Question:
What will be the projection of vector
\(\vec{A} = \hat{i} + \hat{j} + \hat{k}\)
on vector
\(\vec{B} = \hat{i} + \hat{j}\) ?
What will be the projection of vector
\(\vec{A} = \hat{i} + \hat{j} + \hat{k}\)
on vector
\(\vec{B} = \hat{i} + \hat{j}\) ?
\(\vec{A} = \hat{i} + \hat{j} + \hat{k}\)
on vector
\(\vec{B} = \hat{i} + \hat{j}\) ?
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The options are vectors, so use the vector projection \(\frac{\vec{A}\cdot\vec{B}}{|\vec{B}|^2}\vec{B}\).
Updated On: Nov 4, 2026
- \(\sqrt{2}\,(\hat{i} + \hat{j} + \hat{k})\)
- \((\hat{i} + \hat{j})\)
- \(\sqrt{2}\,(\hat{i} + \hat{j})\)
- \(2(\hat{i} + \hat{j} + \hat{k})\)
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The Correct Option is B
Solution and Explanation
\(\vec{A}\cdot\vec{B} = 1 + 1 + 0 = 2\) and \(|\vec{B}|^2 = 2\).
The vector projection of \(\vec{A}\) on \(\vec{B}\) is \(\frac{\vec{A}\cdot\vec{B}}{|\vec{B}|^2}\,\vec{B} = \frac{2}{2}(\hat{i} + \hat{j}) = \hat{i} + \hat{j}\).
Answer: (2) \((\hat{i} + \hat{j})\)
The vector projection of \(\vec{A}\) on \(\vec{B}\) is \(\frac{\vec{A}\cdot\vec{B}}{|\vec{B}|^2}\,\vec{B} = \frac{2}{2}(\hat{i} + \hat{j}) = \hat{i} + \hat{j}\).
Answer: (2) \((\hat{i} + \hat{j})\)
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