If the direction cosines (l, m, n) of two lin | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given the relations for the direction cosines $(l, m, n)$ of the two lines: $l+m+n=0$ and $lm=0$ From $lm=0$, we have two cases: $l=0$ or $m=0$. C

Vedclass · Accepted Answer

$\frac{\pi}{3}$

Vedclass · Answer

(A) Given the relations for the direction cosines $(l, m, n)$ of the two lines: $l+m+n=0$ and $lm=0$ From $lm=0$, we have two cases: $l=0$ or $m=0$. Case $1$: If $l=0$, then $0+m+n=0 \Rightarrow n=-m$. The direction ratios are $(0, m, -m)$, which simplifies to $(0, 1, -1)$. Case $2$: If $m=0$, then $l+0+n=0 \Rightarrow n=-l$. The direction ratios are $(l, 0, -l)$, which simplifies to $(1, 0, -1)$. Let the direction vectors be $\vec{a} = (0, 1, -1)$ and $\vec{b} = (1, 0, -1)$. The cosine of the angle $	heta$ between the two lines is given by: $\cos 	heta = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| |\vec{b}|} = \frac{|(0)(1) + (1)(0) + (-1)(-1)|}{\sqrt{0^2+1^2+(-1)^2} \cdot \sqrt{1^2+0^2+(-1)^2}}$ $\cos 	heta = \frac{|0 + 0 + 1|}{\sqrt{2} \cdot \sqrt{2}} = \frac{1}{2}$ Therefore, $	heta = \cos^{-1}(\frac{1}{2}) = \frac{\pi}{3}$.

If the direction cosines $(l, m, n)$ of two lines are connected by the relations $l+m+n=0$ and $lm=0$, then the angle between those lines is

Similar Questions

If the direction cosines of two lines are such that $l+m+n=0$ and $l^2+m^2-n^2=0$, then the angle between them is

The lines whose direction cosines are given by the relations $a l+b m+c n=0$ and $m n+n l+l m=0$ are

If the direction cosines of two lines satisfy the equations $2l+m-n=0$ and $l^2-2m^2+n^2=0$, and $\theta$ is the angle between the lines, then $\cos \theta=$

The direction cosines of the line $x = y = z$ are

The number of straight lines that are equally inclined to the three-dimensional coordinate axes is

Vedclass Test Series

Exam Paper Generator

Online Exam Module