Let `lambda` be any non - zero scalar. Then for what possible values of x, y and z given below, the vectors `2hat(i)-3hat(j)+4hat(k)` and `x hat(i)-y hat(j)-z hat(k)` are perpendicular :

Options

A. `x=2lambda, y=lambda, z =lambda`
B. `x = lambda, y=2lambda, z=-lambda`
C. `x=-lambda, y=2lambda, z=lambda`
D. `x=-lambda, y=-2lambda, z=lambda`

Answer

To determine the values of \( x, y, \) and \( z \) such that the vectors \( \mathbf{A} = 2\hat{i} - 3\hat{j} + 4\hat{k} \) and \( \mathbf{B} = x\hat{i} - y\hat{j} - z\hat{k} \) are perpendicular, we need to use the property that two vectors are perpendicular if their dot product is zero.

### Step-by-Step Solution:

1. **Write the dot product of the vectors**:
\[ \mathbf{A} \cdot \mathbf{B} = (2\hat{i} - 3\hat{j} + 4\hat{k}) \cdot (x\hat{i} - y\hat{j} - z\hat{k}) \]

2. **Calculate the dot product**:
\[ \mathbf{A} \cdot \mathbf{B} = 2x + (-3)(-y) + 4(-z) = 2x + 3y - 4z \]

3. **Set the dot product equal to zero**:
\[ 2x + 3y - 4z = 0 \]

4. **Substitute the values of \( x, y, z \) from the options into the equation**:
- **Option 1**: \( x = 2\lambda, y = \lambda, z = \lambda \)
\[ 2(2\lambda) + 3(\lambda) - 4(\lambda) = 4\lambda + 3\lambda - 4\lambda = 3\lambda \neq 0 \quad (\text{Not valid}) \]

- **Option 2**: \( x = \lambda, y = 2\lambda, z = \lambda \)
\[ 2(\lambda) + 3(2\lambda) - 4(\lambda) = 2\lambda + 6\lambda - 4\lambda = 4\lambda \neq 0 \quad (\text{Not valid}) \]

- **Option 3**: \( x = -\lambda, y = 2\lambda, z = \lambda \)
\[ 2(-\lambda) + 3(2\lambda) - 4(\lambda) = -2\lambda + 6\lambda - 4\lambda = 0 \quad (\text{Valid}) \]

- **Option 4**: \( x = -\lambda, y = -2\lambda, z = \lambda \)
\[ 2(-\lambda) + 3(-2\lambda) - 4(\lambda) = -2\lambda - 6\lambda - 4\lambda = -12\lambda \neq 0 \quad (\text{Not valid}) \]

5. **Conclusion**:
The only valid option where the vectors are perpendicular is **Option 3**: \( x = -\lambda, y = 2\lambda, z = \lambda \).