The line L1 has vector equation,  L1 = (  6, 1 ,-1  ) + λ ( 2, 1, 0). The line L2 passes through the points (2, 3, −1) and (4, −1, 1). i) find vector equation of L2 ii)show L2 and L1 are perpendicular.

i)L(r) vector line equations in general are in the form L(r) = p1 + λ (p2) where p1 is any ponit on the line and p2 is the vector direction of the line(unsure how to get the whiteboard up or i would describe this with a diagram)to find the vector diction on the line you work out the vector between two points on the line this is done by subtacting one from the other. (again would show this on a diagram) in this instance for L2, p2 = (2, 3, -1) - (4, -1, 1) = (-2, 4 ,-2) Therefore L2 can either be L2= (2, 3, −1) + λ(-2, 4 ,-2) or L2= (4, −1, 1) + λ(-2, 4 ,-2)ii) for vectors to be perpendicular the vector directions dot products must equal to 0L1 . L2 = 0(2, 1 , 0) . (-2, 4 , -2)= -4 + 4 = 0 

CE

Related Maths A Level answers

All answers ▸

Using the trigonometric identity for tan(A + B), prove that tan(3x)=(3tan(x)-tan^3(x))/(1-3tan^2(x))


The gradient of a curve is defined as Dy/dx = 3x^2 + 3x and it passes through the point (0,0), what is the equation of the curve


The curve C has equation y=2x^2 -11x +13. (a) The point P has coordinates (2, – 1) and lies on C. Find the equation of the tangent to C at P.


Write 9sin(x) + 12 cos(x) in the form Rsin(x+y) and hence solve 9sin(x) + 12 cos(x) = 3