Question

The angle between vectors $\overrightarrow{A}=4\hat{i}+3\hat{j}$ & $\overrightarrow{B}=3\hat{i}-4\hat{j}$ is

• A. ${90}^o$
• B. $0^o$
• C. ${120}^o$
• D. ${60}^o$

Answer 1

Answer: (A)

${90}^o$

Given$,$

$\overrightarrow{A}=4\hat{i}+3\hat{j}$

$\overrightarrow{B}=3\hat{i}-4\hat{j}$

$or,\overrightarrow{A}.\overrightarrow{B}=\left|\overrightarrow{A}\right|.\left|\overrightarrow{B}\right|\cos \theta =A_x{B}_x+A_y+B_y$

$or,\left|\overrightarrow{A}\right|.\left|\overrightarrow{B}\right|\cos \theta =A_x{B}_x+A_y+B_y$

$or,\sqrt{4^2+3^2}\left(\sqrt{3^2+\left(4^2\right)}\right)\cos \theta =4\cdot 3+3\cdot \left(-4\right)$

$or,\sqrt{25}\cdot \sqrt{25}\cos \theta =12-12$

$or,25\cos \theta =0$

$or,\cos \theta =0$

$\therefore \theta ={\cos }^{-1}0={90}^0$

Hence,

option $\left(A\right)$ is correct answer.