Find a vector that is normal to lines L1 and L2 and passes through their common point of intersection where L1 is the line r = (3,1,1) + u(1,-2,-1) and L2 is the line r = (0,-2,3) + v(-5,1,4) where u and v are scalar values.

We will use the cross product of these two vectors to help us find the perpendicular vector. The cross product works by finding a vector which has no effect on the position of a point with respect the the directions of the other 2 vectors. Its like looking down at a ball on top of a table. Moving the ball up and down won't affect its movement in the plane of the table. We use a 3x3 matrix to find the cross product of the two vectors and find the determinant of that matrix. We define it as: (i j k a b c d e f) where abc is the direction vector of L1 and def is the direction vector of L2. The determinant is then i(bf-ce) -j(af-cd) + k(ae-bd) or in the case of the question i(-7) -j(-1) + k(-9) or -7i + j -9k. To solve for their point of intersection we set the r values equal as this will find the common point. We will solve this equation simultaneously. Therefore (3+u) = (-5v), (1-2u) = (-2+v) and (1-u) = (3+4v) solving by adding equations one and two gives us 4 = 3-v so v = -1. Given we are told the lines intersect we can now use this value to find the point of intersection which will be where v = -1 in the second equation. So the point of intersection is (5,-3,-1). Therefore our final answer is (5,-3,-1) + w((-7,1,-9) where w is a scalar value.

Related Further Mathematics A Level answers

All answers ▸

Find the square roots of 2 + isqrt(5)


Differentiate w.r.t x the expression arccos(x).


How do you find the derivative of arcsinx?


Why does matrix multiplication seem so unintuitive and weird?!


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences