How to find a vector in terms of m?

Question

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

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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.!