Question

The triangle formed by the three points whose position vectors are $2i+4j-k,4i+5j+k$ and $3i+6j-3k$, is

• A. An equilateral triangle
• B. A right angled triangle but not isosceles
• C. An isosceles triangle but not right angled triangle
• D. A right angled isosceles triangle
• E. A scalene triangle

Answer 1

The correct option is D A right angled isosceles triangle

Let $OA=2i+4j-k,OB=4i+5j+k$
and $Oc=3i+6j-3k$
Now, $AB=OB-OA=2i+j+2k$
$\therefore AB=|2i+j+2k|=\sqrt{4+1+4}=\sqrt{9}=3$
$BC=OC-OB=-i+j-4k$
$\therefore BC=|-i+j-4k|=\sqrt{1+1+16}=\sqrt{18}=3\sqrt{2}$
and $CA=OA-OC=-i-2j+2k$
$\therefore CA=|-i-2j+2k|=\sqrt{1+4+4}=\sqrt{9}=3$
Now, $B{C}^2=A{B}^2+C{A}^2$ (by Pythagoras theorem)
$\Rightarrow (3\sqrt{2}{)}^2=(3{)}^2+(3{)}^2\Rightarrow 18=18$
So, triangle is isosceles right angled triangle.