Question:
If \(\vec{a} = \hat{i} + \hat{j} + \hat{k}\) and \(\vec{b} = \hat{i} - \hat{j} + \hat{k}\), then the projection of \(\vec{a}\) on \(\vec{b}\) is:
If \(\vec{a} = \hat{i} + \hat{j} + \hat{k}\) and \(\vec{b} = \hat{i} - \hat{j} + \hat{k}\), then the projection of \(\vec{a}\) on \(\vec{b}\) is:
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Projection is scalar unless vector projection is asked.
Updated On: Apr 30, 2026
- \(\sqrt{3} \)
- \(\frac{1}{\sqrt{3}} \)
- \(-\frac{1}{\sqrt{3}} \)
- \(-\sqrt{3} \)
- \(\frac{1}{3} \)
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The Correct Option is B
Solution and Explanation
Concept:
Projection formula:
\[
\text{Projection} = \frac{\vec{a}\cdot\vec{b}}{|\vec{b}|}
\]
Step 1: Find dot product.
\[ \vec{a}\cdot\vec{b} = 1(1) + 1(-1) + 1(1) = 1 \]
Step 2: Find magnitude of $\vec{b}$.
\[ |\vec{b}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3} \]
Step 3: Compute projection.
\[ \frac{1}{\sqrt{3}} \]
Step 1: Find dot product.
\[ \vec{a}\cdot\vec{b} = 1(1) + 1(-1) + 1(1) = 1 \]
Step 2: Find magnitude of $\vec{b}$.
\[ |\vec{b}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3} \]
Step 3: Compute projection.
\[ \frac{1}{\sqrt{3}} \]
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