If c is a given non - zero scalar, and `vecA and vecB` are given non- zero , vectors such that `vecA bot vecB` . Then find vector, `vecX` which satisfies the equations `vecA.vecX =c and vecA xx vecX =vecB`.

Explanation

To find the vector \(\vec{X}\) that satisfies the equations \(\vec{A} \cdot \vec{X} = c\) and \(\vec{A} \times \vec{X} = \vec{B}\), we can follow these steps:

### Step 1: Understand the Given Information
We are given:
- \(\vec{A}\) and \(\vec{B}\) are non-zero vectors.
- \(\vec{A} \perp \vec{B}\) (which means \(\vec{A} \cdot \vec{B} = 0\)).
- \(c\) is a non-zero scalar.

### Step 2: Write the Equations
The equations we need to solve are:
1. \(\vec{A} \cdot \vec{X} = c\) (Equation 1)
2. \(\vec{A} \times \vec{X} = \vec{B}\) (Equation 2)

### Step 3: Use the Cross Product Identity
From Equation 2, we can use the vector triple product identity:
\[ \vec{A} \times (\vec{A} \times \vec{X}) = \vec{A} (\vec{A} \cdot \vec{X}) - \vec{X} (\vec{A} \cdot \vec{A}) \]
Substituting Equation 1 into this identity gives:
\[ \vec{A} \times \vec{B} = \vec{A} \cdot \vec{X} \cdot \vec{A} - \vec{X} \cdot |\vec{A}|^2 \]
Since \(\vec{A} \cdot \vec{X} = c\), we can rewrite it as:
\[ \vec{A} \times \vec{B} = c \cdot \vec{A} - |\vec{A}|^2 \cdot \vec{X} \]

### Step 4: Rearranging the Equation
Now, we can rearrange this equation to solve for \(\vec{X}\):
\[ |\vec{A}|^2 \cdot \vec{X} = c \cdot \vec{A} - \vec{A} \times \vec{B} \]
Thus, we have:
\[ \vec{X} = \frac{c \cdot \vec{A} - \vec{A} \times \vec{B}}{|\vec{A}|^2} \]

### Step 5: Final Expression for \(\vec{X}\)
The final expression for the vector \(\vec{X}\) is:
\[ \vec{X} = \frac{c \cdot \vec{A} - \vec{A} \times \vec{B}}{|\vec{A}|^2} \]