direction cosines

Eliminating " l " from the two given equations , we obtain : -

( m + n ) ² + m ² - n ² = 0

Or , 2 m ² + 2 m n = 0

Or , ( mn ) ² + mn = 0

Or , K ² + K = 0

So , K = 0 , - 1

Now , we assume that the two lines have direction ratios " l₁ , m₁ , n₁ " and " l₂ , m₂ , n₂ " corresponding to the two values of " K " .

1 . K = 0 → m₁ = 0

So , from the first equation , l₁ = - n₁

Hence , if we let " a = l₁ " , we must have : -

n₁ = - a

m₁ = 0

2 . K = - 1 → m₂ + n₂ = 0

So , l₂ = 0 .

Now , if we let " b = m₂ " , then : -

n₂ = - b

l₂ = 0

Let the angle between the lines be " θ " . Then : -

cos θ = l₁ l₂ + m₁ m₂ + n₁ n₂( l₁² + m₁² + n₁² )^{1 / 2} ( l₁² + m₁² + n₁ )^{1 / 2}

= 0 + 0 + a b( a ² + a ² + 0 )^{1 / 2} ( b ² + b ² + 0 )^{1 / 2}

= 12

So , θ = 60 ° .