Find the value of `lamda` such that the vectors `vec(a)=2hat(i)+lamda hat(j)+hat(k)` and `vec(b)=hat(i)+2hat(j)+3hat(k)` are orthogonal.

Options

A. 0
B. 1
C. `3/2`
D. `(-5)/(2)`

Answer

To find the value of \( \lambda \) such that the vectors \( \vec{a} = 2\hat{i} + \lambda \hat{j} + \hat{k} \) and \( \vec{b} = \hat{i} + 2\hat{j} + 3\hat{k} \) are orthogonal, we need to use the property that two vectors are orthogonal if their dot product is zero.

### Step-by-step Solution:

1. **Write down the vectors**:
\[ \vec{a} = 2\hat{i} + \lambda \hat{j} + \hat{k} \]
\[ \vec{b} = \hat{i} + 2\hat{j} + 3\hat{k} \]

2. **Calculate the dot product**:
The dot product \( \vec{a} \cdot \vec{b} \) is calculated as follows:
\[ \vec{a} \cdot \vec{b} = (2\hat{i} + \lambda \hat{j} + \hat{k}) \cdot (\hat{i} + 2\hat{j} + 3\hat{k}) \]

Using the distributive property of the dot product:
\[ \vec{a} \cdot \vec{b} = 2 \cdot 1 + \lambda \cdot 2 + 1 \cdot 3 \]

3. **Simplify the expression**:
\[ \vec{a} \cdot \vec{b} = 2 + 2\lambda + 3 \]
\[ \vec{a} \cdot \vec{b} = 5 + 2\lambda \]

4. **Set the dot product to zero** (since the vectors are orthogonal):
\[ 5 + 2\lambda = 0 \]

5. **Solve for \( \lambda \)**:
\[ 2\lambda = -5 \]
\[ \lambda = -\frac{5}{2} \]

### Final Answer:
The value of \( \lambda \) is \( -\frac{5}{2} \).

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