hello again. problem with vectors this time.
this is a CIE paper from October/November 2007, paper 3 and is question 10.
the paper is attached but here it is anyway
The straight line l has equation r = i + 6j - 3k + s(i - 2j + 2k). The plane p has equation
(r - 3i).(2i - 3j + 6k) = 0. The line l intersects the plane p at the point A.
(i) Find the position vector of A. [3]
(ii) Find the acute angle between l and p. [4]
(iii) Find a vector equation for the line which lies in p, passes through A and is perpendicular to l. [5]
for part (i) I have put in r into the plane equation and completed the dot product to work out s = 2.
But then i get lost, the answers suggest putting s = 2 into the equation for l to find A but i don't understand why that is can somebody explain??
for part (ii) I know i have to use r.N = cos x / (|r|*|N|) where N is the normal to the plane but what is the normal to the plane?? the answers suggest 2i ? 3j + 6k (the second part from the plane equation) but i don't know why.
for part (iii) i think i may have to use the dot product = 0 but I'm not really sure how??
Thank you to whoever helps me