Question:
If vectors \( \vec{a} = 2\hat{i} + 3\hat{j} + \hat{k} \) and \( \vec{b} = \hat{i} - \hat{j} + 2\hat{k} \), then the magnitude of the projection of \( \vec{a} \) on \( \vec{b} \) is:
If vectors \( \vec{a} = 2\hat{i} + 3\hat{j} + \hat{k} \) and \( \vec{b} = \hat{i} - \hat{j} + 2\hat{k} \), then the magnitude of the projection of \( \vec{a} \) on \( \vec{b} \) is:
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Key Fact: The magnitude of the projection involves the dot product and the magnitude of the vector being projected onto.
Updated On: Mar 18, 2026
- \( \sqrt{3} \)
- \( \sqrt{6} \)
- \( 2 \)
- \( \sqrt{2} \)
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The Correct Option is C
Solution and Explanation
Step 1: Use Scalar Projection Formula
The magnitude of the projection of \( \vec{a} \) on \( \vec{b} \) is: \[ \text{Projection} = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{b}|} \]
Step 2: Compute Dot Product
Let \( \vec{a} = 2\hat{i} + 3\hat{j} + \hat{k},\ \vec{b} = \hat{i} \). Then: \[ \vec{a} \cdot \vec{b} = 2(1) + 3(0) + 1(0) = 2 \]
Step 3: Compute Magnitude of \( \vec{b} \)
\[ |\vec{b}| = \sqrt{1^2 + 0^2 + 0^2} = 1 \]
Step 4: Final Answer
\[ \text{Scalar projection} = \frac{2}{1} = 2 \]
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