Question

A vector is given as $\vec{A} = 3\hat{j} - 4\hat{k}$. The vector parallel to $\vec{A}$ and magnitude the same as that of the vector $\hat{i} - 2\hat{j}$ is

• A. $\frac{1}{3} (3\hat{i} - 4\hat{k})$
• B. $\sqrt{5} (3\hat{i} - 2\hat{j})$
• C. $\frac{1}{5} (3\hat{i} - 2\hat{j})$
• D. $\frac{1}{\sqrt{5}} (3\hat{j} - 4\hat{k})$
• E. $\sqrt{5} (3\hat{j} - 4\hat{k})$

Hint:

Remember that a parallel vector is simply (unit vector of target) $\times$ (desired magnitude). For standard triplets like (3,4,5), the magnitude is usually easy to compute.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
To find a vector parallel to $\vec{A}$ with a specific magnitude $M$, we first find the unit vector of $\vec{A}$ and then multiply it by the required magnitude $M$.

Step 2: Detailed Explanation:
1. Magnitude of vector $\hat{i} - 2\hat{j}$:
$M = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}$.
2. Unit vector parallel to $\vec{A} = 3\hat{j} - 4\hat{k}$:
Magnitude of $\vec{A} = \sqrt{0^2 + 3^2 + (-4)^2} = \sqrt{25} = 5$.
Unit vector $\hat{u}_A = \frac{3\hat{j} - 4\hat{k}}{5}$.
3. Required vector $\vec{V}$:
$\vec{V} = M \cdot \hat{u}_A = \sqrt{5} \cdot \frac{3\hat{j} - 4\hat{k}}{5}$.
4. Simplify:
Since $5 = \sqrt{5} \cdot \sqrt{5}$, we have:
$\vec{V} = \frac{\sqrt{5}}{\sqrt{5} \cdot \sqrt{5}} (3\hat{j} - 4\hat{k}) = \frac{1}{\sqrt{5}} (3\hat{j} - 4\hat{k})$.

Step 3: Final Answer:
The required vector is $\frac{1}{\sqrt{5}} (3\hat{j} - 4\hat{k})$.