A level H2 Maths: Vectors

viopurp SG

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. >.<

viopurp SG

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! =)

eagle SG

can anyone try? I very very very long never touch vectors... a bit tired now...

If not solved by tmr noon, then I attempt...

eagle SG

bump...

i too busy to do recently...

16/f/lonely SG

NS made me stupid.

Uncertain SG

-- remove --

jiaxing2 SG

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...

^tamago^ SG

i'm also not free sia. i shld be able to do it tonight for u. (:

eagle SG

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

^tamago^ SG

jiaxing’s one seems correct liao… i thought Uncertain’s is the model answer. =.=

Uncertain SG

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

Uncertain SG

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)

jiaxing2 SG

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...

Uncertain SG

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.

eagle SG

there's no sry in homework forum for extra workings or information

we are all sharing our knowledge with each other

viopurp SG

thanks a lot for the help! =)

sorry if this seems a bit late, haven't logged in for quite some time. lol.