If the vectors `veca=hati-hatj+2hatk.vecb=2hati+4hatj+hatk and veccc=lambdahati+hatj+muhatk` are mutually orthogonal, then `(lambda, mu)`

Options

A. `(-3, 2)`
B. `(2,-3)`
C. `(-2, 3)`
D. `(3, -2)`

Answer

To solve the problem, we need to find the values of \( \lambda \) and \( \mu \) such that the vectors \( \vec{a} \), \( \vec{b} \), and \( \vec{c} \) are mutually orthogonal. This means that the dot products \( \vec{a} \cdot \vec{b} = 0 \), \( \vec{b} \cdot \vec{c} = 0 \), and \( \vec{c} \cdot \vec{a} = 0 \).

### Step 1: Write down the vectors
Given:
- \( \vec{a} = \hat{i} - \hat{j} + 2\hat{k} \)
- \( \vec{b} = 2\hat{i} + 4\hat{j} + \hat{k} \)
- \( \vec{c} = \lambda \hat{i} + \hat{j} + \mu \hat{k} \)

### Step 2: Find \( \vec{a} \cdot \vec{c} = 0 \)
Calculate the dot product:
\[ \vec{a} \cdot \vec{c} = (\hat{i} - \hat{j} + 2\hat{k}) \cdot (\lambda \hat{i} + \hat{j} + \mu \hat{k}) \]
\[ = 1 \cdot \lambda + (-1) \cdot 1 + 2 \cdot \mu \]
\[ = \lambda - 1 + 2\mu \]
Setting this equal to zero gives us:
\[ \lambda + 2\mu - 1 = 0 \quad \text{(Equation 1)} \]

### Step 3: Find \( \vec{b} \cdot \vec{c} = 0 \)
Calculate the dot product:
\[ \vec{b} \cdot \vec{c} = (2\hat{i} + 4\hat{j} + \hat{k}) \cdot (\lambda \hat{i} + \hat{j} + \mu \hat{k}) \]
\[ = 2\lambda + 4 \cdot 1 + 1 \cdot \mu \]
\[ = 2\lambda + 4 + \mu \]
Setting this equal to zero gives us:
\[ 2\lambda + \mu + 4 = 0 \quad \text{(Equation 2)} \]

### Step 4: Solve the system of equations
We now have two equations:
1. \( \lambda + 2\mu = 1 \)
2. \( 2\lambda + \mu = -4 \)

From Equation 1, we can express \( \lambda \) in terms of \( \mu \):
\[ \lambda = 1 - 2\mu \]

Substituting this into Equation 2:
\[ 2(1 - 2\mu) + \mu = -4 \]
\[ 2 - 4\mu + \mu = -4 \]
\[ 2 - 3\mu = -4 \]
\[ -3\mu = -6 \]
\[ \mu = 2 \]

### Step 5: Substitute back to find \( \lambda \)
Now substitute \( \mu = 2 \) back into Equation 1:
\[ \lambda + 2(2) = 1 \]
\[ \lambda + 4 = 1 \]
\[ \lambda = 1 - 4 = -3 \]

### Final Result
Thus, the values of \( \lambda \) and \( \mu \) are:
\[ (\lambda, \mu) = (-3, 2) \]