How to find a vector in terms of m?

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

1 Answer
Dec 20, 2017

Part (a)(i): vec(OS)=ka-2kb.

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

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

Explanation:

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)=kvec(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)=kvec(OP) rArr vec(OS)=k(a-2b)=ka-2kb.

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

It is given that vec(RS)=mvec(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)].

:. =mvec(OQ)-mvec(OR),

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

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

=a+4b+3ma+mb-ma-4mb,

=a+4b+2ma-3mb.

rArr vec(OS)=(1+2m)a+(4-3m)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) :

ka-2kb=vec(OS)=(1+2m)a+(4-3m)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.!