The scalars l and m such that `lveca + m vecb =vecc, " where " veca, vecb and vecc` are given vectors, are equal to

Options

A. `l= ((veccxxvecb).(vecaxxvecb))/((vecaxxvecb)^(2))`
B. `l = ((vecc xx veca). (vecb xxveca))/((vecbxxveca))`
C. `m = ((veccxx veca). (vecbxxveca))/((vecb xx veca)^(2))`
D. `m = ((veccxx veca). (vecbxxveca))/((vecb xx veca))`

Explanation

To solve the problem of finding the scalars \( l \) and \( m \) such that \( l \vec{a} + m \vec{b} = \vec{c} \), we can use the properties of vector cross products. Here’s a step-by-step solution:

### Step 1: Set up the equation
We start with the equation given in the problem:
\[ l \vec{a} + m \vec{b} = \vec{c} \]

### Step 2: Cross with vector \( \vec{b} \)
To isolate \( l \), we can cross both sides of the equation with vector \( \vec{b} \):
\[ (l \vec{a} + m \vec{b}) \times \vec{b} = \vec{c} \times \vec{b} \]
Using the distributive property of the cross product, this simplifies to:
\[ l (\vec{a} \times \vec{b}) + m (\vec{b} \times \vec{b}) = \vec{c} \times \vec{b} \]
Since \( \vec{b} \times \vec{b} = \vec{0} \), we have:
\[ l (\vec{a} \times \vec{b}) = \vec{c} \times \vec{b} \]

### Step 3: Solve for \( l \)
Now, we can express \( l \) as:
\[ l = \frac{\vec{c} \times \vec{b}}{\vec{a} \times \vec{b}} \]

### Step 4: Cross with vector \( \vec{a} \)
Next, we cross both sides of the original equation with vector \( \vec{a} \):
\[ (l \vec{a} + m \vec{b}) \times \vec{a} = \vec{c} \times \vec{a} \]
This simplifies to:
\[ l (\vec{a} \times \vec{a}) + m (\vec{b} \times \vec{a}) = \vec{c} \times \vec{a} \]
Since \( \vec{a} \times \vec{a} = \vec{0} \), we have:
\[ m (\vec{b} \times \vec{a}) = \vec{c} \times \vec{a} \]

### Step 5: Solve for \( m \)
Now, we can express \( m \) as:
\[ m = \frac{\vec{c} \times \vec{a}}{\vec{b} \times \vec{a}} \]

### Final Result
Thus, the scalars \( l \) and \( m \) are given by:
\[ l = \frac{\vec{c} \times \vec{b}}{\vec{a} \times \vec{b}} \quad \text{and} \quad m = \frac{\vec{c} \times \vec{a}}{\vec{b} \times \vec{a}} \]