MCQ
What will be the projection of vector $A=\hat{i}+\hat{j}+\hat{k}$ on vector $\vec{B}=\hat{i}+\hat{j}$.
- A$\sqrt{2}(\hat{i}+\hat{j})$
- ✓$(\hat{i}+\hat{j})$
- C$\sqrt{2}(\hat{i}+\hat{j}+\hat{k})$
- D$2(\hat{i}+\hat{j}+\hat{k})$
✓
Answer
Correct option: B.
$(\hat{i}+\hat{j})$
b
Projection of vector $A$ on vector $B$
Projection of vector $A$ on vector $B$
$(A \cos \theta) \hat{B}=A\left(\frac{\bar{A} \cdot \bar{B}}{A B}\right) \hat{B}=\frac{\bar{A} \cdot \bar{B}}{B} \hat{B}$
$=\frac{2}{\sqrt{2}}\left(\frac{\hat{i}+\hat{j}}{\sqrt{2}}\right)=\hat{i}+\hat{j}$
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