Determine whether the given line and the given plane are parallel :
a.) x = 1 + t, y=-1, z=-2t and x = 2y +3z - 9 =0,
b.) <0, 1, 2> +t <3,2,-1> and 4x - 2z +1 = 0
(a)(a)x=1+t, y=−1−t, z=−2t,a⃗=(1,−1,−2)(α)x+2y+3z−9=0,n⃗=(1,2,3)n⃗⋅a⃗=1⋅1+(−1)⋅2+(−2)⋅3=−7≠0(\text a)\\(\text a) x=1+t,\ y=-1-t,\ z=-2t, \\\vec{a}=(1,-1,-2)\\ (\alpha) x+2y+3z-9=0, \vec{n}=(1,2,3)\\ \vec{n}\cdot\vec{a}=1\cdot1+(-1)\cdot2+(-2)\cdot3=-7\neq 0(a)(a)x=1+t, y=−1−t, z=−2t,a=(1,−1,−2)(α)x+2y+3z−9=0,n=(1,2,3)n⋅a=1⋅1+(−1)⋅2+(−2)⋅3=−7=0
Hence, we can say that the given line and the given plane are not parallel.
(b)(a)<0,1,2> +t <3,2,−1>,a⃗=(3,2,−1)(α)4x−y+2z+1=0,n⃗=(4,−1,2)n⃗⋅a⃗=3⋅4+2⋅(−1)+(−1)⋅2=8≠0(\text b)\\(a) <0,1,2>\ +t\ <3,2,-1>, \\\vec{a}=(3,2,-1)\\ (\alpha) 4x-y+2z+1=0, \\\vec{n}=(4,-1,2)\\ \vec n\cdot\vec a=3\cdot4+2\cdot(-1)+(-1)\cdot 2=8\neq0(b)(a)<0,1,2> +t <3,2,−1>,a=(3,2,−1)(α)4x−y+2z+1=0,n=(4,−1,2)n⋅a=3⋅4+2⋅(−1)+(−1)⋅2=8=0
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