Find the angle between two lines with vector equations r1 = (2i+j+k)+t(3i-5j-k) and r2 = (7i+4j+k)+s(2i+j-9k)

Explain the dot product and the relation : a.b= ¦a¦¦b¦cos(x) where a.b is the dot product of two vectors, ¦a¦ and ¦b¦ are the magnitude of a and b respectively and x is the angle between them. Since the angle between two lines only depends on their gradients, we only need to consider the direction product for this question.Therefore let a = (3i-5j-k) and b = (2i+j-9k) 1 - Find the dot product of vectors a and b: (3i-5j-k).(2i+j-9k) = (32)+(-51)+(-1*-9) = 6-5+9 = 10 2 - Find the magnitude of vectors a and b: ¦a¦ = ¦3i-5j-k¦ = sqrt(3²+(-5)²+(-1)²) = sqrt(35) ¦b¦ = ¦2i+j-9k¦ = sqrt(2²+1²+(-9)²) = sqrt(86) Substitute the values of a.b ¦a¦ and ¦b¦ into a.b= ¦a¦¦b¦cos(x) and simplify: 10 = sqrt(35)*sqrt(86)cos(x) 10 = sqrt(3586)*cos(x) Rearrange for cos(x): cos(x) = 10/sqrt(3010) x = cos^-1(10/sqrt(3010)) x = 79.5^o (t 3s.f)