Question

if the dcs of two lines are connected by the relation 2l+m+2n=0 and 3l2+5m2-11n2=0 then angle between the lines

Answer 1

Direction cosines

Solution:

The given relations are

$\quad 2l+m+2n=0\quad ...(1)$

$\quad 3l^{2}+5m^{2}-11n^{2}=0\quad ...(2)$

From (1), we get

$\quad m=-2l-2n$ and substituting this in (2), we find

$\quad 3l^{2}+5(-2l-2n)^{2}-11n^{2}=0$

$\Rightarrow 3l^{2}+20l^{2}+40ln+20n^{2}-11n^{2}=0$

$\Rightarrow 23l^{2}+40ln+9n^{2}=0$

$\Rightarrow 23(\frac{l}{n})^{2}+40\frac{l}{n}+9=0$

If $l_{1},\:m_{1},\:n_{1}$ and $l_{2},\:m_{2},\:n_{2}$ be direction cosines of the given lines, then we can write

$\quad \frac{l_{1}l_{2}}{n_{1}n_{2}}=\frac{9}{23}$

$\Rightarrow \frac{l_{1}l_{2}}{9}=\frac{n_{1}n_{2}}{23}$

Again, from (1), we get

$\quad l=-\frac{m+2n}{2}$ and substituting this in (2), we find

$\quad 3(-\frac{m+2n}{2})^{2}+5m^{2}-11n^{2}=0$

$\Rightarrow 3\frac{m^{2}+4mn+4n^{2}}{4}+5m^{2}-11n^{2}=0$

$\Rightarrow 3m^{2}+12mn+12n^{2}+20m^{2}-44n^{2}=0$

$\Rightarrow 23m^{2}+12mn-32n^{2}=0$

$\Rightarrow 23(\frac{m}{n})^{2}+12\frac{m}{n}-32=0$

Here we get

$\quad \frac{m_{1}m_{2}}{n_{1}n_{2}}=-\frac{32}{23}$

$\Rightarrow \frac{m_{1}m_{2}}{-32}=\frac{n_{1}n_{2}}{23}$

We have found the relation:

$\quad \frac{l_{1}l_{2}}{9}=\frac{m_{1}m_{2}}{-32}=\frac{n_{1}n_{2}}{23}=k$ (say)

Now, $l_{1}l_{2}+m_{1}m_{2}+n_{1}n_{2}$

$=9k-32k+23k=0$

So, the given lines are perpendicular to each other.

Answer: The angle between the given lines is 90°.