C4 vectors
Can someone please help me on these Vectors questions from C4?? I am majorly stuck!! 
Q1) Referred to a fixed origin, the line L1 has the equation r=12i+15j+k+s(2i+j+k) and the line L2 has the equation r=i+j-2k+t(3i-k)
a) Show that the lines L1 and L2 intersect and find the position vector of A, their point of intersection
b) Find, to the nearest degree, the acute angle between the lines L1 and L2
The point B with position vector 16i+7j+3k lies on the line L1. The point C with position vector 22i+j-9K and the point D both lie on the line L2.
c)Given that BD is perpendicular to AC, find the position vector of D
d)Hence, or otherwise, prove that ?ABC is isosceles
Thank you for the help in advance....I REALLY APPRECIATE IT!!
Q1) Referred to a fixed origin, the line L1 has the equation r=12i+15j+k+s(2i+j+k) and the line L2 has the equation r=i+j-2k+t(3i-k)
a) Show that the lines L1 and L2 intersect and find the position vector of A, their point of intersection
b) Find, to the nearest degree, the acute angle between the lines L1 and L2
The point B with position vector 16i+7j+3k lies on the line L1. The point C with position vector 22i+j-9K and the point D both lie on the line L2.
c)Given that BD is perpendicular to AC, find the position vector of D
d)Hence, or otherwise, prove that ?ABC is isosceles
Thank you for the help in advance....I REALLY APPRECIATE IT!!
(a) If the lines intersect, there must be a point that lies on both of them. That point will lie on L1 for some value of s and L2 for some value of t. So set up simultaneous equations for the coefficients of I J and K.
(b) have you just done scalar product?
(b) have you just done scalar product?
Original post by ian.slater
Original post by ian.slater
(a) If the lines intersect, there must be a point that lies on both of them. That point will lie on L1 for some value of s and L2 for some value of t. So set up simultaneous equations for the coefficients of I J and K.
(b) have you just done scalar product?
(b) have you just done scalar product?
Thanks, I've managed to do parts a) and b) but can't do c) and d)
Yes the scalar product --> a.b= [modulas of a x modulas of b] cos theta ??
Original post by Air18
Thanks, I've managed to do parts a) and b) but can't do c) and d)
Yes the scalar product --> a.b= [modulas of a x modulas of b] cos theta ??
Yes the scalar product --> a.b= [modulas of a x modulas of b] cos theta ??
So if the scalar product of two vectors (worked out by multiplying coefficients and adding) is zero, then the lines are perpendicular.
You can work out AC. You know D lies on L2 and can work out BD - which will have a 't' in it. If you set AC.BD to zero you can find t and hence D.
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