equation \( 2\lambda^2 + 3\lambda - 5 = 0 \), we can use the factorization method or the quadratic formula. Here, we will factor it.

Rearranging gives:
\[ 2\lambda^2 + 5\lambda - 2\lambda - 5 = 0 \]

Grouping terms:
\[ (2\lambda^2 + 5\lambda) + (-2\lambda - 5) = 0 \]

Factoring by grouping:
\[ \lambda(2\lambda + 5) - 1(2\lambda + 5) = 0 \]

Factoring out \( (2\lambda + 5) \):
\[ (2\lambda + 5)(\lambda - 1) = 0 \]

### Step 7: Find the values of \( \lambda \)
Setting each factor to zero gives:
1. \( 2\lambda + 5 = 0 \) → \( \lambda = -\frac{5}{2} \)
2. \( \lambda - 1 = 0 \) → \( \lambda = 1 \)

Thus, the values of \( \lambda \) are:
\[ \lambda = -\frac{5}{2} \quad \text{or} \quad \lambda = 1 \]