Question

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

`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`.

Answer 1

`2/5`

Explanation:

`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`