Given vectors $\vec A = \hat i + \hat j + \hat k$ $\vec B = \hat i + \hat j$

Projection of $\vec A$ on $\vec B$

The vector projection of $\vec A$ on $\vec B$ is $\operatorname{proj}_{\vec B} \vec A = \frac{\vec A \cdot \vec B}{|\vec B|^2}\,\vec B$

Find the dot product $\vec A \cdot \vec B$ $\vec A \cdot \vec B = (\hat i + \hat j + \hat k)\cdot(\hat i + \hat j)$ $= 1 + 1 + 0 = 2$

Find $|\vec B|^2$ $|\vec B|^2 = (\hat i + \hat j)\cdot(\hat i + \hat j) = 1+1 = 2$

Compute the projection $\operatorname{proj}_{\vec B} \vec A = \frac{2}{2}(\hat i + \hat j)$ $= \hat i + \hat j$

Match with the options This corresponds to: $\boxed{(\hat i + \hat j)}$ So the correct option is B.

Verification with stored answer Stored correct answer: B

Our derived answer also gives B, so they agree.