Question:
What will be the projection of vector $\overrightarrow{\mathrm{A}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$ on vector $\overrightarrow{\mathrm{B}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}$ ?
Correct Option: , 4
Solution:
$(A \cos \theta) \hat{B}=A\left(\frac{\vec{A} \cdot \vec{B}}{A B}\right) \hat{B}=\frac{\vec{A} \cdot \vec{B}}{B} \hat{B}$
$=\frac{2}{\sqrt{2}}\left(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}}{\sqrt{2}}\right)=\hat{\mathrm{i}}+\hat{\mathrm{j}}$
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