`veca` and `vecc` are unit collinear vectors and `|vecb|=6`, then `vecb-3vecc = lambda veca`, if `lambda` is:

Options

A. `-9,3`
B. 9,3
C. `3,-3`
D. None of these

Answer

To solve the problem, we start with the given information and work step by step.

### Step 1: Understand the Given Information
We have two unit collinear vectors, \(\vec{a}\) and \(\vec{c}\), and a vector \(\vec{b}\) such that \(|\vec{b}| = 6\). The equation given is:
\[ \vec{b} - 3\vec{c} = \lambda \vec{a} \]

### Step 2: Rearranging the Equation
Rearranging the equation gives us:
\[ \vec{b} = \lambda \vec{a} + 3\vec{c} \]

### Step 3: Squaring Both Sides
To find \(\lambda\), we will take the dot product of both sides with themselves:
\[ \vec{b} \cdot \vec{b} = (\lambda \vec{a} + 3\vec{c}) \cdot (\lambda \vec{a} + 3\vec{c}) \]

### Step 4: Expanding the Right Side
Expanding the right-hand side using the distributive property of the dot product:
\[ |\vec{b}|^2 = \lambda^2 (\vec{a} \cdot \vec{a}) + 2\lambda \cdot 3 (\vec{a} \cdot \vec{c}) + 9 (\vec{c} \cdot \vec{c}) \]

### Step 5: Substitute Known Values
We know that \(|\vec{b}| = 6\), so:
\[ |\vec{b}|^2 = 36 \]
Also, since \(\vec{a}\) and \(\vec{c}\) are unit vectors:
\[ \vec{a} \cdot \vec{a} = 1 \quad \text{and} \quad \vec{c} \cdot \vec{c} = 1 \]
Thus, we can substitute these values into the equation:
\[ 36 = \lambda^2 + 6\lambda(\vec{a} \cdot \vec{c}) + 9 \]

### Step 6: Simplifying the Equation
Rearranging gives us:
\[ 36 = \lambda^2 + 6\lambda(\vec{a} \cdot \vec{c}) + 9 \]
Subtracting 9 from both sides:
\[ 27 = \lambda^2 + 6\lambda(\vec{a} \cdot \vec{c}) \]

### Step 7: Finding \(\vec{a} \cdot \vec{c}\)
Since \(\vec{a}\) and \(\vec{c}\) are collinear, the angle between them is 0 degrees, thus:
\[ \vec{a} \cdot \vec{c} = |\vec{a}| |\vec{c}| \cos(0) = 1 \cdot 1 \cdot 1 = 1 \]

### Step 8: Substitute Back into the Equation
Substituting \(\vec{a} \cdot \vec{c} = 1\) back into the equation:
\[ 27 = \lambda^2 + 6\lambda \]

### Step 9: Rearranging into Standard Form
Rearranging gives us:
\[ \lambda^2 + 6\lambda - 27 = 0 \]

### Step 10: Solving the Quadratic Equation
We can solve this quadratic equation using the quadratic formula:
\[ \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \(a = 1\), \(b = 6\), and \(c = -27\):
\[ \lambda = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot (-27)}}{2 \cdot 1} \]
\[ = \frac{-6 \pm \sqrt{36 + 108}}{2} \]
\[ = \frac{-6 \pm \sqrt{144}}{2} \]
\[ = \frac{-6 \pm 12}{2} \]

### Step 11: Finding the Values of \(\lambda\)
Calculating the two possible values:
1. \(\lambda = \frac{6}{2} = 3\)
2. \(\lambda = \frac{-18}{2} = -9\)

Thus, the possible values of \(\lambda\) are \(3\) and \(-9\).

### Final Answer
The values of \(\lambda\) are \(3\) and \(-9\).

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