A level H2 Maths: Vectors
-
Hello, I've got two questions here which i can't solve. >.<
First question: [SOLVED]
The plane π passes through the origin O and is perpendicular to 2i +2j + k. The line L has cartesian equation
(x-4)/1 = (y-8)/3 = (z-4)/2.
The points A and B have position vectors i + 2j - 3k and 3i + 5j + 2k respectively.
(i) State a cartesian equation for the plane π.
(ii) The point C on L has position vector c, where c.k = 0. Find c, and hence state an alternative equation for L of the form
(x-p)/l = (y-q)/m = z/n,
giving numerical values for p, q, l, m and n.
I've found the answer to part (i) of the question and it's 2x + 2y + z = 0.
The problem is with part (ii) of the question. I can get all the correct numerical values for the letters except for p and l, which are supposed to be 2 and 1 respectively (according to the answers given). I've gotten -8 and 4 for those two letters instead. Oh and for position vector c, what I've gotten was -8i + 2j. Not sure whether that's correct though, since the answer to that was not given.
So can anyone kindly enlighten me on this question? Thanks! =)
**Updates**: This question has been solved. Realised that it was due to a careless mistake at the start. Oops. >.<
-
Second question:
With respect to an origin O, the point A has position vector i + 3j - 4k and the point B has position vector -12i + 8j + 4k. The point P is on the line OA. The points Q and R are such that vector PQ = 1/3 vector QB and vector OR = 5 vector OQ.
(i) Show that R is on the line l whose vector equation is
r = -15i + 10j + 5k + θ( i + 3j - 4k ).
(ii) The plane π contains the line l and is parallel to the vector 11i + 7j + 8k. The line through A perpendicular to π meets π at L, and M is such that vector AM = 1/5 vector AL. Find the position vector of M.
Again, I have problems with part (ii) of this question as well. >.<
Would really appreciate if anyone can help me out with these. Thanks a lot! =)
-
can anyone try? I very very very long never touch vectors... a bit tired now...
If not solved by tmr noon, then I attempt...
-
bump...
i too busy to do recently...
-
NS made me stupid.
-
-- remove --
-
express the plane in normal product form...makes it alot easier to see
(i + 3j - 4k ) x (11i + 7j + 8k)...u get 52,-52,-26
Thus the normal vector is (2i -2j -k)
π: r.(2i - 2j - k )=(-15i + 10j + 5k). (2i - 2j - k ) = -55
Now to find the intersection between line through A perpendicular to the plane,
L can be expressed as (i + 3j - 4k )+ t(2i - 2j - k ) where t is a real value such that:
[(i + 3j - 4k )+ t(2i - 2j - k ) ].(2i - 2j - k ) = -55
solving the eqn t=-55/9
Therefore OL=OA+AL
=(i + 3j - 4k ) - 11/9 (2i - 2j - k )
=1/9 ( -13i + 49j - 27k).......cant be sure though...answer seems like a bomb...
-
i'm also not free sia. i shld be able to do it tonight for u. (:
-
Work is always busy...
Other than my full-time work, I'm also starting to give grp tuition (at a tuition centre) and will be looking at part time 1 to 1 home tuition soon at the same time
And looking at investments too :D
-
jiaxing's one seems correct liao... i thought Uncertain's is the model answer. =.=
-
Originally posted by ^tamago^:
jiaxing’s one seems correct liao… i thought Uncertain’s is the model answer. =.=
-- see my latter post for the suggested answer --A bit late cos i kinda busy lately
-
Ok, i solve the question ... seems correct though.
(i + 3j - 4k ) x (11i + 7j + 8k)...u get 52,-52,-26
Thus the normal vector is (2i -2j -k) <--- same as jiaxing2
Line through OA = (1, 3, -4) + b (2, 2, -1) , b is a constant
Since (2, 2, -1) is perpendicular to plane π ,
its equation becomes r. (2, 2, -1) = (-15, 10, 5) . (2, 2, -1) ---- [1]
Then sub r = (1+2b, 3+2b, -4-b) <-- eqn of line OA into [1] to find OL,
u will get b = -3 , OL = (-5, -3, -1)
AM = 1/5 (AL)
OM - OA = 1/5 (OL - OA)
OM = 1/5 OL + 4/5OA = 1/5 (-1, 9, -17)
-
the methods are the same....we have different normal vectors thus different answers...
if (2 -2 -1) is correct then 1/9 ( -13i + 49j - 25k) should be correct...
-
Originally posted by jiaxing2:
the methods are the same....we have different normal vectors thus different answers...
if (2 -2 -1) is correct then 1/9 ( -13i + 49j - 25k) should be correct...
i agree with u. sry for the duplicate then. -
there's no sry in homework forum for extra workings or information
we are all sharing our knowledge with each other
-
thanks a lot for the help! =)
sorry if this seems a bit late, haven't logged in for quite some time. lol.