Angle between Two Intersecting Lines

Let v = a1 + λb1 and v = a2 + λb2 be the equations of two straight lines. If θ is the angle between them, then cos θ = \(\frac{{{b}_{1}}-{{b}_{2}}}{\left| {{b}_{1}} \right|\left| {{b}_{2}} \right|}\),

Also if θ is the angle between \(\frac{x-{{x}_{1}}}{{{a}_{1}}}=\frac{y-{{y}_{1}}}{{{b}_{1}}}=\frac{z-{{z}_{1}}}{{{c}_{1}}}\) and \(\frac{x-{{x}_{1}}}{{{a}_{2}}}=\frac{y-{{y}_{1}}}{{{b}_{2}}}=\frac{z-{{z}_{1}}}{{{c}_{2}}}\),

Then \(\cos \theta =\frac{{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}\).Angle between Two Intersecting LinesCondition of perpendicularity: The lines are perpendicular, if b1 – b2 = 0 or a₁a₂ + b₁b₂ + c₁c₂ = 0

Condition of parallelism: The lines are parallel, if b₁ = λb₂ for some scalar λ or \(\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{{{c}_{1}}}{{{c}_{2}}}\).

Example: Find the angle between the pair of lines

v = 3i + 2j – 4k + λ (I + 2j + 2k)

And v = 5i – 2k + µ (3i + 2j + 6k)

Solution: Let the angle is θ then \(\cos \theta =\frac{{{b}_{1}}{{b}_{2}}}{\left| {{b}_{1}} \right|\left| {{b}_{2}} \right|}\),

= \(\frac{\left( i+2j+2k \right)\left( 3i+2j+6k \right)}{\sqrt{1+4+4}\sqrt{9+4+36}}\),

= \(\frac{3+4+12}{\sqrt{9}\sqrt{49}}\),

= \(\frac{19}{21}\),

θ = cos⁻¹ \(\left( \frac{19}{21} \right)\).