Three Dimensional Geometry Part 2

Given,
2l + 2m − n = 0 ....(i)
mn + nl + lm = 0 ...(ii)
From Eq. (i), we get
n = 2l + 2m ...(iii)
Substituting, n = 2l + 2m in Eq. (ii), we have
m(2l + 2m) + l(2l + 2m) + lm = 0
⇒2‌Im+2m2+2I2+2‌Im+Im=0
⇒2I2+4‌Im+Im+2m2=0
⇒ 2l(l + 2m) + m(l + 2m) = 0
⇒ (2l + m) (l + 2m) = 0
When
2l = − m
From Eq (iii),
n = m
⇒

2I

−1

=

m

1

=

n

1

⇒

I

− 1 2

=

m

1

=

n

1

or

I

1

=

m

−2

=

n

−2

⇒ (l, m, n) = (1, − 2, − 2)
When
l = − 2m
From Eq. (iii),
n = − 2m
⇒ l = − 2m = n
⇒

I

−2

=

m

1

=

n

−2

⇒ (l, m, n) = (− 2,1, − 2)
∴ Angles between straight lines cos‌θ=

−2−2+4

9

=0
⇒θ=

π

2