this is really simple but the solution is not coming to me.

Question

zephyr141 · Answer

find and equation of a plane through the point (2,0,1) and perpendicular to the line x=3t ; y=2-t ; z=3+4t.

zephyr141 · Answer

so far this is what i did:
$$(0-3t)+(2-t)+(3+4t)=r$$$$r=(0i+2j+3k)+(3i-j+4k)t$$$$r=2j+3k $$$$v=3i-j+4k$$so i got that... am I on the right path so far? where did i go wrong?

academicgurusinc · Answer

To write the equation of this plane, we need a point on it and a vector perpendicular to it. We're already given that the point (2,0,1) is on the plane. The vector <3,-1,4> is parallel to the line x=4t, y=2-t, z=3+4t, and because this line is perpendicular to the plane we're dealing with, the vector <3,-1,4> is perpendicular to this plane. An equation for this plane is therefore <3,-1,4> times < - <2,0,1>>=0 If you found this helpful, I encourage to subscribe to our youtube channel https://www.youtube.com/channel/UCYiI7SmkU4_vhdSzKBWsifg, to stay current with all of our new videos. Regards, Academic Gurus Inc. Twitter (@Academic_Gurus) Facebook (AcademicGurusInc) Youtube (Academic Gurus Inc)

zephyr141 · Answer

so it's $$3x-2-y+4y-1=0$$$$3x-y+4y-3=0$$$$3x-y+4y=3$$

zephyr141 · Answer

my final answer will be $$3x-y+4y=3$$

zephyr141 · Answer

oh i made a mistake in my second comment. 3x needs to be positive in the first equation and negative in the vector at the end and in every following instance. so i guess my answer will be$$-3x-y+4y=3$$right?

alekos · Answer

I thought it would have been -3x - y + 4z = 3

zephyr141 · Answer

whoops. yeah i wrote y twice. sorry. multiple variables can mix me up a bit.