Find parametric equations for the line through the point that is parallel to the plane and perpendicular to the line

Question

Find parametric equations for the line through the point  that is parallel to the plane  and perpendicular to the line

Quizplus · Accepted Answer

To find the line parallel to the given plane, we can use the normal vector of the plane. Therefore, the direction vector of the line will be perpendicular to the normal vector of the plane, and parallel to the line given.

From the given point, we can read off the coordinates:

$P = \begin{pmatrix} 11 \ ea \ a3e5 \ 5501 \ 064a \ 9f8f \ b1ae \ 2ab8 \ 0935 \ TB \ 5971 \ 11 \end{pmatrix}$

The given plane has the following normal vector:

$\vec{n} = \begin{pmatrix} 11 \ ea \ a3e5 \ 5501 \ 064a \ 9f8f \ b1ae \ 2ab8 \ 0935 \ TB \ 5971 \ 11 \end{pmatrix}$

To find a direction vector in the plane, we can use the cross product of the given direction vector and the normal vector of the plane:

$\vec{v} = \vec{n} 	imes \begin{pmatrix} 11 \ ea \ a3e5 \ 5501 \ 2d5b \ 9f8f \ bb7b \ 4233 \ 6c6e \ TB \ 5971 \ 11 \end{pmatrix}$

To compute the cross product, we can first convert the vectors to component form, and then use the formula for the cross product:

$\vec{n} = \begin{pmatrix} 11 \ ea \ a3e5 \ 5501 \ 064a \ 9f8f \ b1ae \ 2ab8 \ 0935 \ 54 \ 9f \ 11 \end{pmatrix}$

$\begin{pmatrix} 11 \ ea \ a3e5 \ 5501 \ 2d5b \ 9f8f \ bb7b \ 4233 \ 6c6e \ 54 \ 9f \ 11 \end{pmatrix}$

$\vec{n} 	imes \begin{pmatrix} 11 \ ea \ a3e5 \ 5501 \ 2d5b \ 9f8f \ bb7b \ 4233 \ 6c6e \ 54 \ 9f \ 11 \end{pmatrix} = \begin{pmatrix} -1f433b7 \ 5dc5f5a \ -2eca874 \ 7f3c3 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \ 0 \end{pmatrix}$

Therefore, the parametric equations of the line passing through the given point and parallel to the plane are:

$x(t) = 11 + (-1f433b7) t$

$y(t) = ea + 5dc5f5a t$

$z(t) = a3e5 + (-2eca874) t$

$w(t) = 5501 + 7f3c3 t$

$u(t) = 064a$

$v(t) = 9f8f$

$r(t) = b1ae$

$s(t) = 2ab8$

$t(t) = 0935$

$n(t) = TB$

$m(t) = 5971$

$p(t) = 11$

Simplifying, we get:

$x(t) = 11 - 508c0f73 t $

$y(t) = ea + 15c2faa1 t$

$z(t) = a3e5 - 1735268 t$