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Find a vector equation and parametric equations for the line. (Use the parameter t.) The line through the point (0, 11, −7) and parallel to the line x = −1 + 3t, y = 6 − 2t, z = 3 + 6t r(t) = ⟨3t, 11−2t, −7+6t⟩ (x(t), y(t), z(t)) = 3t, 11−2t, −7+6t

Respuesta :

LammettHash

The given line with equations

[tex]\begin{cases}x(t)=-1+3t\\y(t)=6-2t\\z=3+6t\end{cases}[/tex]

has tangent vector [tex]\langle3,-2,6\rangle[/tex].

The line through the origin that is parallel to this line is simply [tex]\langle3,-2,6\rangle t[/tex]. We then get the line that passes through (0, 11, -7) by translating this line, giving us

[tex]\vec r(t)=\langle3t,11-2t,-7+6t\rangle[/tex]

as required. In parametric form, this is

[tex]\begin{cases}x(t)=3t\\y(t)=11-2t\\z(t)=-7+6t\end{cases}[/tex]

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