Question

The three lines with direction cosiners

$(\frac{12}{13},\frac{-3}{13},\frac{-4}{13});(\frac{4}{13},\frac{12}{13},\frac{3}{13}):(\frac{3}{13},\frac{-4}{13},\frac{12}{13})$ are mutually perpendicular.

• A. True
• B. False

Answer 1

The correct option is A True

DCs of the line $(\frac{12}{13},\frac{-3}{13},\frac{-4}{13});(\frac{4}{13},\frac{12}{13},\frac{3}{13});(\frac{3}{13},\frac{-4}{13},\frac{12}{13})$

$\therefore$ DR of the lines

$<12,-3,-4>,<4,12,3>,<3,-4,12>$

$\therefore 12\times 4+(-3)\times 12+(-4)\times 3=0$

$4\times 3+12\times (-4)+3\times 12=0$

$12\times 3+(-3)\times (-4)+(-4)\times 12=0$

$\therefore$ They are mutually perpendicular to each other.