If `vec(c) =2lambda (vec(a)timesvec(b))+3mu (vec(b)times vec(a)),vec(a)xx vec(b) ne0,vec(c).(vec(a)times vec(b))=0 `then

Options

A. `lambda = 3 mu`
B. `2lambda = 3 mu`
C. `lambda + mu = 0 `
D. None of these

Answer

To solve the problem step by step, we start with the given vector equation and conditions.

### Step 1: Write down the given equation for vector \( \vec{c} \)
We have:
\[ \vec{c} = 2\lambda (\vec{a} \times \vec{b}) + 3\mu (\vec{b} \times \vec{a}) \]
Using the property of cross products, we know that:
\[ \vec{b} \times \vec{a} = -(\vec{a} \times \vec{b}) \]
Thus, we can rewrite \( \vec{c} \) as:
\[ \vec{c} = 2\lambda (\vec{a} \times \vec{b}) - 3\mu (\vec{a} \times \vec{b}) = (2\lambda - 3\mu)(\vec{a} \times \vec{b}) \]

### Step 2: Use the condition \( \vec{c} \cdot (\vec{a} \times \vec{b}) = 0 \)
We know that:
\[ \vec{c} \cdot (\vec{a} \times \vec{b}) = 0 \]
Substituting our expression for \( \vec{c} \):
\[ ((2\lambda - 3\mu)(\vec{a} \times \vec{b})) \cdot (\vec{a} \times \vec{b}) = 0 \]

### Step 3: Simplify the dot product
The dot product \( (\vec{a} \times \vec{b}) \cdot (\vec{a} \times \vec{b}) \) gives us the square of the magnitude of \( \vec{a} \times \vec{b} \):
\[ |\vec{a} \times \vec{b}|^2 \cdot (2\lambda - 3\mu) = 0 \]

### Step 4: Analyze the equation
Since we know that \( \vec{a} \times \vec{b} \neq 0 \), it follows that:
\[ |\vec{a} \times \vec{b}|^2 \neq 0 \]
Thus, the only way for the product to equal zero is if:
\[ 2\lambda - 3\mu = 0 \]

### Step 5: Solve for the relationship between \( \lambda \) and \( \mu \)
From the equation \( 2\lambda - 3\mu = 0 \), we can rearrange it to find:
\[ 2\lambda = 3\mu \]
This gives us the relationship between \( \lambda \) and \( \mu \).

### Final Answer
The relationship derived from the given conditions is:
\[ \boxed{2\lambda = 3\mu} \]