Vectors Given that #vec(OP)=((-3),(-2))#, #vec(OQ)=((5),(-2))# and #T# is a point on the line #vec(PQ)# such that #5vec(PT)=3vec(TO)# express as column vectors? (I) #vec(PQ)# (II) #vec(TQ)#

Question

2069 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-04T12:08:19

# (I) : vec(PQ)=((8),(0)).#

# (II) : vec(TQ)=((55/8),(-3/4)).#

For ease of writing, let us denote the position vectors of #P and Q#

by #(-3,-2) and (5,-2)#, resp.

Let, #T=T(x,y)" be on line "PQ," such that, "5vec(PT)=3vec(TO)#.

Now, #5vec(PT)=5((x,y)-(-3,-2))=5(x+3,y+2),#

#"&, "3vec(TO)=3(-x,-y)#.

#:.5vec(PT)=3vec(TO) rArr 5(x+3,y+2)=3(-x,-y)#.

# rArr 5x+15=-3x, and, 5y+10=-3y#.

# rArr x=-15/8, and, y=-10/8#.

Hence, #T=T(-15/8,-10/8)#.

So, # (I) : vec(PQ)=(5,-2)-(-3,-2)=(8,0).#

# (II) : vec(TQ)=(5,-2)-(-15/8,-10/8)=(55/8,-3/4).#

Answer 2 · 2018-07-04T12:14:57

The vectors are #vec(PQ)=((8),(0))# and #vec(TQ)=((31/8),(-3/4))#

Apply Chasles' relation

#vec(OP)=((-3),(-2))#

#vec(OQ)=((5),(-2))#

#vec(PO)=-vec(OP)#

#vec(PQ)=vec(PO)+vec(OQ)#

#=vec(OQ)-vec(OP)#

#=((5),(-2))-((-3),(-2))#

#=((5+3),(-2+2))#

#=((8),(0))#

#5vec(PT)=3vec(TO)#

#5(vec(PQ)+vec(QT))=3(vec(TQ)+vec(QO))#

#5vec(PQ)-5vec(TQ)=3vec(TQ)-3vec(OQ)#

#8vec(TQ)=5vec(PQ)+3vec(OQ)#

#vec(TQ)=1/8(5vec(PQ)+3vec(OQ))#

#=1/8(5*((8),(0))+3*((5),(-2)))#

#=1/8((55),(-6))#

#=((55/8),(-3/4))#