Question

Three lines with direction ratios $⩽1,1,2><\sqrt{3}-1,-\sqrt{3}-1,4>$ and $<-\sqrt{3}-1,\sqrt{3}-1,4⟩$ 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 $\overrightarrow{AB}=<1,1,2>$;
$\overrightarrow{AC}<\sqrt{3}-1,-\sqrt{3}-1,4>$
$\overrightarrow{BC}=<-\sqrt{3}-1,\sqrt{3}-1,4>$
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.