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UNDERGRAD: CALCULUS 3 LEVEL QUESTION

I've read this but can't seem to reason it around for the reverse case: finding equation of a plane given a perpendicular line.

I'm looking for an equation for a plane perpendicular to the line $L(t) =(4+3t, 2t, 4t-1)$ that passes through $(5, -4, 2)$.

I tried to find two vectors $u, v$ whose cross product equals $L(t)$ or a vector $u$ whose dot product to $L(t)$ is zero and passes through $(5,-4,2)$ but couldn't (maybe the t is throwing me off). How should I approach this?

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  • $\begingroup$ Do you know how to write the equation of a plane, given a normal vector? Hint: by looking at the $t$-coefficients, you can see that the vector $v = (3,2,4)$ is a normal vector to the plane. $\endgroup$
    – Nick
    Commented Oct 13, 2019 at 22:24
  • $\begingroup$ I know it as $0=i(x-x_0)+j(y-y_0)+k(z-z_0)$ for the normal $n=(i,j,k)$. Working on understanding the rest of your hint now... $\endgroup$
    – Five9
    Commented Oct 13, 2019 at 22:29
  • $\begingroup$ Correct. In this case $i=3$, $j=2$, $k=4$, and $x_0=5$, $y_0=-4$, $z_0=2$. $\endgroup$
    – Nick
    Commented Oct 13, 2019 at 22:31
  • $\begingroup$ Omg it's so simple that way...I was too thrown off by the fact that there was a point I think somehow...So to recap the overall approach was to realize they gave us a line with direction vector $(3,2,4)$ (we don't care about it's $(4,0,-1)$ point) and that since it's the normal to our desired plane i just plug it into the "plane equation"? And from my understanding any point in a plane can go into the $x_0, y_0, z_0$ spots so we just use the 5,-4,2 it gave us. EDIT: I can't +1 comments but bless u for answering in a quick and clear way...post ur comment as an answer so I can vote it! $\endgroup$
    – Five9
    Commented Oct 13, 2019 at 22:40
  • $\begingroup$ Correct! All that matters is the direction vector of the line, and any point on the plane will work. $\endgroup$
    – Nick
    Commented Oct 13, 2019 at 22:41

1 Answer 1

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Since $L(t) = (4,0,-1) + (3,2,4)t$, the vector $v = (3,2,4)$ is a normal vector to the plane. Then using the given point $(5,-4,2)$, the equation of the plane is

$$0 = 3(x-5) + 2(y+4) + 4(z-2) $$

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