Find the measure of the angle between two lines if their direction cosines $ℓ, m, n$ satisfy $ℓ + m - n = 0, ℓ^{2} + m^{2} - n^{2} = 0$ .

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Solution

given ocs of two line satisfy

l + m - n = 0 equation (1)
$l^{2} + m^{2} - n^{2} = 0 e q u a t i o n (2)$

to add angle between from (1) l = n - m in equation (2)

$\begin{matrix} \Rightarrow {(n - m)}^{2} + m^{2} - n^{2} = 0 n^{2} + m^{2} - 2 n m + m^{2} - n^{2} = 0 2 m^{2} - 2 n m = 0 2 m (m - n) = 0 \end{matrix}$

m = 0 equation (3) and m - n = 0 equation (4)

l + m - n = 0

0 + m - n = 0

$\frac{l}{0 + 1} = \frac{m}{0} = \frac{n}{1}$

l:m:n = 1 : 0 : 1
equation 1 and 4
l + m - n = 0

0 + m - n = 0
$\frac{l}{0} = \frac{m}{- (- 1)} = \frac{n}{1}$
l : m : n = 0 : 1 : 1

$cos θ = \frac{a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}}{\sqrt{{a_{1}}^{2} + {b_{1}}^{2} + {c_{1}}^{2}} \sqrt{{a_{2}}^{2} + {b_{2}}^{2} + {c_{2}}^{2}}}$

$= \frac{(1) (0) + (0) (0) + (1) (1)}{\sqrt{1 + 0 + 1} \sqrt{0 + 1 + 1}} = \frac{1}{2}$

$cos θ = \frac{1}{2} \Rightarrow θ = \frac{π}{2}$

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