Question

If $\hat{a}$ is a unit vector such that $(\vec{x} - \hat{a}) \cdot (\vec{x} + \hat{a}) = 8$, then $|\vec{x}| =$

• A. $\pm 3$
• B. $2\sqrt{2}$
• C. 3
• D. $\pm \sqrt{7}$

Hint:

Vector dot products expand exactly like standard scalar algebraic difference-of-squares configurations: $(x-a)(x+a) = x^2 - a^2$. Since $a$ is a unit vector ($a^2=1$), the expression simplifies instantly to $|\vec{x}|^2 - 1 = 8 \implies |\vec{x}|^2 = 9$. Remember that a vector norm can never be negative, so discard $-3$ immediately and pick $+3$.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given a dot product equation involving an unknown vector $\vec{x}$ and a known unit vector $\hat{a}$. We need to compute the absolute magnitude of vector $\vec{x}$, denoted as $|\vec{x}|$.

Step 2: Key Formula or Approach:
1. Apply the distributive property of the vector dot product: $$(\vec{u} - \vec{v}) \cdot (\vec{u} + \vec{v}) = \vec{u} \cdot \vec{u} + \vec{u} \cdot \vec{v} - \vec{v} \cdot \vec{u} - \vec{v} \cdot \vec{v}$$ 2. Because the dot product is commutative ($\vec{u} \cdot \vec{v} = \vec{v} \cdot \vec{u}$), the middle terms cancel out perfectly, leaving: $$|\vec{u}|^2 - |\vec{v}|^2$$ 3. For a unit vector $\hat{a}$, its magnitude is by definition equal to 1 ($|\hat{a}| = 1$).

Step 3: Detailed Explanation:
4. Expand the given algebraic vector expression: $$(\vec{x} - \hat{a}) \cdot (\vec{x} + \hat{a}) = 8$$ $$\vec{x} \cdot \vec{x} + \vec{x} \cdot \hat{a} - \hat{a} \cdot \vec{x} - \hat{a} \cdot \hat{a} = 8$$ 5. Substitute $\vec{x} \cdot \vec{x} = |\vec{x}|^2$ and $\hat{a} \cdot \hat{a} = |\hat{a}|^2$. Since $\vec{x} \cdot \hat{a} = \hat{a} \cdot \vec{x}$, they drop out: $$|\vec{x}|^2 - |\hat{a}|^2 = 8$$ 6. Since $\hat{a}$ is explicitly stated to be a unit vector, substitute $|\hat{a}| = 1$: $$|\vec{x}|^2 - (1)^2 = 8$$ $$|\vec{x}|^2 - 1 = 8$$ $$|\vec{x}|^2 = 8 + 1 = 9$$ 7. Take the square root on both sides to solve for the magnitude. Because the magnitude of any vector represents a physical length or norm, it must strictly be a non-negative real number ($|\vec{x}| \ge 0$): $$|\vec{x}| = \sqrt{9} = 3$$

Step 4: Final Answer:
The magnitude $|\vec{x}|$ is 3, which corresponds to option (C).