The vector has components 2p and 1 with respect to a rectangular Cartesian system. The axes rotated through system. The axes are rotated through an a about the origin in the anticlockwise sense. If the vector has components (p+1) and 1 unit respect to the new system, then

Options

A. p=1,`-1//3`
B. `p=0`
C. `p=-1,1//3`
D. `p=1,-1`

Answer

To solve the problem step-by-step, we will analyze the given information and apply the properties of vectors and coordinate transformations.

### Step 1: Understand the initial vector components
The vector has components \(2p\) and \(1\) in the original Cartesian coordinate system. This means:
- The x-component of the vector is \(2p\).
- The y-component of the vector is \(1\).

### Step 2: Determine the new vector components after rotation
After rotating the axes through an angle \(a\) in the anticlockwise direction, the new components of the vector are given as \(p + 1\) and \(1\). This means:
- The new x-component is \(p + 1\).
- The new y-component is \(1\).

### Step 3: Use the rotation transformation formulas
The transformation of the components of a vector due to rotation can be expressed as:
\[ x' = x \cos a - y \sin a \]
\[ y' = x \sin a + y \cos a \]
where \((x, y)\) are the original components and \((x', y')\) are the new components.

Substituting the original components \(x = 2p\) and \(y = 1\), we have:
\[ p + 1 = 2p \cos a - 1 \sin a \quad (1) \]
\[ 1 = 2p \sin a + 1 \cos a \quad (2) \]

### Step 4: Rearranging and simplifying the equations
From equation (2):
\[ 1 - \cos a = 2p \sin a \quad (3) \]

From equation (1):
\[ p + 1 + \sin a = 2p \cos a \quad (4) \]

### Step 5: Solve for \(p\)
Now, we need to eliminate \(\sin a\) and \(\cos a\) from equations (3) and (4).

From equation (3):
\[ \sin a = \frac{1 - \cos a}{2p} \quad (5) \]

Substituting equation (5) into equation (4):
\[ p + 1 + \frac{1 - \cos a}{2p} = 2p \cos a \]

Multiply through by \(2p\) to eliminate the fraction:
\[ 2p(p + 1) + 1 - \cos a = 4p^2 \cos a \]

Rearranging gives:
\[ 2p^2 + 2p + 1 = (4p^2 + 1) \cos a \]

### Step 6: Solve for \(\cos a\)
Now we can express \(\cos a\) in terms of \(p\):
\[ \cos a = \frac{2p^2 + 2p + 1}{4p^2 + 1} \quad (6) \]

### Step 7: Substitute back to find \(p\)
Since \(\sin^2 a + \cos^2 a = 1\), we can substitute equation (6) into the identity to find \(p\).

After substituting and simplifying, we will arrive at a quadratic equation in \(p\):
\[ 3p^2 - 2p - 1 = 0 \]

### Step 8: Solve the quadratic equation
Using the quadratic formula:
\[ p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \(a = 3\), \(b = -2\), and \(c = -1\):
\[ p = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 3 \cdot (-1)}}{2 \cdot 3} \]
\[ p = \frac{2 \pm \sqrt{4 + 12}}{6} \]
\[ p = \frac{2 \pm 4}{6} \]

Thus, we get two possible values for \(p\):
1. \(p = 1\)
2. \(p = -\frac{1}{3}\)

### Final Answer:
The values of \(p\) are \(1\) and \(-\frac{1}{3}\).