IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => GCE AS & A2 Level => Math => Topic started by: T.Q on April 20, 2010, 10:56:36 am
-
The line l1 has vector equation
r = 8i + 12j + 14k + t(i + j – k),
where t is a parameter.
The point A has coordinates (4, 8, a), where a is a constant. The point B has coordinates
(b, 13, 13), where b is a constant. Points A and B lie on the line l1.
(a) Find the values of a and b.
(3)
Given that the point O is the origin, and that the point P lies on l1 such that OP is
perpendicular to l1,
(b) find the coordinates of P.
(5)
i need help in part B with explanation plz
-
Sorry Cant help ya :(
I am still doing my A-Ls this year..
Maybe Astar or nid can :(
-
Sorry Cant help ya :(
I am still doing my A-Ls this year..
Maybe Astar or nid can :(
no problem
i will wait for them :)
-
Can u post your answers for the first part...
-
Can u post your answers for the first part...
a=18
b=9
-
I need to know how u got it....
I can't help otherwise...if i get this i can help using my iit books
Cuz evn im doing AS
-
I need to know how u got it....
I can't help otherwise...if i get this i can help using my iit books
Cuz evn im doing AS
if ur doing AS, then i guess u can do nth in this q. :-\
-
if ur doing AS, then i guess u can do nth in this q. :-\
I'm doing iit as well...that's engineering entrance stuff...
so if he can tell how he derived the ans to the first part...i maybe able to do some good..hopefully
-
I need to know how u got it....
I can't help otherwise...if i get this i can help using my iit books
Cuz evn im doing AS
to find a
4=8+t
t= -4
a=14-t
a=18
to find b
13=12+t
t= 1
b=8+t
b=9
-
I'm doing iit as well...that's engineering entrance stuff...
so if he can tell how he derived the ans to the first part...i maybe able to do some good..hopefully
its okay nid
i will wait for astar
-
Here
-
The line l1 has vector equation
r = 8i + 12j + 14k + t(i + j – k),
where t is a parameter.
The point A has coordinates (4, 8, a), where a is a constant. The point B has coordinates
(b, 13, 13), where b is a constant. Points A and B lie on the line l1.
(a) Find the values of a and b.
(3)
Given that the point O is the origin, and that the point P lies on l1 such that OP is
perpendicular to l1,
(b) find the coordinates of P.
(5)
i need help in part B with explanation plz
a)to find a 8i + 12j + 14k + t(i + j – k) =4i+8j+ak
equate the i's to get 8+t=4 so t=-4
equate the j's to get 12+t=8 so again t=-4
equate the k's to get 14-t=14--4=18=a
To find b 8i + 12j + 14k + t(i + j – k) =bi+13j+13k
equate the i's to get 8+t=b (1)
equate the j's to get 12+t=13 so t=1
equate the k's to get14-1=13 so t=1
then from (1) b=9
The tangent vector of the line is (1,1,-1)=v
OP.v=0 since they are perpendicular
OP=8i + 12j + 14k + t(i + j – k)=(8+t)i+(12+t)j+(14-1)k
OP.v=8+t+12+t-(14-t)=6+3t=0 t=-2 then P=8i + 12j + 14k -2(i + j – k)=6i+10j+16k
-
The line l1 has vector equation
r = 8i + 12j + 14k + t(i + j – k),
where t is a parameter.
The point A has coordinates (4, 8, a), where a is a constant. The point B has coordinates
(b, 13, 13), where b is a constant. Points A and B lie on the line l1.
(a) Find the values of a and b.
(3)
Given that the point O is the origin, and that the point P lies on l1 such that OP is
perpendicular to l1,
(b) find the coordinates of P.
(5)
i need help in part B with explanation plz
a)to find a 8i + 12j + 14k + t(i + j – k) =4i+8j+ak
equate the i's to get 8+t=4 so t=-4
equate the j's to get 12+t=8 so again t=-4
equate the k's to get 14-t=14--4=18=a
To find b 8i + 12j + 14k + t(i + j – k) =bi+13j+13k
equate the i's to get 8+t=b (1)
equate the j's to get 12+t=13 so t=1
equate the k's to get14-1=13 so t=1
then from (1) b=9
The tangent vector of the line is (1,1,-1)=v
OP.v=0 since they are perpendicular
OP=8i + 12j + 14k + t(i + j – k)=(8+t)i+(12+t)j+(14-1)k
OP.v=8+t+12+t-(14-t)=6+3t=0 t=-2 then P=8i + 12j + 14k -2(i + j – k)=6i+10j+16k
thanx astar :)
+rep