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it gives 2 vector eqn's of lines L1 and L2 as follows:

L1 = r = (2i+4j)+k(-1i+3j)

L2= r = (-3i+1j)+m(6i+2j)

(ie k= lambda and m=mew- also they are inform matrice form but i didnlt now how to type i into that format so sorry for that)

i was converting them to cartesian eqn y=mx+c, then the gradients must multiply =-1......howveer i suspect u can use the scalar product rule (ie a.b = 0 for perpendicular lines) but i can seem to work the equations to give me = 0....any assistance or hints would be greatly appreciated!!!

thanks