MCQ
The angle between the lines whose direction cosines satisfy the equations $l + m + n = 0$ and ${l^2} = {m^2} + {n^2}$ is
- A$\frac{\pi }{6}$
- B$\frac{\pi }{2}$
- ✓$\frac{\pi }{3}$
- D$\frac{\pi }{4}$
Eliminationg $n$ from both the equations, we have
${l^2} + {m^2} - {\left( {l + m} \right)^2} = 0$
$ \Rightarrow {l^2} + {m^2} - {l^2} - {m^2} - 2ml = 0$
$ \Rightarrow 2lm = 0$
$ \Rightarrow lm = 0$
$ \Rightarrow l = 0\,\,\,\,or\,\,\,m = 0$
If $l=0$, we have $m+n=0$ and ${m^2} - {n^2} = 0$
$ \Rightarrow l = 0,m = \lambda ,n = - \lambda $
If $m=0$, we have $l+m=0$ and ${l^2} - {m^2} = 0$
$ \Rightarrow l = - \lambda ,m = 0,n = \lambda $
So, the vector parallel to these given lines
are $\vec a = \hat j - \hat k\,$ and $\,\,\vec b = - \hat i + \hat k$
If angle between the lines is $'\theta ',$ then
$\cos \theta = \frac{{\left| {\vec a.\vec b} \right|}}{{\left| {\vec a} \right|\left| {\vec b} \right|}} = \frac{1}{{\sqrt 2 .\sqrt 2 }}$
$ \Rightarrow \cos \theta = \frac{1}{2}$
$\therefore \theta = \frac{\pi }{3}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$(A)$ The length of the line segment $O A_1$ is $4 \sqrt{3}$
$(B)$ The length of the line segment $A_1 B_1$ is 16
$(C)$ The orthocenter of the triangle $A_1 B_1 C_1$ is $(0,0)$
$(D)$ The orthocenter of the triangle $A_1 B_1 C_1$ is $(1,0)$