Question

If $\vec{a}$ is a nonzero vector of magnitude $a$ and $\lambda$ is a nonzero scalar, then $\lambda \vec{a}$ is a unit vector if

• A. $\lambda = 1$
• B. $\lambda = -1$
• C. $a = |\lambda|$
• D. $a = \dfrac{1}{|\lambda|}$

Hint:

Magnitude of a scalar multiple of a vector follows the rule $|\lambda \vec{a}| = |\lambda|\,|\vec{a}|$.

Answer 1

The Correct Option is D

Step 1: Recall the definition of a unit vector.
A vector is called a unit vector if its magnitude is equal to 1. Thus if a vector $\vec{v}$ is a unit vector, then \[ |\vec{v}| = 1 \]
Step 2: Determine the magnitude of $\lambda \vec{a$.

For any scalar $\lambda$ and vector $\vec{a}$, the magnitude satisfies \[ |\lambda \vec{a}| = |\lambda| \, |\vec{a}| \] Since the magnitude of $\vec{a}$ is given as $a$, we obtain \[ |\lambda \vec{a}| = |\lambda| a \]
Step 3: Apply the unit vector condition.
For $\lambda \vec{a}$ to be a unit vector, its magnitude must be 1. Thus \[ |\lambda| a = 1 \]
Step 4: Solve for $a$.
\[ a = \frac{1}{|\lambda|} \]
Step 5: Conclusion.
Thus $\lambda \vec{a}$ becomes a unit vector only when \[ a = \frac{1}{|\lambda|} \] Final Answer: $\boxed{a = \dfrac{1}{|\lambda|}}$