Discussion
Let \(\vec{a} = \alpha\hat{i} - 3\hat{j} - 2\hat{k}\) and \(\vec{c} = \hat{i} - 2\hat{j} + 2\hat{k}\) . If the projection of \(\vec{a}\) on \(\vec{c}\) is 2, then the value of \(\alpha\) is equal to
Answer: B
\(\alpha =\) 4.
The projection of \(\vec{a}\) on \(\vec{c}\) is \(\frac{\vec{a}\cdot\vec{c}}{|\vec{c}|}\), with \(|\vec{c}| = \sqrt{1 + 4 + 4} = 3\):
\[ \begin{aligned} \vec{a}\cdot\vec{c} &= \alpha + 6 - 4 = \alpha + 2 \\ \frac{\alpha + 2}{3} &= 2 \;\Rightarrow\; \alpha = 4 \end{aligned} \]
Forgetting to divide by \(|\vec{c}|\) gives \(\alpha = 0\); the other values do not make the projection 2.
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