Vector discussion...

[daredevil]

here u go full solution... rem me in yah prayers...

shouldnt it be -28 instead of 28 on page 3 last step? because u took the minus common ryt??

Thanks a tonn for ur help!! May you get the highest of grades and get into the uni of ur dreams

 

[daredevil]

if you dont stop posting question after question your gona scare me to the moon -_-

hahha i'm already there so ....

 

[A star]

hahha i'm already there so ....

haha you acted as a alarm clock now i am freakin out too thanks alot -_-

 

[daredevil]

haha you acted as a alarm clock now i am freakin out too thanks alot -_-

hahahhaa sorrryy??

 

[Thought blocker]

Nice thread

 

[ZaqZainab]

ZaqZainab

 

[saadgujjar]

Plz explain vector part 2

 

[shinnyyy]

thanks i had it written down but couldnt remember what D was xD

np , D is the distance of one plane from the origin , , If you'll have the distances of each plane from the origin you can calculate the distance between planes.
There's also a direct formula to calculate the distance but i wont tell you as it can confuse you -_-

 

[A star]

np , D is the distance of one plane from the origin , , If you'll have the distances of each plane from the origin you can calculate the distance between planes.
There's also a direct formula to calculate the distance but i wont tell you as it can confuse you -_-

yeah thanks. dont worry you cannot confuse me more than i already a

 

[A star]

Plz explain vector part 2View attachment 42676

its like this bro
solve the two vector equations simultaneously and obtain the coordinate of a point od intersection( its on the line of intersection)
for example taking z=0 in both cases the equations become
x+ 2y=7
2x+y=5
solve them and find the value of a and b then this will be a
a=(x,y,0)
equation of a line is r=a+ lamdam
to find out m = vector product of n1 and n2
find out m and place it in equation
p.s instead of z=0 you can take any value 2,1,3,5) it doesnt matter but use the same value for both equations

 

[saadgujjar]

its like this bro
solve the two vector equations simultaneously and obtain the coordinate of a point od intersection( its on the line of intersection)
for example taking z=0 in both cases the equations become
x+ 2y=7
2x+y=5
solve them and find the value of a and b then this will be a
a=(x,y,0)
equation of a line is r=a+ lamdam
to find out m = vector product of n1 and n2
find out m and place it in equation
p.s instead of z=0 you can take any value 2,1,3,5) it doesnt matter but use the same value for both equations

we can also take x or y=0?

 

[A star]

we can also take x or y=0?

yes you can. its up to your choice but for practice if you take z=0 your ans matches the ms and its in my opinion the easiest option

 

[Thought blocker]

u posted a question in the math thread and its answer was here

Which question

 

[Rutzaba]

Which question

oops wrong quote srry

 

[Rutzaba]

u posted a vector question in the math thread dint u?

 

[Thought blocker]

oops wrong quote srry

Oki Okay...

 

[ZaqZainab]

u posted a vector question in the math thread dint u?

I am already done with my maths exam and i don't remember posting a vector question

 

[A star]

found it!
done by sumone else
We have to find the point p where the perpendicular distance to the two planes is same.


first we will for an equation the way the paper gave.
therefore
for plane m
|x+2y-2z-1|/sqrt((1^2) + (2^2) + (-2^2) = |x+2y-2z-1|/3 [P.S, sqrt is square root]
for plane n
|2x-2y+z-7|/sqrt((2^2)+(-2^2)+(1^2) = |2x-2y+z-7|/3


since you are finding a point where both distance is same , therefore.
|x+2y-2z-1|/3 = |2x-2y+z-7|/3 (3 can be cancelled both sides)
=
|x+2y-2z-1| = |2x-2y+z-7|

now we will sub the line values in the equation.
the x component (1+2t) [P.S, t is lamda ]
the y comp. (1+t)
the z comp. (-1+2t)

therefore.
| (1+2t) +2 ( 1+t) -2(-1+2t) -1 | = |2(1+2t) - 2(1+t) + (-1+2t) - 7 |
simplifying it you get
|4| = |-8+4t|
now to solve modulus, we do squaring method
16 = 64-64t+16t^2
simplify
2 = 8 - 8t +2t^2
2t^2 - 8t + 6 = 0
solve it and get
t = 3 or t= 1
when t=3 the position of point is [P.S, u do this by sub t value in line equation)
(7, 4, 5) => OA
when t = 1 the position of point is
(3, 2, 1) => OB

BA (or AB, same thing) = OA - OB
= (4 , 2, 4)

now find the mod
sqrt(4^2 + 2^2 + 4^2)
= 6

all this for 6 marks

 

[Thought blocker]

all this for 6 marks

Is A level very tough than As level ?

 

[Rutzaba]

all this for 6 marks

pehli dafa aya tha ppr me... i did half of it