The point of intersection of the lines `r times a=b times a, r times b=a times b` is

Options

A. a
B. b - a
C. a - b
D. a + b

Answer

To find the point of intersection of the lines given by the equations \( r \times a = b \times a \) and \( r \times b = a \times b \), we can follow these steps:

### Step 1: Rewrite the equations
We start with the two equations:
1. \( r \times a = b \times a \)
2. \( r \times b = a \times b \)

### Step 2: Rearranging the equations
From the first equation, we can rearrange it as:
\[ r \times a - b \times a = 0 \]
This can be rewritten as:
\[ r \times a = b \times a \]

From the second equation, we can rearrange it similarly:
\[ r \times b - a \times b = 0 \]
This can be rewritten as:
\[ r \times b = a \times b \]

### Step 3: Expressing in terms of cross products
We can express these equations in terms of the cross product:
1. \( r - b = \lambda a \) (for some scalar \( \lambda \))
2. \( r - a = \mu b \) (for some scalar \( \mu \))

### Step 4: Equating the two expressions for \( r \)
Now we have two expressions for \( r \):
1. \( r = b + \lambda a \)
2. \( r = a + \mu b \)

Setting these equal to each other gives:
\[ b + \lambda a = a + \mu b \]

### Step 5: Rearranging the equation
Rearranging this equation, we get:
\[ b - \mu b = a - \lambda a \]
This simplifies to:
\[ (1 - \mu)b = (1 - \lambda)a \]

### Step 6: Comparing coefficients
For this equation to hold for arbitrary vectors \( a \) and \( b \), the coefficients must be equal:
1. \( 1 - \mu = 0 \) (implying \( \mu = 1 \))
2. \( 1 - \lambda = 0 \) (implying \( \lambda = 1 \))

### Step 7: Finding the intersection point
Substituting \( \lambda = 1 \) and \( \mu = 1 \) back into either expression for \( r \):
\[ r = b + 1 \cdot a = b + a \]
or
\[ r = a + 1 \cdot b = a + b \]

### Conclusion
Thus, the point of intersection of the two lines is:
\[ \boxed{a + b} \]