Question

Three lines with direction ratios $\leqslant 1,1,2 > < \sqrt{3}-1,-\sqrt{3}-1,4 > $ and $ < -\sqrt{3}-1, \sqrt{3}-1,4\rangle$ form

• A. a right angled triangle
• B. a scalene triangle
• C. an equilateral triangle
• D. None of these

Answer 1

Answer: (C)

Let D.R.'s of $\overrightarrow{A B}=<1,1,2>$;
$\overrightarrow{A C}<\sqrt{3}-1,-\sqrt{3}-1,4>$
$\overrightarrow{B C}=< -\sqrt{3}-1, \sqrt{3}-1,4 >$
So $\cos A=\frac{\sqrt{3}-1-\sqrt{3}-1+8}{\sqrt{6} \sqrt{24}}=\frac{1}{2} $
$\Rightarrow A=60^{\circ}$
and $\cos C=\frac{-3+1-3+1+16}{\sqrt{24} \sqrt{24}}=\frac{12}{24}$
$ \Rightarrow C=60^{\circ}$
Obviously $B=60^{\circ}$
$\therefore $ Triangle will be an equilateral.