Question

Find the components of vector $\overrightarrow{a}=3\hat{i}+4\hat{j}$ along the direction of vectors $\hat{i}+\hat{j}$ & $\hat{i}-\hat{j}$.

• A. $\frac{7}{2}\left(\begin{array}{c}\hat{i}+\hat{j}\end{array}\right),-\frac{1}{2}\left(\begin{array}{c}\hat{i}-\hat{j}\end{array}\right)$
• B. $\frac{1}{2}\left(\begin{array}{c}\hat{i}+\hat{j}\end{array}\right),-\frac{7}{2}\left(\begin{array}{c}\hat{i}-\hat{j}\end{array}\right)$
• C. $\frac{-7}{2}\left(\begin{array}{c}\hat{i}+\hat{j}\end{array}\right),-\frac{1}{2}\left(\begin{array}{c}\hat{i}-\hat{j}\end{array}\right)$
• D. $\frac{7}{2}\left(\begin{array}{c}\hat{i}+\hat{j}\end{array}\right),\frac{1}{2}\left(\begin{array}{c}\hat{i}-\hat{j}\end{array}\right)$

Answer 1

The correct option is A $\frac{7}{2}\left(\begin{array}{c}\hat{i}+\hat{j}\end{array}\right),-\frac{1}{2}\left(\begin{array}{c}\hat{i}-\hat{j}\end{array}\right)$

Component along the vector $\hat{i}+\hat{j}$

$=\left(Acos\theta \right)\hat{B}=\frac{(\overrightarrow{A}.\overrightarrow{B})}{B^2}\overrightarrow{B}=\frac{(3\hat{i}+4\hat{j}).(\hat{i}+\hat{j})}{(\sqrt{2}{)}^2}(\hat{i}+\hat{j})$

$=\frac{3+4}{2}(\hat{i}+\hat{j})=\frac{7}{2}(\hat{i}+\hat{j})$

Component along the vector $\hat{i}-\hat{j}$

$=\left(Acos\theta \right)\hat{B}=\frac{(\overrightarrow{A}.\overrightarrow{B})}{B^2}\overrightarrow{B}=\frac{(3\hat{i}+4\hat{j}).(\hat{i}-\hat{j})(\hat{i}-\hat{j})}{(\sqrt{2}{)}^2}$

$=\frac{(3-4)}{2}(\hat{i}-\hat{j})=-\frac{1}{2}(\hat{i}-\hat{j})$