Question

Consider the lines
$L_1:x=3-t,y=2+t,z=5t$ intersecting the plane $x-y+2z=9$ at the point $A$
$L_2:x=1+2t,y=4t,z=2-3t$ intersecting the plane $x+2y-z+1=0$ at the point $B$ and
$L_3:x=y-1=2z$ intersecting the plane $4x-y+3z=8$ at the point $C$. Then the points $A,B,C$

• A. do not form a triangle.
• B. constitute an obtuse triangle.
• C. constitute an acute triangle.
• D. constitute a right triangle.

Answer 1

The correct option is C constitute an acute triangle.

Given:
$L_1:x=3-t,y=2+t,z=5t$ intersecting the plane $x-y+2z=9$
$\Rightarrow (3-t)-(2+t)+2\left(5t\right)=9$
$\Rightarrow t=1$
$\therefore$ Intersecting point $A(2,3,5)$

$L_2:x=1+2t,y=4t,z=2-3t$ intersecting the plane $x+2y-z+1=0$
$\Rightarrow (1+2t)+2\left(4t\right)-(2-3t)+1=0$
$\Rightarrow t=0$
$\therefore$ Intersecting point $B(1,0,2)$ and

$L_3:\frac{x}{2}=\frac{y-1}{2}=\frac{z}{1}=k$ (say)
$\Rightarrow x=2k,y=2k+1,z=k$
$\Rightarrow 4\left(2k\right)-(2k+1)+3k=8$
$\Rightarrow k=1$
$\therefore$ Intersecting point $C(2,3,1)$

Now, $AB=\sqrt{19},BC=\sqrt{11}$ and $CA=4$

Figure of trinagle :

By cosine rule
$\cos C=\frac{11+16-19}{2\cdot \sqrt{11}\cdot 4}=\frac{1}{\sqrt{11}}$
$\cos B=\frac{19+11-16}{2\cdot \sqrt{19}\cdot \sqrt{11}}=\frac{7}{\sqrt{19}\cdot \sqrt{11}}$
$\cos A=\frac{19+16-11}{2\cdot \sqrt{19}\cdot 4}=\frac{3}{\sqrt{19}}$

Hence, points $A,B,C$ constitute an acute triangle.