Painful vector problem (please see below - thank you!!). Can you find lambda?

Question

#OABC# is a square with #A# and #C# having position vectors of #a = -4i + 3j# and #c = 3i + 4j#.

#OB = -i + 7j#
#AC = 7i + j#
#OD = -2i + 17/3j# where #D# is a point along the line #AB#, such that #BD = 1/3BA#.

#OD# intersects with #AC# at a point #E# such that #OE = (1-lambda)OA + lambdaOC#.

Find the value of #lambda#.

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Answer 1 · 2018-03-10T14:25:12

#2/5#

#A=(-4,3)#
#C=(3,4)#

and now

#1/2(A+C)=1/2(B+O) rArr B + O = A+C#

also

#B - O = bar(OB)#

Solving now

#{( B + O = A+C),(B - O = bar(OB)):}#

we have

#B = 1/2(A+C + bar(OB)) = (-1,7)#
#O = 1/2(A+C-bar(OB))=(0,0)#

Now

#D = A + 2/3(B-A) = (-2,17/3)#

#E# is the intersection of segments

#s_1 = O + mu(D-O)#
#s_2 = C + rho(A-C)#

with #{mu, rho} in [0,1]^2#

then solving

#O + mu(D-O) = C + rho(A-C)#

we obtain

#mu = 3/5, rho = 3/5#

#E = O+3/5(D-O) = (-6/5,17/5)#

and finally from

#bar(OE) = (1-lambda)bar(OA) + lambdabar(OC) rArr lambda = abs(bar(OE)-bar(OA))/abs(bar(OC)-bar(OA)) = 2/5#

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