If x y z is on the plane π we have x 1 3 t 3 s y 1 t

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If (x, y, z) is on the planeπ, we havex=1+ 3t+ 3sy=1t2sz=25t. The third equa-tion givest=2z5.If we substitute this into the second equation we gety= 12z52swhich givess=12y+310+z10. We substitute the values fortandsinto the first equation givingx=1 + 3parenleftbigg2z5parenrightbigg+ 3parenleftbigg12y+310+z10parenrightbigg.Rewriting we get 10x+ 15y+ 3z= 11 as an equation forπ.(An alternate approach would be to findn=v×w.)(c) A direction vector for this line is a normal vector for the planeπ,n= (10,15,3).A parametric description of the line is (0,1,0) +t(10,15,3),tR.To see where the line and the plane meet we must find atsuch that (10t,1+15t,3t)satisfies the equation of the plane.For this we need 10(10t) + 15(1 + 15t) +3(3t) = 11=334t=4=t=2167.The point of intersection isparenleftbigg20167,137167,6167parenrightbigg.
MATB41HSolutions # 4page33. Let (p, q) be the point where the perpendicularfrom (x1, y1) meets the line. A normal forisn= (a, b). Hence (px, qy) =t(a, b), for sometR.The required distance isbardblt(a, b)bardbl.Wecan also write (p, q) as (p, q) = (x1+t a, y1+t b)and asa p+b q=c(since (p, q) is a point on).
5. Thetangentplanetothegraphofz=f(x, y) atthepoint(a, b, f(a, b))isz=f(a, b) +∂f∂x(a, b) (xa) +∂f∂y(a, b) (yb).

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Term
Fall
Professor
moore
Tags
XY sex determination system

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