Question

Unit orthogonal vector to the vector $(3,4)$ is ______.

• A. $(\frac{3}{5},\frac{4}{5})$
• B. $(\frac{4}{5},\frac{3}{5})$
• C. $(-\frac{4}{5},\frac{3}{5})$
• D. $(-\frac{3}{5},\frac{4}{5})$

Answer 1

Answer: (C)

$(-\frac{4}{5},\frac{3}{5})$

Let $\overline{y}=(a,b)$ is unit orthogonal vector to $\overline{x}=(3,4)$
$\overline{y}=(a,b)$ unit vector $\left|\overline{y}\right|=1$
$\therefore \sqrt{a^2+b^2}=1$
$\therefore a^2+b^2=\mathrm{1...}\left(1\right)$
Now $\overline{x}\perp \overline{y}$
$\therefore \overline{x}\overline{y}=0$
$\therefore (3,4).(a,b)=0$
$\therefore 3a+4b=0$
$\therefore a=-\frac{4}{3}b...\left(2\right)$
Now subtitute value of (2) in (1)
$\therefore \frac{16}{9}{b}^2+b^2=1$
$\therefore 16b^2+b^2=1$
$\therefore 25b^2=9$
$\therefore b^2=\frac{9}{25}$
$\therefore b=\pm \frac{3}{5}$
If $b=\frac{3}{5}$ then from (2) we get
$a=-\frac{4}{5}$
$\therefore$ Required vector $(a,b)=(-\frac{4}{5},\frac{3}{5})$
And if $b=-\frac{3}{5}$ then from (2) we get
$a=\frac{4}{5}$
$\therefore$ Required vector is $(\frac{4}{5},-\frac{3}{5})$
Here $\overline{y}=(-\frac{4}{5},\frac{3}{5})$