A2 vectors Question

[loggerhead]

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

[Ghost Of Highbury]

dont just copy pste it from the paper...u gotta correct all the "?" marks displayed...

check theQ again..many tiny errors

e.g "he 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."

[loggerhead]

thanks, i didn't notice all the '-' went to '?' but i attached the paper anyway.

[nid404]

I'm still doing AS level....will try this tho

[astarmathsandphysics]

Each value of s labels a unique pount on the line. Putting s=2 will give you the unique point on the line where the line intersects the plane.

[astarmathsandphysics]

Use cos x =u>v/|u||v|
Take u ast 2i-3j+6k and v as 1-2j+2k
This will give you the angle between the normal and sthe line. To find the angle plane and the line find 90-x

[astarmathsandphysics]

For part 3 solve(r-A).(i-2j+2k)=0
i-2j+2k is the tangent vector of the line. any vector perpendiculr to this will give zero when we take the dot product.

[loggerhead]

awesome thanks astar. i think i figured out another way for part three and that is find the cross product of the tangent vectors of l and the normal of p which finds a vector on p perpendicular to l.