Question

If for a unit vector \(\vec a\), \((\vec x-\vec a)\cdot(\vec x+\vec a)=12\), then find \(|\vec x|\).

Hint:

Expand as (x-a).(x+a) = |x|^2 - |a|^2, and use |a| = 1.

Answer 1

Step 1: Expand the dot product:
Use the distributive property of the dot product, similar to the algebraic identity \((A-B)\cdot(A+B) = A\cdot A - B\cdot B\) for vectors.
\[ (\vec x - \vec a)\cdot(\vec x + \vec a) = \vec x\cdot\vec x + \vec x\cdot\vec a - \vec a\cdot\vec x - \vec a\cdot\vec a \]
Since the dot product is commutative, \(\vec x\cdot\vec a = \vec a\cdot\vec x\), so these two middle terms cancel.

Step 2: Simplify using magnitudes:
What remains is:
\[ (\vec x - \vec a)\cdot(\vec x + \vec a) = \vec x\cdot\vec x - \vec a\cdot\vec a = |\vec x|^2 - |\vec a|^2 \]
Since \(\vec a\) is a unit vector, \(|\vec a| = 1\), so \(|\vec a|^2 = 1\).

Step 3: Solve for the magnitude of x:
Substitute the given value 12 and \(|\vec a|^2 = 1\) into the simplified equation:
\[ |\vec x|^2 - 1 = 12 \quad \Rightarrow \quad |\vec x|^2 = 13 \quad \Rightarrow \quad |\vec x| = \sqrt{13} \]

Final Answer:
The magnitude of vector x is root 13. \[ \boxed{|\vec x| = \sqrt{13}} \]