Find an equation of the plane. the plane through the point (3, 0, 7) and perpendicular to the line x = 5t, y = 2 − t, z = 8 + 4t

1 answer:

4 0

Let's write the parametric line

(x,y,z) = (0,2,8) + t(5,-1,4)

(5,-1,4) is called the direction vector of the line. It's normal to the plane

5x - y + 4z = constant

We get the constant by substituting the point (3,0,7),

5x - y +4z = 5(3) - 0 + 4(7) = 43

5x - y +4z = 43

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$\bf \textit{hyperbolas, horizontal traverse axis }\\\\
\cfrac{(x- h)^2}{ a^2}-\cfrac{(y- k)^2}{ b^2}=1
\qquad 
\begin{cases}
center\ ( h, k)\\
vertices\ ( h\pm a, k)\\
c=\textit{distance from}\\
\qquad \textit{center to foci}\\
\qquad \sqrt{ a ^2+ b ^2}\\
\textit{asymptotes}\quad 
y= k\pm \cfrac{b}{a}(x- h)
\end{cases}\\\\
-------------------------------$

$\bf \begin{cases}
a=9\\
c=15
\end{cases}\implies c=\sqrt{a^2+b^2}\implies \sqrt{c^2-a^2}=b
\\\\\\
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-------------------------------\\\\
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