If a and b are non-collinear, then the point of intersectioon of the lines `r = 6a -b +lambda (2b-a)` and `r = a-b + mu(a+3b)` is `3a +4b`. Is it true or false ?

Answer

To determine if the statement is true or false, we need to find the intersection point of the two lines given by the equations:

1. \( r = 6a - b + \lambda(2b - a) \)
2. \( r = a - b + \mu(a + 3b) \)

We want to check if this intersection point equals \( 3a + 4b \).

### Step 1: Set the two equations equal to each other

Since both equations represent the same point at the intersection, we can set them equal to each other:

\[ 6a - b + \lambda(2b - a) = a - b + \mu(a + 3b) \]

### Step 2: Rearrange the equation

Rearranging gives us:

\[ 6a - b + \lambda(2b - a) - a + b - \mu(a + 3b) = 0 \]

This simplifies to:

\[ (6 - \lambda - 1 + \mu)a + (-1 + 2\lambda - 3\mu)b = 0 \]

### Step 3: Collect coefficients of \( a \) and \( b \)

From the equation, we can separate the coefficients of \( a \) and \( b \):

1. Coefficient of \( a \): \( 6 - \lambda - 1 + \mu = 0 \)
2. Coefficient of \( b \): \( -1 + 2\lambda - 3\mu = 0 \)

### Step 4: Form the system of equations

This gives us the following system of equations:

1. \( \lambda + \mu = 5 \) (Equation 1)
2. \( 2\lambda - 3\mu = 1 \) (Equation 2)

### Step 5: Solve the system of equations

From Equation 1, we can express \( \lambda \) in terms of \( \mu \):

\[ \lambda = 5 - \mu \]

Substituting this into Equation 2:

\[ 2(5 - \mu) - 3\mu = 1 \]

Expanding and simplifying:

\[ 10 - 2\mu - 3\mu = 1 \]
\[ 10 - 5\mu = 1 \]
\[ 5\mu = 9 \implies \mu = \frac{9}{5} \]

Now substituting back to find \( \lambda \):

\[ \lambda = 5 - \frac{9}{5} = \frac{25}{5} - \frac{9}{5} = \frac{16}{5} \]

### Step 6: Find the intersection point

Now we substitute \( \lambda \) back into either line equation to find the intersection point. We will use the first line:

\[ r = 6a - b + \frac{16}{5}(2b - a) \]

Calculating this gives:

\[ r = 6a - b + \frac{32}{5}b - \frac{16}{5}a \]
\[ = \left(6 - \frac{16}{5}\right)a + \left(-1 + \frac{32}{5}\right)b \]
\[ = \left(\frac{30}{5} - \frac{16}{5}\right)a + \left(-\frac{5}{5} + \frac{32}{5}\right)b \]
\[ = \frac{14}{5}a + \frac{27}{5}b \]

### Step 7: Compare with the given point

The calculated intersection point is \( \frac{14}{5}a + \frac{27}{5}b \), which we need to compare with \( 3a + 4b \):

Converting \( 3a + 4b \) to a fraction gives:

\[ 3a + 4b = \frac{15}{5}a + \frac{20}{5}b \]

### Conclusion

Since \( \frac{14}{5}a + \frac{27}{5}b \) is not equal to \( \frac{15}{5}a + \frac{20}{5}b \), the statement is **false**.