Question

If $ \vec{a} $ is a non-zero vector and $ k $ is a scalar such that $ |k \vec{a}| = 1 $, then $ k $ is equal to

• A. \( |\vec{a}| \)
• B. 1
• C. \( \frac{1}{|\vec{a}|} \)
• D. \( \pm \frac{1}{|\vec{a}|} \)

Hint:

When given a scalar multiple of a vector, the magnitude of the result is the product of the scalar's absolute value and the magnitude of the vector.

Answer 1

The Correct Option is D

Step 1: Understanding the Problem We are given that \( |k \vec{a}| = 1 \), where \( \vec{a} \) is a non-zero vector and \( k \) is a scalar.
The magnitude of a scalar multiple of a vector is given by: \[ |k \vec{a}| = |k| \cdot |\vec{a}| \] We are told that \( |k \vec{a}| = 1 \), so: \[ |k| \cdot |\vec{a}| = 1 \]
Step 2: Solve for \( k \) Solving for \( |k| \), we get: \[ |k| = \frac{1}{|\vec{a}|} \] Thus, \( k \) can be either \( \frac{1}{|\vec{a}|} \) or \( -\frac{1}{|\vec{a}|} \), since the absolute value of \( k \) is \( \frac{1}{|\vec{a}|} \).
Step 3: Conclusion Thus, \( k \) is \( \pm \frac{1}{|\vec{a}|} \).