Question

When vector \(\overrightarrow A = 2\hat i+3\hat J+2\hat k\) is subtracted from vector \(\overrightarrow B\)  , it gives a vector equal to \(2\hat j\) . Then the magnitude of vector \(\overrightarrow B\) will be :

• A. √6
• B. √5
• C. 3
• D. √13

Answer 1

The Correct Option is C

Given:

• \( \vec{A} = 2\hat{i} + 3\hat{j} + 2\hat{k} \)
• \( \vec{B} - \vec{A} = 2\hat{j} \)

Step 1: Solve for Vector \( \vec{B} \)

From the given equation:

\[ \vec{B} - \vec{A} = 2\hat{j}. \]

Add \( \vec{A} \) to both sides:

\[ \vec{B} = 2\hat{j} + \vec{A}. \]

Substitute \( \vec{A} = 2\hat{i} + 3\hat{j} + 2\hat{k} \):

\[ \vec{B} = 2\hat{j} + (2\hat{i} + 3\hat{j} + 2\hat{k}). \]

Simplify:

\[ \vec{B} = 2\hat{i} + 5\hat{j} + 2\hat{k}. \]

Step 2: Calculate the Magnitude of \( \vec{B} \)

The magnitude of a vector \( \vec{B} = x\hat{i} + y\hat{j} + z\hat{k} \) is given by:

\[ |\vec{B}| = \sqrt{x^2 + y^2 + z^2}. \]

For \( \vec{B} = 2\hat{i} + 5\hat{j} + 2\hat{k} \):

\[ |\vec{B}| = \sqrt{2^2 + 5^2 + 2^2}. \]

Simplify the expression:

\[ |\vec{B}| = \sqrt{4 + 25 + 4} = \sqrt{33}. \]

Final Answer:

The magnitude of vector \( \vec{B} \) is \( \sqrt{33} \).