Question

The projection of the vector $\hat{i}+\hat{j}+\hat{k}$ along the vector of $\hat{j}$ is
(a) 1
(b) 0
(c) 2
(d) −1
(e) −2

Answer 1

(a) 1

$$\begin{gathered} \mathrm{Let}\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}\mathrm{and}\overrightarrow{b}=\hat{j} \\ \text{The projection of}\overrightarrow{a}\text{on}\overrightarrow{b}\text{is} \\ \frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{b}\right|} \\ =\frac{\left(\hat{i}+\hat{j}+\hat{k}\right).\hat{j}}{\left|\hat{j}\right|} \\ =\frac{0+1+0}{1} \\ =1 \end{gathered}$$