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# Help me in vectors further maths student !! watch

1. line equation, L: (x, y, z) = (1, 1, -3) + t(1, 2, 3)
plane equation, P: -10z - y + 4z = 0

find distance between these 2
2. (Original post by MalaysianDude)
line equation, L: (x, y, z) = (1, 1, -3) + t(1, 2, 3)
plane equation, P: -10z - y + 4z = 0

find distance between these 2
hmm i hope this helps

first change the equation of the plane
r.(-10i - j + 4k) = a.n = (i +2j + 3k).(-10i -j +4k)
which implies r.(-10i - j + 4k) = -10 -2 + 12 = 0
hmm tricky
iv confused myself
see what u can do from here
3. (Original post by kriztinae)
hmm i hope this helps

first change the equation of the plane
r.(-10i - j + 4k) = a.n = (i +2j + 3k).(-10i -j +4k)
which implies r.(-10i - j + 4k) = -10 -2 + 12 = 0
hmm tricky
iv confused myself
see what u can do from here
nopes
that doesnt help at all
i got stuck there as well
so can someone help me out here?
4. (Original post by MalaysianDude)
nopes
that doesnt help at all
i got stuck there as well
so can someone help me out here?
check the question again, sometimes i copy things out wrong
5. (Original post by kriztinae)
check the question again, sometimes i copy things out wrong
no
its not wrong
its the same question
and its very small
they just want you to find the distance
thats all they gave
6. (Original post by MalaysianDude)
no
its not wrong
its the same question
and its very small
they just want you to find the distance
thats all they gave
well do u have the answer?
maybe the distance is zero coz the line intersects with the plane?
7. As they are asking for the distance between the two, they must be parallel (as otherwise they would intersect).

Therefore, just find the distance between the point on the line (1,1,-3) and the line using
Code:
```ae+bf+cg-d
d= -----------------
sqrt(a^2+b^2+c^2)```

Of course it could be a trick question as kriztinae suggests, whereby they actually intersect giving d=0.
8. My other post was wrong - I was using the wrong angle - here's the right way.

The line and the plane don't intersect (fairly obvious from the question but you can check by subbing the general point on the line into the equation of the plane), so the line will be parallel to the plane and therefore a fixed distance from it. The shortest distance will be perpendicular to the plane.

If you then draw a sketch diagram (attached), you can see that h = |a|cosx where a is just any vector joining a point on the plane and a point on the line. The dot product of a.n where n is the normal vector to the plane is a.n = |a||n|cosx so |a|cosx = h = (a.n)/|n| = a.m where m is unit vector perpendicular to the plane.

The plane passes through the origin so you can take the vector a to be (1,1,-3) - the vector from O to the point on the line with t=0. m = (-10,-1,4)/(sqrt127) so h = (-10 - 1 - 12)/(sqrt127) = -23/(sqrt127) As h is a distance it should obviously be positive so distance of line from plane is 23/(sqrt127) which is the same answer as zaphods formula gives. My method is basically the working behind this formula.
Attached Images

9. (Original post by Bezza)
My other post was wrong - I was using the wrong angle - here's the right way.

The line and the plane don't intersect (fairly obvious from the question but you can check by subbing the general point on the line into the equation of the plane), so the line will be parallel to the plane and therefore a fixed distance from it. The shortest distance will be perpendicular to the plane.

If you then draw a sketch diagram (attached), you can see that h = |a|cosx where a is just any vector joining a point on the plane and a point on the line. The dot product of a.n where n is the normal vector to the plane is a.n = |a||n|cosx so |a|cosx = h = (a.n)/|n| = a.m where m is unit vector perpendicular to the plane.

The plane passes through the origin so you can take the vector a to be (1,1,-3) - the vector from O to the point on the line with t=0. m = (-10,-1,4)/(sqrt127) so h = (-10 - 1 - 12)/(sqrt127) = -23/(sqrt127) As h is a distance it should obviously be positive so distance of line from plane is 23/(sqrt127) which is the same answer as zaphods formula gives. My method is basically the working behind this formula.
anyways,did you use the formula given by Zaphod?
10. For OCR at least, the formula I quoted is given (albeit in a form I can't easily reproduce here). So you'd have to check whether or not it would be given on your syllabus as to whether it's useable like that or not.
11. (Original post by MalaysianDude)
anyways,did you use the formula given by Zaphod?
I didn't use the formula, I just thought about it and drew a diagram. I checked in the formula though and realised I'd done it wrong the first time - I'd used x as the angle between the vector a and the plane, rather than a and the normal vector to the plane.

The formula is given in the edexcel formula book too.
12. the easiest way to solve vector problems like this is to draw a diagram.

then you can see what you have and hence whether to use the cross or dot product

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