Question

Let \( \vec{a} \) be a unit vector. If \( (\vec{x} - \vec{a}) \cdot (\vec{x} + \vec{a}) = 12 \), then the magnitude of \( \vec{x} \) is:

• A. \( \sqrt{8} \)
• B. \( \sqrt{9} \)
• C. \( \sqrt{10} \)
• D. \( \sqrt{13} \)
• E. \( \sqrt{12} \)

Hint:

Remember the vector identity \( (\vec{u} - \vec{v}) \cdot (\vec{u} + \vec{v}) = |\vec{u}|^2 - |\vec{v}|^2 \). It is the vector equivalent of the algebraic difference of squares formula \( (a-b)(a+b) = a^2 - b^2 \).

Answer 1

The Correct Option is D

Concept: The dot product of two vectors follows the distributive property, similar to algebraic expansion. For any vector \( \vec{v} \), the dot product \( \vec{v} \cdot \vec{v} \) is equal to the square of its magnitude, \( |\vec{v}|^2 \). A unit vector is defined as a vector whose magnitude is exactly 1.

Step 1: Expand the dot product expression.
Using the distributive property of the dot product: \[ (\vec{x} - \vec{a}) \cdot (\vec{x} + \vec{a}) = \vec{x} \cdot \vec{x} + \vec{x} \cdot \vec{a} - \vec{a} \cdot \vec{x} - \vec{a} \cdot \vec{a} \] Since the dot product is commutative (\( \vec{x} \cdot \vec{a} = \vec{a} \cdot \vec{x} \)), the middle terms cancel out: \[ |\vec{x}|^2 - |\vec{a}|^2 = 12 \]

Step 2: Substitute the value of the unit vector magnitude.
Since \( \vec{a} \) is a unit vector, \( |\vec{a}| = 1 \), which implies \( |\vec{a}|^2 = 1 \). Substituting this into the equation: \[ |\vec{x}|^2 - 1 = 12 \] \[ |\vec{x}|^2 = 13 \]

Step 3: Solve for the magnitude of \( \vec{x} \).
Taking the square root of both sides: \[ |\vec{x}| = \sqrt{13} \]