Question

The angle between the lines whose dcs are given by th lines l+3m+5n and 5lm-2mn+6nl

Answer 1

Answer:

Step-by-step explanation:

Given l + 3m + 5n = 0 and 5lm - 2mn + 6nl = 0

l = -3m - 5n

l(5m + 6n) = 2mn

(3m + 5n)(5m + 6n) = -2mn

15m² + 45mn + 15n² = 0

m² + 3mn + n² = 0

(m + 2n)(m + n) = 0

=> m = -2n or m = -n

If m = -2n, we get l = n. Hence dc's would be (n, -2n, n)

Also we know for any dc's (l, m, n), l²+m²+n²=1-----------(1)

Hence, we have n² + (-2n)² + n² = 1

we get 6n² = 1

n = 1/√6,

dc's of line 1 are (1/√6, -2/√6, 1/√6)

Similarly, if we consider m = -n, we get

l = -2n

so dc's are ( -2n, -n, n) and from equation(1) ,

we get dc's of 2nd line as (-2/√6, -1/√6, 1/√6).

If (l₁, m₁, n₁) and (l₂, m₂, n₂) are dc's of any two given lines , then the angle between lines is given by cos(Ф) = l₁l₂ + m₁m₂ +n₁n₂

   = 1/6

Thu, angle between the lines Ф is given by cos⁻¹(1/6).