Question

Show that the lines whose d.c's are given by $2l+2m-n=0,mn+nl+lm=0$ are perpendicular to each other.

Answer 1

Consider the given equation.

$2l+2m-n=0$

$n=2m+2n$

Substitute this value in the second equation.

$m(2l+2m)+l(2l+2m)+lm=0$

$2lm+2m^2+2l^2+2lm+lm=0$

$2m^2+5lm+2l^2=0$

$(m+2l)(2m+l)=0$

$m=-2lor-\frac{l}{2}$

Now, put $l=1$.

$m=-2or-\frac{1}{2}$

$n=-2or1$

Therefore,

$d_1=(1,-2,-2)$

$d_2=(1,-\frac{1}{2},1)$

Now,

$d_1\cdot d_2=1\times 1+(-2)\times (-\frac{1}{2})+(-2)\times \left(1\right)$

$d_1\cdot d_2=1+1-2$

$d_1\cdot d_2=0$

Hence, both the lines are perpendicular to each other.