Discussion
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
Answer: B
The projection of \(\vec{a}\) on \(\vec{b}\) is \(\frac{1}{\sqrt{3}}\).
The scalar projection of \(\vec{a}\) on \(\vec{b}\) is \(\frac{\vec{a}.\vec{b}}{|\vec{b}|}\).
\[ \begin{aligned} \vec{a}.\vec{b} &= 1 - 1 + 1 = 1 \\ |\vec{b}| &= \sqrt{1 + 1 + 1} = \sqrt{3} \\ \text{projection} &= \frac{1}{\sqrt{3}} \end{aligned} \]
\(\frac{1}{3}\) divides by \(|\vec{b}|^{2}\) instead of \(|\vec{b}|\), and \(\sqrt{3}\) is \(|\vec{b}|\) itself. The dot product is positive, so the negative options are wrong.
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