Question

Find the equation for a plane containing 3 points: A(2, 2,1) in the form: ax+by+cz+d = 0 C(0, -2,1). Put the plane equation B(3,1, 0) х — 3 z+2 = y+5 = 2 L: Find the intersection point between 2 lines whose symmetric equations are: 4 х-2 L, : у-2 = z-3 -3 Find the parametric equation for a line that is going through point A(2,4,6) and perpendicular to the plane 5х-3у+2z-4%3D0. Name: x-3y4z 10 Find the distance between 2 parallel planes x-3y+4z =6 x+z-1 0 Find the angle between the 2 planes: x+y+z-6=0 2x+5z+3 0 Find the equation for the line of intersection between the 2 planes: х-Зу+2+2-0

Answer 1

A C2,2,1), 6(3,1,0) aud C0,-2,) Dyeven points nctoining these 3 paints To fndhpa so lution equation o lausart by+czid poopundicular vectov tothe plare is ABXBC So So A 1 AB XBC ,2,- So a = C = cquatim ths is aala-ko)tblay)+tlz--0

-2)3-4. (t-h1e+ (t-x)h plana dtan botan 2 paralll ies Jo 9/2thg 9-2nth So areordbg fermla to the d- S-S1 t 1 b 6 VI+9ナ16 26

betueen blana tz10 az C So Cyazt bb2+C CoS Avaittaje 2N3 1 : 35.264 line ntosad 2 tS2+30 the equakion of 16 betaveant 2 plaues: 1

The plane haue movwal vectors 2,S,3 11-3,2 The l be the ds of the tine so axb L14,-1, So To find a So 2a t S2 2 2n tz 27 1V13 2 Se equation f line is 11