• We need your support!

    We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
    Click here to Donate Now (View Announcement)

Maths P3( Vectors)

Messages
45
Reaction score
0
Points
0
When finding vector equation of line of intersection of planes.....in the markscheme they are putting y=0....but in the book they are putting z=0
The answers will be different for both so what should i do?
 
Messages
47
Reaction score
0
Points
0
guys i also need help in P3 vectors...if anyone could provide me consolidated notes of vectors it would be of much help..!!
 
Messages
122
Reaction score
2
Points
0
sourgho16 said:
When finding vector equation of line of intersection of planes.....in the markscheme they are putting y=0....but in the book they are putting z=0
The answers will be different for both so what should i do?

can you be more specific please
 
Messages
502
Reaction score
8
Points
28
no matter....u can keep anyone 0...either its x or y or z..answers are different..but doesnot matter..!! b sure u r correct in ur methods and calculations..!!
 
Messages
45
Reaction score
0
Points
0
Yeah....but if the answers are not in the markscheme(in the markscheme only one shown is y=0) won't they cut marks???
 
Messages
760
Reaction score
10
Points
0
as mrpaudel said u may put any1 as zero as long as ur direction vector is the same as the one in the markscheme then it is correct!
 
Messages
621
Reaction score
31
Points
28
Okay let me explain this.

When two planes intersect they form a line. To get the equation of this line we apply the elimination method on the equations of the two planes. Now we have to eliminate two of the constants, it can either be x, y or x, z or y, z. It doesn't really matter which two you eliminate. Once you have eliminated two of them correctly, you will have formed a cartesian equation of the required line in the following format:

(x-1)/2 : (y-2)/3 : (z-3)/4 = X

Where 1, 2, 3 is a fixed point on the required line and 2, 3 , 4 is the direction ratio (make sure it is in the simplest ratio). Now this direction ratio should match with the marking scheme and the fixed point could be different depending on which two constants you eliminated.

Now if required simply write the vector equation of this line from the above cartesian equation:

r = (1i, 2j, 3k) + X (2i, 3j, 4k)

Hope that helps!
 
Top