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 :
- √6
- √5
- 3
- √13
The Correct Option is C
Solution and Explanation
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 \( \sqrt{33} \).
Top Questions on Vectors
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