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# P6-CRAZY-vectors watch

1. looks like i seem to be the only dumb one around with all the questions.

4 questions appended in one document. ur help would be appreciated. its from the review exercise modular books for edexcel. but i have typed the whole questions and my areas of difficulty within a question are in blue.thankyou
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2. P6 Vectors.pdf (28.7 KB, 196 views)
3. Q53(d)
The equation of the plane can be written as: r.(2i+j+2k)=6
A line perpendicular to the plane and passing through the origin can be written as r=t(2i+j+2k)=i(2t)+j(t)+k(2k).
The line passes through the origin at t=0.
The line passes through the plane at:
[i(2t+j(t)+k(2t)].[2i+j+2k]=6
4t+t+4t=6
9t=6
t=2/3.
Hence the point R is at t=2.(2/3)=(4/3) as R lies on the same line and the same distance away from the origin, but on the other side of the plane.
Hence R is at i(8/3)+j(4/3)+k(8/3)=(4/3)[2i+j+2k]
4. Q92: A vector perpendicular to the plane is:
[i... j.. k]
[1 -2 -2]
[3..7..-6]
=i(12+14)-j(0)+k(13)
=26i+13k=13(2i+k)
So a vector perpendicular to the plane is 2i+0j+k.
The plane has equation r.(2i+k)=(0i+0j+ok).(2i+0j+k).
Taking r=(xi+yj+zk) we have 2x+z=0.

R1 can be written as r=i(3+s)+j(3-s)+k(-1-2s).
'2x+z'=2(3+s)-1-2s=6+2s-1-2s=5, which does not satisfy 2x+z=0 so the line does not lie in the plane.
R1 is parallel to the plane so for any line you draw through R1 and the plane, the distance is the same.
For this perpendicular line through O we have: r=i(2p)+j(0)+k(p), where p is our variable.
We want the intersection of: r=i(3+s)+j(3-s)+k(-1-2s) and r=i(2p)+j(0)+k(p). Clearly 3-s=0 so s=3 and the point of intersection is (6,0,-7)
The distance is that between the origin and (6,0,-7)=sqrt[36+49]=sqrt[85]

R2 can be written as r=i(4+3t)+j(5+7t)+k(-8-6t).
'2x+z'=2(4+3t)+(-8-6t)=2(4+3t)-2(4+3t)=0, which does satisfy the equation of the plane so R2 lies in the plane.
5. For 104:
(d) Find the two cartesian equations of the planes. Eliminate one the variables between then so you have a relationship between 2 variables, say x and y. Substitute this relationship back into one of the equations so you have a relationship between say x and z. You can then say x=f(y)=f(z)=t, where t is the parameter. Find x,y,z in terms of t and hence write the equation of the plane as r=i[f(t)]+j[g(t)]+k[h(t)] and rearrange to put it in the correct form.
(e) If x is on L then you can rearrange L to put it in the form X=i[..]+j[..]+k[..] then dot product it with the normal directional vector of L which is 0, and so find the value of the parameter at X, and hence the position vector of X.

Q116: N is simply where the perpendicular from P to the plane intersects the plane. You can find an equation of the line easily as you know a vector perpendicular to the plane and the position vector P. You then rearrange the line in the form r=i[..]+j[..]+k[..] and set that as 'r', a point in the plane. Solve to find t and hence N.
6. (Original post by Gaz031)
For 104:
(d) Find the two cartesian equations of the planes. Eliminate one the variables between then so you have a relationship between 2 variables, say x and y. Substitute this relationship back into one of the equations so you have a relationship between say x and z. You can then say x=f(y)=f(z)=t, where t is the parameter. Find x,y,z in terms of t and hence write the equation of the plane as r=i[f(t)]+j[g(t)]+k[h(t)] and rearrange to put it in the correct form.
(e) If x is on L then you can rearrange L to put it in the form X=i[..]+j[..]+k[..] then dot product it with the normal directional vector of L which is 0, and so find the value of the parameter at X, and hence the position vector of X.

Q116: N is simply where the perpendicular from P to the plane intersects the plane. You can find an equation of the line easily as you know a vector perpendicular to the plane and the position vector P. You then rearrange the line in the form r=i[..]+j[..]+k[..] and set that as 'r', a point in the plane. Solve to find t and hence N.
gaz u dunno who much u have taken the weight off my shoulders thanks a heap.

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