Determine whether u and v are orthogonal, parallel, or neither: u=(1,7), v=(-42,3)
do you know their dot product?
i think that their dot product is -63
so it's not orthagonal...
the dot product is not -63
dot product = 1*(-42)+7*3 = -21
but because the dot product isn't 0, the two vectors are not orthogonal (they don't form a 90 degree angle)
oh, sorry i accidentally typed the wrong number... but its still not orthogonal
yes in both cases they're not orthogonal
but how do i know if it is parallel or not?
parallel vectors are two vectors u = (a,b) and v = (c,d) such that v = k*u = k(a,b) for some scalar k
So if you can show that (c,d) = (ka,kb), then the two vectors are parallel
what is k?
k is any scalar number
so in this case (-42,3) = (k*1, k*7) can you find a single value of k that works here?
dot product = uvcosA A is the angle b/w u and v if orthogonal, A=90 so dot product(DP) = 0 since cos90 = 0 if they are parallel, A=0 DP = uv, since cos0 = 1 and else they are neither orthogonal nor parallel
i don't think so
so because there are no solutions to (-42,3) = (k*1, k*7), the two vectors are NOT parallel
Ok, I think I get how you found that!
ok heres the deal with this problem, I will tell you exactly how to do this! Ok so in parallel the answer is either 1 or -1 if its orthogonal the answer is 0 and neither would be Anything that isnt 1,-1, or 0. To do this problem V*W ------- IIVII*IIUII Lets first find v*u 1*(-42)+7*3 = -21 now lets find magnitude v * magnitude u to find magnitude V we just take that square root of the numbers in v so..... sqr((-42)^2+(3)^2 =42.10 now lets find the magnitude of W sqr((1)^2+(7)^2 =7.07 plug this into our original equation
sorry that is V*U ------- IIVII*IIUII
V*U ------- IIVII*IIUII -21 ------ (42.10)(7.07 =-0.07 so its neither!
@violin37 why wats the problem...???
don't worry about it, i got my answer
Please use my way as it is much easier to understand!