1. Referred to the origin O, the position vectors of the 2 points A and V are given by vector OA = 3i + j + 3k and vecotr OB = 5i - 4j +3k.
(i) show that the length of the proj of vector OB onto vector OA is 20/ sq rt 19.
(ii) Deduce from (i) the perp. dist from B to the line joining O and A.
(iii) find the proj of vector OA onto OB and hence write down the position vector P on vector OB such that vector AP is perpendicular to vector OB.
(iv) Find the position vector of A', the reflection of A in the line OB.
(v) Find a vector of magnitude sq.rt. 22 units which is perpe. to vector OA and vector OB.
Only need help for (iv) and (v)
2. W.R.T the origin O, the position vectors of 2 points A and B are given by a and b respectively where |a| = 1, |b| =2 and a.b = 1. C is the point on AB such that AC = 2CB.
(i) find position vector of point C in terms of a and b.
(ii) Show that the length of proj of OC onto OB is 3/2 and deduce the position vector of the foot N of the perpendicular from C to the line OB
Thanks.
Q1
from part iii you have the direction vector of the line perpendicular to OB
point A also lies on this perpendicular line
with a point and direction vector you can work out the vector equation of the perpendicular line
the intersection between line OB and the perpendicular line is the midpt of A and A'
from there you can A'
hope that answers part iv
for part v
you can use cross product of OA and OB to find vector perpendicular to OA and OB (normal vector)
OA + k (normal vector) will give a vector perpendicular to OA and OB
with the magnitude of root 22 you can solve for k
Hi,
I will help with Q2 :)
Recall that length of projection of vector OC onto vector OB is | (c.b)/|b| |, where
c.b = 1/3 (2b + a).b,
which can be expanded to make use of given information.
The next part is easy, since O, N and B are collinear. Thanks.
Cheers,
Wen Shih
Originally posted by quailmaster:Q1
from part iii you have the direction vector of the line perpendicular to OB
point A also lies on this perpendicular line
with a point and direction vector you can work out the vector equation of the perpendicular line
the intersection between line OB and the perpendicular line is the midpt of A and A'
from there you can A'
hope that answers part iv
for part v
you can use cross product of OA and OB to find vector perpendicular to OA and OB (normal vector)
OA + k (normal vector) will give a vector perpendicular to OA and OB
with the magnitude of root 22 you can solve for k
Hi quailmaster,
I dont understand this particular part where
OA + k (normal vector) will give a vector perpendicular to OA and OB
with the magnitude of root 22 you can solve for k
Is it not sufficient to just use the cross pdt of OA and OB to find the direction of the vector divided by the magnitude of the vector sq.rt 22? the OA + k steps appear foreign to me and I have no idea why they are introduced for. Why is there a need to do addition involving "OA" vector? Thank u!
the cross pdt of OA and OB to find the direction of the vector
divided by the magnitude of the direction vector
multiplied by sq.rt 22
yeah think you are right
Saw Q1 in NYJC tutorial question
For those who have SRJC2007 H2 Maths Prelims, take a look at their question for part (b), where vectors is used to prove that angle at circumference = 90 degrees