How to find a vector in terms of m?

Question

Can someone please explain to me how to do question aii)? Thanks!

1368 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-20T12:25:21

Part (a)(i): #vec(OS)=#k#a#-#2#k#b#.

Part (a)(ii): #vec(OS)=(1+2#m#)a+(4-3#m#)b.#

Part (b): k=-11, ad m=-6.

Recall that, #vec(AB)=vec(OB)-vec(OA)...........(star)#.

Part a(i) : Express #vec(OS)# in terms of k, #a and b#.

It is given that, #vec(OS)=#k#vec(OP)#.

The p.v. of #P,# relative to an Origin #O," is "a-2b#, which we will

denote by #P=P(a-2b) :. vec(OP)=a-2b#.

#:. vec(OS)=#k#vec(OP) rArr vec(OS)=#k#(a-2b)=#k#a#-#2#k#b#.

Part a(ii): Express #vec(OS)# in terms of m, #a and b#.

It is given that #vec(RS)=#m#vec(RQ)#.

It is known that, #R=R(a+4b), and Q=Q(3a+b)#.

#:. vec(OS)-vec(OR)=#m#[vec(OQ)-vec(OR)]......[because, (star)]#.

#:. =#m#vec(OQ)-#m#vec(OR)#,

#:. vec(OS)=vec(OR)+#m#vec(OQ)-#m#vec(OR), i.e.#,

#:. vec(OS)=(a+4b)+#m#(3a+b)-#m#(a+4b)#

#=a+4b+3#m#a+#m#b-#m#a-4#m#b#,

#=a+4b+2#m#a-3#m#b#.

#rArr vec(OS)=(1+2#m#)a+(4-3#m#)b.#

Part (b): Hence evaluate k and m.

Thus, we have the following #2# expressions for #vec(OS)#, one

from Part (a) and the other from Part (b) :

# #k#a#-#2#k#b#=#vec(OS)#=(1+2#m#)a+(4-3#m#)b.#

Obviously, we must have,

k=1+2m, and -2k=4-3m.

Sub.ing k from the #1^(st)# eqn. into the #2^(nd)#, we get,

-2(1+2m)=4-3m# rArr # m=-6, so that, k=1+2m=1-12=-11.

Enjoy Maths.!