Question

The vector $\overrightarrow{b}=3\hat{i}+4\hat{k}$ is to be written as the sum of a vector $\overrightarrow{\mathrm{\alpha }}$parallel to $\overrightarrow{a}=\hat{i}+\hat{j}$ and a vector $\overrightarrow{\mathrm{\beta }}$perpendicular to $\overrightarrow{a}$. Then $\overrightarrow{\mathrm{\alpha }}=$
(a) $\frac{3}{2}\left(\hat{i}+\hat{j}\right)$

(b) $\frac{2}{3}\left(\hat{i}+\hat{j}\right)$

(c) $\frac{1}{2}\left(\hat{i}+\hat{j}\right)$

(d) $\frac{1}{3}\left(\hat{i}+\hat{j}\right)$

Answer 1

(a) $\frac{3}{2}\left(\hat{i}+\hat{j}\right)$

$$\begin{gathered} \text{Let:} \\ \overrightarrow{\alpha }\text{=}{a}_1\hat{i}+a_2\hat{j}+a_3\hat{k} \\ \overrightarrow{\beta }=b_1\hat{i}+b_2\hat{j}+b_3\hat{k} \\ \text{Now,} \\ \overrightarrow{b}\text{=3}\hat{i}\text{+4}\hat{k}\text{=}\overrightarrow{\alpha }\text{+}\overrightarrow{\beta }\text{(Given)} \\ \Rightarrow 3\hat{i}+0\hat{j}+4\hat{k}=\left(a_1+b_1\right)\hat{i}+\left(a_2+b_2\right)\hat{j}+\left(a_3+b_3\right)\hat{k} \\ \Rightarrow a_1+b_1=3;a_2+b_2=0;a_3+b_3=4 \\ \Rightarrow a_1+b_1=3;a_2=-b_2;a_3+b_3=4...\left(1\right) \\ \overrightarrow{a}\text{=}\hat{i}\text{+}\hat{j}\left(\mathrm{Given}\right) \\ \text{Also,}\overrightarrow{\alpha }\text{is parallel to}\overrightarrow{a}\text{.} \\ \Rightarrow \overrightarrow{\alpha }\times \overrightarrow{a}=\overrightarrow{0} \\ \Rightarrow \left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ a_1& a_2& a_3\\ 1& 1& 0\end{array}\right|=\overrightarrow{0} \\ \Rightarrow -a_3\hat{i}+a_3\hat{j}+\left(a_1-a_2\right)\hat{k}=0\hat{i}+0\hat{j}+0\hat{k} \\ \Rightarrow a_3=0;a_1-a_2=0 \\ \Rightarrow a_3=0;a_1=a_2...\left(2\right) \\ \text{Since}\overrightarrow{\beta }\text{is perpendicular to}\overrightarrow{a}\text{, we get} \\ \Rightarrow \overrightarrow{\beta }.\overrightarrow{a}=0 \\ \Rightarrow \left(b_1\hat{i}+b_2\hat{j}+b_3\hat{k}\right).\left(\hat{i}\text{+}\hat{j}\right)=0 \\ \Rightarrow b_1+b_2=0 \\ \Rightarrow b_1=-b_2...\left(3\right) \\ \text{Solving (1), (2) and (3), we get} \\ {a}_1=\frac{3}{2};a_2=\frac{3}{2};a_3=0 \\ \text{∴}\overrightarrow{\alpha }\text{=}{a}_1\hat{i}+a_2\hat{j}+a_3\hat{k} \\ =\frac{3}{2}\hat{i}+\frac{3}{2}\hat{j}+0\hat{k} \\ =\frac{3}{2}\left(\hat{i}+\hat{j}\right) \end{gathered}$$