1a)Three points X,Y Z have position vectors pa,qb,ra+sb with respect to a point respectively,where a and b are non zero vectors which are not parallel and p,q,r,s are scalars.If the points are collinear ,prove that ps+qr=pq
[hint: express vector XY and vector YZ in terms of a and b ]
1b)The positiom vectors of points E,F ,G,H with repect to a point O are e,f,5e,3f respectively,where e and f are non-zero vctors which are not parallel .If EF and GH intersect at K,find the position vector of K with respect to O in terms of e and f
[hint:let the position vector of K be re+sf and use the result of (a)]
做到一條都得!
Amaths:vector(Urgent!)
2008-05-09 7:08 am
回答 (2)
2008-05-09 11:25 pm
✔ 最佳答案
a)[First of all, we can use the idea of position vector(?).]
vector XY = vector OY - vector OX = q(vector b) - p(vector a)******(1)
vector YZ = vector OZ - vector OY = [r(vector a) + s(vector b)] - q(vector b) = r(vector a) + (s - q)(vector b)******(2)
from (1) and (2),
because the pts are collinear,
q = s - q******(3) and -p = r******(4)
(3)/(4)
q/-p = (s - q)/r
qr = (-p)(s - q)
thus, ps +qr = pq
[I am skipping the word VECTOR in the following to save time.]
b)
Let the position vector of K be (re + sf).
Because K is a pt on line EF, i.e. E, K, F are collinear,
By a),
ps + qr = pq
put p = 1 and q = 1 [OE = e, OF = f]
s + r = 1******(1)
Similarly, G, K, H are collinear,
By a),
put p = 5, q = 3,
5s + 3r = 15******(2)
from (1),
s = 1 - r******(3)
put (3) into (2),
5(1 - r) + 3r = 15
-2r = 10
r = -5
s - 5 = 1
s = 6
thus, OK = (re + sf) = -5e + 6f
參考: 做過，而且我有偉大的solution guide（但我的做法和solution guide裡的有點出入）
2008-05-09 8:23 am
答你1 (a) 先
XY = qb - pa
YZ = ra + sb - qb = ra + (s-q)b
如果佢地 3 點 collinear, 即係 XY // YZ
咁我地可以寫成
XY = k YZ where k is a scalar
qb - pa = k [ ra + (s-q)b ] = kra + k(s-q)b
由於 a 同 b 唔係 paraelle, so
-pa = kra and qb = k(s-q)b
得出
-p = kr ---(1) and q = k(s-q) --- (2)
將 (1) / (2)
-p / q = ( kr ) / [ k(s-q) ]
-p / q = r / (s-q)
-ps + pq = rq
pq = ps + qr
1(b) 有待補充
XY = qb - pa
YZ = ra + sb - qb = ra + (s-q)b
如果佢地 3 點 collinear, 即係 XY // YZ
咁我地可以寫成
XY = k YZ where k is a scalar
qb - pa = k [ ra + (s-q)b ] = kra + k(s-q)b
由於 a 同 b 唔係 paraelle, so
-pa = kra and qb = k(s-q)b
得出
-p = kr ---(1) and q = k(s-q) --- (2)
將 (1) / (2)
-p / q = ( kr ) / [ k(s-q) ]
-p / q = r / (s-q)
-ps + pq = rq
pq = ps + qr
1(b) 有待補充
收錄日期: 2021-04-20 15:13:33
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