Math P3 Urgent help.

[fluffycloud]

Hey!
Got a few doubts in VECTORS.
I just hate this topic. :'(
May/June 2003 Q9(II)
May June 2002 Q8 (II)

Urgent Help please.
I am drowning.

 

[ahmed t]

q9ii) if u know how to find a common perpendicular then do it that way i cant do it any other way so if someone else could help u
q8ii)i also would do this using common perpendicular so tell me if u know common per so i should solve it or not

 

[WellWIshER]

GUYS HOW SHOULD WE KNOW WHICH AREA to shade

like when its < > i dont get it???

 

[fluffycloud]

q9ii) if u know how to find a common perpendicular then do it that way i cant do it any other way so if someone else could help u
q8ii)i also would do this using common perpendicular so tell me if u know common per so i should solve it or not

Could you solveeeit?

 

[fluffycloud]

bump

 

[WellWIshER]

:crazy: :O: :O:

 

[jejunestar]

For May/June 2003 Q9(II)
solve the planes to find the intersecting point....the planes are x+2y-2z= 2 and 2x-3y+6z= 3

To solve the equation suppose one of the variable as 0 , i suppose z = 0 so the equations now become x + 2y = 2 and 2x - 3y = 3, now solve them both using the linear equation solving method, u'll get x = 12/7 and y = 1/7 SO the intersecting point is (12/7,1/7,0)
[ Note the intersecting point depends upon the value which you supposed as zero, so different points of intersection of points can be obtained with one value as O]

Next you should find a common perpendicular between the normal of planes to find the direction vector of the required line.

n1 = i + 2j -2k and n2 = 2i -3k +6j so the direction vector of line is given by

p = [(2*6) - (-2*-3)] i + [(-2*2) - (1*6)] j + [(1*-3) - (2*2)] k
this gives us p = 6i -10j -7k
now you have the point and direction vector, you get the equation of line r = 12/7 i +1/7 j+ t(6i -10j -7k)


For May June 2002 8 ii
many of the vector question are simple but the condition that they give u can make your teeth sweat lol. In this question, we are to find the equation of the plane which is perpendicular to a plane p, and the line l that we got in the question (i)

we need some parts of question no (i) here, like the direction vector of line l which i got (3i - j +2) [denoted as d]

now we should get the normal of plane p from its equation which i got (i + 3j -2k) [denoted as n1]

since the plane p and our required plane (lets say b) are perpendicular so the normal of plane p is parallel to b.

So in order to find normal of b we have direction vector which is parallel to the line (that lies on b) and normal of p (which is parallel to the plane b)

So the both direction vector and normal of plane p are parallel to the required plane b, We can use them both to find a common perpendicular which is the normal of the required plane b

Normal of Required Plane b [n2]
n1 = (i + 3j -2k) d = (3i - j +2)

n 2 = [(3*2) - (-2*-1)]i +[(-2*3) - (1*2)]j +[(1*-1)-(3*3)]k
thus n 2 = 4i -8j -10k

For the plane's equation,
r.n2 = n2.a
for a use any value from the question i.e either i +j or 4i -j +3k

i used i + j
so the equation is
r(4i -8j -10k )= (4i -8j -10k) . (i+j)
4x-8y-10z = 4*1-10
or, 4x-8y-10z = -6 but the question specifies you to leave the answer as ax+by+cz = 1 so the equation becomes

-2/3x +4 /3y +5/3z = 1
hope this helped

 

[ahmed t]

thanks for taking the time and tipeing it up, i was asleep

 

[srukhan]

Guys can anyone help me in Nov/10 varaint 3 Q7)iii) I don't get what is going to be 'n' and 'a' in this to be put in the formula(r.n=a.n) to be applied?