Let `vecx and vecy` are 2 non - zero and non - collinear vectors, then the largest value of k such that the non - zero vectors `(k^(2)-5k+6)vecx+(k-3)vecy` and `2vecx+5vecy` are collinear is

Options

A. 3
B. 6
C. `(12)/(5)`
D. `-1`

Answer

To solve the problem, we need to find the largest value of \( k \) such that the vectors \( (k^2 - 5k + 6) \vec{x} + (k - 3) \vec{y} \) and \( 2\vec{x} + 5\vec{y} \) are collinear.

### Step 1: Set up the collinearity condition
Two vectors \( \vec{a} \) and \( \vec{b} \) are collinear if there exists a scalar \( \lambda \) such that:
\[ \vec{a} = \lambda \vec{b} \]
In our case, we can write:
\[ (k^2 - 5k + 6) \vec{x} + (k - 3) \vec{y} = \lambda (2\vec{x} + 5\vec{y}) \]

### Step 2: Compare coefficients
From the equation above, we can compare the coefficients of \( \vec{x} \) and \( \vec{y} \):
1. Coefficient of \( \vec{x} \):
\[ k^2 - 5k + 6 = 2\lambda \quad \text{(1)} \]
2. Coefficient of \( \vec{y} \):
\[ k - 3 = 5\lambda \quad \text{(2)} \]

### Step 3: Solve for \( \lambda \)
From equation (2), we can express \( \lambda \) in terms of \( k \):
\[ \lambda = \frac{k - 3}{5} \]

### Step 4: Substitute \( \lambda \) into equation (1)
Substituting \( \lambda \) into equation (1):
\[ k^2 - 5k + 6 = 2\left(\frac{k - 3}{5}\right) \]
Multiplying through by 5 to eliminate the fraction:
\[ 5(k^2 - 5k + 6) = 2(k - 3) \]
Expanding both sides:
\[ 5k^2 - 25k + 30 = 2k - 6 \]

### Step 5: Rearranging the equation
Bringing all terms to one side:
\[ 5k^2 - 25k - 2k + 30 + 6 = 0 \]
This simplifies to:
\[ 5k^2 - 27k + 36 = 0 \]

### Step 6: Solve the quadratic equation
Now we can solve the quadratic equation using the quadratic formula:
\[ k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 5 \), \( b = -27 \), and \( c = 36 \):
\[ k = \frac{27 \pm \sqrt{(-27)^2 - 4 \cdot 5 \cdot 36}}{2 \cdot 5} \]
Calculating the discriminant:
\[ k = \frac{27 \pm \sqrt{729 - 720}}{10} \]
\[ k = \frac{27 \pm \sqrt{9}}{10} \]
\[ k = \frac{27 \pm 3}{10} \]
This gives us two possible values for \( k \):
\[ k = \frac{30}{10} = 3 \quad \text{and} \quad k = \frac{24}{10} = \frac{12}{5} \]

### Step 7: Determine the largest value of \( k \)
The two values we found are \( k = 3 \) and \( k = \frac{12}{5} \). The largest value is:
\[ \boxed{3} \]