#### Show that the straight lines whose direction cosines are given by 2l + 2m - n = 0 and mn + nl + lm = 0 are at right angles.

We have given

$2l + 2m - n = 0...(i)\\ \Rightarrow n = 2(l + m)...(ii)$
, and

$mn + nl + lm = 0\\ \Rightarrow 2m(l + m) + 2(l + m)l + lm = 0\\ \Rightarrow 2lm + 2m^{2} + 2l^{2} + 2lm + lm = 0\\ \Rightarrow 2m^{2} + 5lm + 2l^{2} = 0\\ \Rightarrow 2m^{2} + 4lm + lm + 2l^{2} = 0\\ \Rightarrow (2m + l)(m + 2l) = 0$

Thus, we get two cases:

$l = -2m$

=> -4m + 2m - n = 0 [from (i)]

=> n = 2m

, and

m = -2l

=> 2l + 2(-2l) - n = 0

=> 2l - 4l = n

=> n = -2l

Hence, the direction ratios of one line is proportional to -2m, m or -2m or direction ratios are (-2, 1, -2) and the direction ratios of another line is proportional to l, -2l, -2l, or direction ratios are (1, -2, -2)

Thus, the direction vectors of two lines are $b_{1}=-2\hat{i}+\hat{j}-2\hat{k} \: and \: b_{2}=\hat{i}-2\hat{j}-2\hat{k}$

Also, the angle between the two lines $\vec{r}=\vec{a_{1}}+\lambda \vec{b_{1}} \: and \: \vec{r}=\vec{a_{2}}+\mu \vec{b_{2}}$ is given by:

$\cos \theta=\left |\frac{\vec{b_{1}}.\vec{b_{2}}}{\left |\vec{b_{1}} \right |\left |\vec{b_{2}} \right |} \right |$

Now,

$\vec{b_{1}}.\vec{b_{2}}=2(1)+1(-2)+(-2)(-2)\\ =2-2+4\\=0\\ \Rightarrow \cos\theta=0\\ \Rightarrow \theta=90^{\circ}$

Therefore, the lines have a 900 angle between them.