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:
We need to create a new vector that has the direction of \(\vec{A}\) but a specific length (magnitude) derived from another vector.
Step 2: Key Formula or Approach:
Find the unit vector of \(\vec{A}\) to isolate its direction: \(\hat{u}_A = \frac{\vec{A}}{|\vec{A}|}\).
Calculate the magnitude of the reference vector \(\vec{B} = \hat{i} - 2\hat{j}\): \(|\vec{B}| = \sqrt{x^2 + y^2}\).
Multiply the unit direction vector by the required magnitude: \(\vec{V}_{new} = |\vec{B}| \cdot \hat{u}_A\).
Step 3: Detailed Explanation:
Given \(\vec{A} = 3\hat{j} - 4\hat{k}\).
Magnitude of \(\vec{A}\):
\[ |\vec{A}| = \sqrt{0^2 + 3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \] Unit vector of \(\vec{A}\):
\[ \hat{u}_A = \frac{3\hat{j} - 4\hat{k}}{5} \] Reference vector is \(\vec{B} = \hat{i} - 2\hat{j}\).
Magnitude of \(\vec{B}\):
\[ |\vec{B}| = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5} \] The required vector must have magnitude \(\sqrt{5}\) and be parallel to \(\vec{A}\):
\[ \vec{V} = \pm |\vec{B}| \hat{u}_A \] \[ \vec{V} = \pm \sqrt{5} \left( \frac{3\hat{j} - 4\hat{k}}{5} \right) \] Since \(5 = \sqrt{5} \times \sqrt{5}\):
\[ \vec{V} = \pm \frac{1}{\sqrt{5}} (3\hat{j} - 4\hat{k}) \] Looking at the options, the positive parallel vector matches exactly with Option D.
Step 4: Final Answer:
The required vector is \(\frac{1}{\sqrt{5}}(3\hat{j} - 4\hat{k})\).