Find the intersection point and the angle between the lines L1 and L2. L1: x = - l + 2t, y = 3 - t, z = 2 + 2t, L2: x = - 2 - t, y = 5 + 2t, z = - 2t.

Answer:(-1,3,2)Step-by-step explanation:For L1:x= -1+2t y= 3-tz= 2+2tand for L2:x= -2-sy= 5+2sz=-2sTo find the intersection point we equalize the x, y and z:-1+2t = -2-s3-t = 5+2s2+2t = -2s⇒ s= -1-tWe replace the s value -1-t in second equation:3-t = 5+2(-1-t)3-t = 5-2-2tt = 0.So, s = -1-t = -1-0 = -1.We replace the s and t value in the parametric equations to find the interception point:x= -1 +2(0) = -1y = 3-0 = 3z = 2 +2(0) = 2.So, the interception point is (-1,3,2). The formula to calculate the angle is in the picture below, where α is the angle, u and v are the parallel vector of each line.For L1: u=(2,-1,2) (the coefficients of t)For L2: v=(-1,2,-2) (the coefficients of s)So, the angle is:cos(α) = [tex]\frac{|2(-1)+(-1)2+2(-2)|}{\sqrt{2^{2}+(-1)^{2}+2^{2}}\sqrt{(-1)^{2}+2^{2}+(-2)^{2}}  }[/tex]cos(α)=  [tex]\frac{|-2-2-4|}{\sqrt{4+1+4}\sqrt{1+4+4}}[/tex]cos(α)= [tex]\frac{|-8|}{\sqrt{9}\sqrt{9}}[/tex]cos(α)= [tex]\frac{8}{9}[/tex]α = [tex]cos^{-1}(\frac{8}{9})= 27.26[/tex]