Key Concepts and Formulas

• Vector Magnitude: For a vector $\overrightarrow{v} = (x, y)$ in a 2D Cartesian system, its magnitude is given by $|\overrightarrow{v}| = \sqrt{x^2 + y^2}$.
• Invariance of Magnitude Under Rotation: The magnitude of a vector remains unchanged when the coordinate system is rotated about the origin. This means $|\overrightarrow{a}|_{old} = |\overrightarrow{a}|_{new}$.
• Quadratic Equation: A quadratic equation of the form $ax^2 + bx + c = 0$ can be solved by factoring or using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Step-by-Step Solution

1. Magnitude of the Vector in the Original System Let the vector be $\overrightarrow{a}$. In the original rectangular Cartesian system, its components are $(3p, 1)$. The square of the magnitude of $\overrightarrow{a}$ in the original system is: $|\overrightarrow{a}|^2_{\text{old}} = (3p)^2 + (1)^2 = 9p^2 + 1$ Explanation: We calculate the square of the magnitude to avoid dealing with square roots, as equating the squares of equal magnitudes is mathematically equivalent.

2. Magnitude of the Vector in the New (Rotated) System After rotation, the components of the same vector $\overrightarrow{a}$ in the new system are $(p+1, \sqrt{10})$. The square of the magnitude of $\overrightarrow{a}$ in the new system is: $|\overrightarrow{a}|^2_{\text{new}} = (p+1)^2 + (\sqrt{10})^2$ Expanding $(p+1)^2$ and simplifying: $|\overrightarrow{a}|^2_{\text{new}} = (p^2 + 2p + 1) + 10 = p^2 + 2p + 11$ Explanation: We expand the squared term $(p+1)^2$ and combine the constant terms.

3. Equating Magnitudes and Forming the Equation Since the magnitude of a vector is invariant under rotation, the square of its magnitude in the old system must equal its square in the new system: $|\overrightarrow{a}|^2_{\text{old}} = |\overrightarrow{a}|^2_{\text{new}}$ Substituting the expressions from Steps 1 and 2: $9p^2 + 1 = p^2 + 2p + 11$ Explanation: This equation directly applies the principle of invariance of vector magnitude.

4. Solving the Quadratic Equation for p Rearrange the equation to form a standard quadratic equation: $9p^2 - p^2 - 2p + 1 - 11 = 0$ $8p^2 - 2p - 10 = 0$ Divide the entire equation by 2 to simplify: $4p^2 - p - 5 = 0$ Now, we solve this quadratic equation. We can factor it by finding two numbers that multiply to $(4)(-5) = -20$ and add to $-1$. These numbers are $-5$ and $4$. $4p^2 - 5p + 4p - 5 = 0$ Factor by grouping: $p(4p - 5) + 1(4p - 5) = 0$ $(4p - 5)(p + 1) = 0$ This gives two possible values for $p$: $4p - 5 = 0 \quad \Rightarrow \quad 4p = 5 \quad \Rightarrow \quad p = \frac{5}{4}$ $p + 1 = 0 \quad \Rightarrow \quad p = -1$ Explanation: We rearranged the equation into standard quadratic form and then solved it by factoring to find the possible values of $p$.

5. Checking the Solution with the Options The possible values for $p$ are $\frac{5}{4}$ and $-1$. Let's examine the given options: (A) 1 (B) $- {5 \over 4}$ (C) ${4 \over 5}$ (D) $-1$

Our derived value $p = -1$ matches option (D).

Common Mistakes & Tips

• Algebraic Errors: Be meticulous with squaring binomials and rearranging terms in the quadratic equation. A small sign error can lead to an incorrect answer.
• Forgetting Both Roots: Quadratic equations often have two distinct solutions. Ensure you find both and check if either fits the problem's context or the given options.
• Magnitude vs. Squared Magnitude: While the magnitude itself is invariant, it's usually easier to work with the squared magnitude to avoid square roots during algebraic manipulations.

Summary

The core principle used to solve this problem is the invariance of a vector's magnitude under the rotation of the coordinate system. By expressing the squared magnitude of the vector in both the original and the rotated coordinate systems and equating them, we formed a quadratic equation in terms of $p$. Solving this quadratic equation yielded two possible values for $p$, one of which matched one of the provided options.

The final answer is $\boxed{-1}$.