The lines whose direction cosines satisfy the | vedclass

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

Question

(A) Given equations are $al + bm + cn = 0$ and $fmn + gnl + hlm = 0$.From the first equation,$n = -\frac{al + bm}{c}$.Substituting this into the secon

Vedclass · Accepted Answer

$\frac{f}{a} + \frac{g}{b} + \frac{h}{c} = 0$

Vedclass · Answer

(A) Given equations are $al + bm + cn = 0$ and $fmn + gnl + hlm = 0$.From the first equation,$n = -\frac{al + bm}{c}$.Substituting this into the second equation: $fm(-\frac{al + bm}{c}) + gl(-\frac{al + bm}{c}) + hlm = 0$.Multiplying by $-c$,we get $fm(al + bm) + gl(al + bm) - chlm = 0$.$aflm + bfm^2 + agl^2 + bglm - chlm = 0$.$agl^2 + (af + bg - ch)lm + bfm^2 = 0$.Dividing by $m^2$,we get $ag(\frac{l}{m})^2 + (af + bg - ch)(\frac{l}{m}) + bf = 0$.Let the roots be $\frac{l_1}{m_1}$ and $\frac{l_2}{m_2}$. Then $\frac{l_1 l_2}{m_1 m_2} = \frac{bf}{ag}$,which implies $\frac{l_1 l_2}{f/a} = \frac{m_1 m_2}{g/b}$.Similarly,by eliminating $l$,we get $\frac{m_1 m_2}{g/b} = \frac{n_1 n_2}{h/c}$.Thus,$\frac{l_1 l_2}{f/a} = \frac{m_1 m_2}{g/b} = \frac{n_1 n_2}{h/c} = k$.For perpendicular lines,$l_1 l_2 + m_1 m_2 + n_1 n_2 = 0$.Substituting the values,$k(\frac{f}{a} + \frac{g}{b} + \frac{h}{c}) = 0$.Since $k 
eq 0$,we have $\frac{f}{a} + \frac{g}{b} + \frac{h}{c} = 0$.

The lines whose direction cosines satisfy the equations $al + bm + cn = 0$ and $fmn + gnl + hlm = 0$ will be perpendicular if...

Similar Questions

If the lines $\frac{x - 1}{-3} = \frac{y - 2}{2k} = \frac{z - 3}{2}$ and $\frac{x - 1}{3k} = \frac{y - 5}{1} = \frac{z - 6}{-5}$ are at right angles,then $k =$

If the lines $x = ay + b, z = cy + d$ and $x = a'z + b', y = c'z + d'$ are perpendicular,then

Let $L$ be the line $\frac{x+1}{2}=\frac{y+1}{3}=\frac{z+3}{6}$ and let $S$ be the set of all points $(a, b, c)$ on $L$, whose distance from the point $P(-1, -1, -9)$ is $7$. Then $\sum_{(a,b,c)\in S} (a+b+c)$ is equal to:

Let the shortest distance between the lines $\frac{x-3}{3}=\frac{y-\alpha}{-1}=\frac{z-3}{1}$ and $\frac{x+3}{-3}=\frac{y+7}{2}=\frac{z-\beta}{4}$ be $3 \sqrt{30}$. Then the positive value of $5 \alpha+\beta$ is

Find the vector equation of the line $\frac{x - 2}{2} = \frac{2y - 5}{-3}, z = -1$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module