Vectors! two lines L1 and L2 are given by r1 = (9, 4, -6) + s(-2, 6,10) and r2 = (1, 20, 2) +t(-6, 10, -2)?

Two lines L1 and L2 are given by r1 = (9, 4, -6) + s(-2, 6,10) and r2 = (1, 20, 2) +t(-6, 10, -2)





a. let (theta) be the acute angle between L1 and L2. show that cos(theta) = 52/140





b. (i) P is the point on L1 when s=1. Find the position vector of P


(ii) SHow that P is also on L2





c. A third line L3 has direction vector (6, x, -30). If L1 and L3 are parallel find the value of x.

Vectors! two lines L1 and L2 are given by r1 = (9, 4, -6) + s(-2, 6,10) and r2 = (1, 20, 2) +t(-6, 10, -2)?
Two lines L1 and L2 are given by


r1 = (9, 4, -6) + s(-2, 6,10) and


r2 = (1, 20, 2) + t(-6, 10, -2)





a. let θ be the acute angle between L1 and L2. show that cosθ = 52/140





Take the dot product of the directional vectors of the two lines.





v1 • v2 = (-2, 6,10) • (-6, 10, -2) = 12 + 60 - 20 = 52





Calculate the magnitude of the two vectors.





|| v1 || = √[(-2)² + 6² + 10²] = √(4 + 36 + 100) = √140


|| v2 || = √[(-6)² + 10² + (-2)²] = √(36 + 100 + 4) = √140





The dot product can also be expressed another way.





v1 • v2 = || v1 || || v2 || cosθ


52 = (√140)*(√140) cosθ


52 = 140 cosθ


cosθ = 52/140


____________





b. (i) P is the point on L1 when s=1. Find the position vector of P





r1(s) = (9, 4, -6) + s(-2, 6,10)


r1(1) = (9, 4, -6) + (-2, 6,10) = (7, 10, 4)





(ii) Show that P is also on L2.





Solve for t. It should be consistent for x, y, and z if P is on the line L2.





L2:


x = 1 - 6t = 7


y = 20 + 10t = 10


z = 2 - 2t = 4





x: -6t = 6


t = -1


____





x = 1 - 6(-1) = 7


y = 20 + 10(-1) = 10


z = 2 - 2(-1) = 4





True for x, y, and z. So the point P is on L2.


____________





c. A third line L3 has direction vector (6, x, -30). If L1 and L3 are parallel find the value of x.





If L1 and L3 are parallel the directional vectors of L1 and L3 should be related by a non-zero constant.





v1 = (-2, 6,10)


v3 = (6, x, -30)





-2a = 6


a = -3





-3(v1) = -3(-2, 6,10) = (6, -18, -30)


x = -18