You are given vector `vec(A)=5hat(i)-6.5hat(j)` and `vec(B)=10hat(i)+7hat(j)`.
A third vector `vec(C )` lies in the `x-y` plane. Vector `(C )` is perpendicular to vector `vec(A)` and the scalar product of `vec(C )` with `vec(B)` is 15. From this information, find the components of `vec(C )`

Explanation

To find the components of vector \( \vec{C} \), we will follow these steps:

### Step 1: Define the vectors
Given:
- \( \vec{A} = 5\hat{i} - 6.5\hat{j} \)
- \( \vec{B} = 10\hat{i} + 7\hat{j} \)

Let \( \vec{C} = x\hat{i} + y\hat{j} \).

### Step 2: Use the condition of perpendicularity
Since \( \vec{C} \) is perpendicular to \( \vec{A} \), their dot product must be zero:
\[ \vec{A} \cdot \vec{C} = 0 \]
Calculating the dot product:
\[ (5\hat{i} - 6.5\hat{j}) \cdot (x\hat{i} + y\hat{j}) = 5x - 6.5y = 0 \]
This gives us our first equation:
\[ 5x = 6.5y \quad \text{(Equation 1)} \]

### Step 3: Express \( x \) in terms of \( y \)
From Equation 1, we can express \( x \):
\[ x = \frac{6.5}{5}y = 1.3y \]

### Step 4: Use the condition of the scalar product
The scalar product of \( \vec{C} \) and \( \vec{B} \) is given as 15:
\[ \vec{C} \cdot \vec{B} = 15 \]
Calculating the dot product:
\[ (x\hat{i} + y\hat{j}) \cdot (10\hat{i} + 7\hat{j}) = 10x + 7y = 15 \]
Substituting \( x = 1.3y \) into this equation:
\[ 10(1.3y) + 7y = 15 \]
This simplifies to:
\[ 13y + 7y = 15 \]
\[ 20y = 15 \]
\[ y = \frac{15}{20} = \frac{3}{4} \]

### Step 5: Find \( x \) using \( y \)
Now substituting \( y \) back into the expression for \( x \):
\[ x = 1.3y = 1.3 \times \frac{3}{4} = \frac{3.9}{4} = \frac{39}{40} \]

### Step 6: Write the final vector \( \vec{C} \)
Now we can write the components of vector \( \vec{C} \):
\[ \vec{C} = \frac{39}{40}\hat{i} + \frac{3}{4}\hat{j} \]

### Final Answer
The components of vector \( \vec{C} \) are:
\[ \vec{C} = \frac{39}{40}\hat{i} + \frac{3}{4}\hat{j} \]

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