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18 votes
18 votes
Find paremetric equations for the line through the point (1, 0, 1) and perpendicular to the vectors (3, 5, −2) and (2, 1, 0).​

User Mcfly
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1 Answer

15 votes
15 votes

Take the cross product of the given vectors to get one that is perpendicular to both of them.


\langle 3, 5, -2 \rangle * \langle 2, 1, 0 \rangle = \langle 2, -4, -7 \rangle

The line parallel to this vector and passing through the origin is given by the vector function


\vec r(t) = \langle 2, -4, -7 \rangle t

where
t\in\Bbb R, and hence parametric equations


\begin{cases}x(t) = 2t \\ y(t) = -4t \\ z(t) = -7t \end{cases}

Translate this line by the vector
\langle1,0,1\rangle to make it pass through the given point. So the line we want has vector function


\vec r(t) = \langle 2, -4, -7 \rangle t + \langle 1, 0, 1 \rangle

and parametric equations


\boxed{\begin{cases} x(t) = 2t + 1 \\ y(t) = -4t \\ z(t) = -7t + 1 \end{cases}}

User Eihwaz
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3.0k points