Question

prove that the straight lines whose direction cosines are given by the relation al+bm+cn=0 and fmn+gnl+hlm=0 are parallel if...

Answer 1

\[\sqrt{af}\pm \sqrt{bg}\pm \sqrt{ch}=0\]

Answer 2

we have,
al+bm+cn=0
fmn+gnl+hlm=0
l^2 + m^2 + n^2 = 1

Answer 3

What is f,g,h ??

Answer 4

variables..

Answer 5

They are not DC' s , right ??

Answer 6

DC-->direction cosine

Answer 7

no, they are not DC's

Answer 8

So. This is how I get the question.

a,b,c--->DC of line 1
l,m,n-->DC of line 2
f,g,h--->Some independent variables

Do correct me if I am wrong because they are essential in solving the question

Answer 9

l, m, n are direction cosines .. a,b,c are some constants and f,g,h are variables

Answer 10

you sure about the if .. case??

Answer 11

LOL, One set has to be DC of a line. Without that How can we prove 2 lines are parallel ??

Answer 12

I meant one more set of variables

Answer 13

the question says so.. and i am sure i haven't made typing mistakes..

Answer 14

\[ mn(af - gb - hc) - gc n^2 - hb m^2 = 0 \]
\[ mn(2bc) + (1+b^2)m^2 + (1+c^2)n^2 - a^2 = 0 \]

These two system should have a singular solution!!

Answer 15

@ujjwal,

It is easy.
1)Take equation al + bm + cn = 0 and write it in terms of n = ...........
2) Then substitute this n in fmn+gnl+hlm=0
3)After simplifying, divide the equation you get by m^2
Therefore you arrive at a quadratic of ax^2 + bx + c = 0
where in your case, x = l/m
a,b,c-->random constants you get while solving.

Now. let
\[\frac{l_1}{m_1} , \frac{l_2}{m_2}\]
be the two roots of the quadratic equation

Now, take product of roots from quadratic equation
\[\frac{l_1}{m_1} . \frac{l_2}{m_2} = -\frac{b}{a}\]
where b, a are the constants you get in quadratic equation while solving

Do till here , after that,I will guide you on the last part of the proving :)

Answer 16

Just tell me the quadratic euqation you are getting finally

Answer 17

*equation

Answer 18

In the end i get values of l1*l2, m1*m2 and n1*n2 but i need l1/l2 , m1/m2 and n1/n2.. do you get their values?

Answer 19

@ujjwal , I don't require any of the values of l1*l2, m1*m2 and n1*n2 or l1/l2 , m1/m2 and n1/n2. My proof is based on different lines
If you are done, can you give me the quadratic equation which is the most important part of this proof, which you get after 3rd step

Answer 20

In my above comment lines means ideas

Answer 21

By a similar process i get a similar equation \[bh \left( m /n\right)^{2}+ (bg+ ch-af)\left( m/n \right)+cg=0\]

Answer 22

Sorry for the delay @ujjwal .
The two lines are parallel if
\[\frac{l_1}{l_2} = \frac{m_1}{m_2}= \frac{n_1}{n_2}\]
Therefore
\frac{m_1}{n_1}= \frac{m_2}{n_2}\

Now with quadratic equation, we can say that two real and equal roots exist for this quadratic equation,

therefore for quad equation of the form ax^2+bx+c=0,
two roots are real and equal if
\[b^2-4ac=0\]
Now substitute b ,a ,c in this equation and simplify. You should get your answer

Answer 23

Typo:
Therefore \[\frac{m_1}{n_1}= \frac{m_2}{n_2}\]

Answer 24

thanks!

Answer 25

Welcome :)